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Pavlo B.'s questions

Martin Hope
Pavlo B.
Asked: 2020-07-28 05:53:39 +0800 CST
  • 4

I have tried to install a mysql server on my local machine but I can't get it to work.

I have downloaded the zip from here: https://dev.mysql.com/downloads/mysql/

Then I followed these steps: http://webdevzoom.com/install-mysql-zip-file-on-windows/

But it gives me the following error:

ERROR 2003 (HY000): Can't connect to MySQL server on 'localhost' (10061)

I have searched for the error but the solutions given on those sites have not worked for me.

When executing mysqld --consoleI get the following:

2018-07-27T13:54:56.349166Z 0 [System] [MY-010116] [Server] c:\mysql-5.7.17\bin\mysqld.exe (mysqld 8.0.11) starting as process 5764

2018-07-27T13:54:56.373933Z 1 [ERROR] [MY-011011] [Server] Failed to find valid data directory.

2018-07-27T13:54:56.376442Z 0 [ERROR] [MY-010020] [Server] Data Dictionary initialization failed.

2018-07-27T13:54:56.379069Z 0 [ERROR] [MY-010119] [Server] Aborting

2018-07-27T13:54:56.382633Z 0 [System] [MY-010910] [Server] c:\mysql-5.7.17\bin\mysqld.exe: Shutdown complete (mysqld 8.0.11) MySQL Community Server - GPL .

start content

[mysqld]

basedir = "/mysql-5.7.17"
datadir = "/mysql-5.7.17/data"
tmpdir = "/mysql-5.7.17/tmp"
port = 3306
server_id = 1
log_error = "mysql_error.log"
pid_file = "mysql.pid"
socket = "/mysql-5.7.17/mysql.sock"

# Optional - Default Configuration
max_allowed_packet = 8M
key_buffer_size=16M

# Where do all the plugins live
plugin_dir = "/mysql-5.7.17/lib/plugin/"
mysql
Martin Hope
Pavlo B.
Asked: 2020-03-21 00:59:25 +0800 CST
  • 4

I have an array like the following:

$lista = [
      '867Ad2' => [
         stock => 2,
         nombre => "cosa1",
      ]
      '838sdff2' => [
         stock => 4,
         nombre => "cosa2",
      ]
      '234hsd4fd' => [
         stock => 6,
         nombre => "cosa3",
      ]  
      '8691240654d' => [
         stock => 8,
         nombre => "cosa4",
      ]
    ]

I order it as I want, by the value of the key 'stock'in this way:

function orderBy($data, $field)
            {
                $code = "return strnatcmp(\$b['$field'], \$a['$field']);";
                sort($data, create_function('$a,$b', $code));
                return $data;
            }

            $sorted_data = orderBy($lista, 'stock');

However, I have noticed that when ordering it, it replaces the indices in the following way:

$lista = [
      '0' => [
         stock => 8,
         nombre => "cosa4",
      ]
      '1' => [
         stock => 6,
         nombre => "cosa3",
      ]
      '2' => [
         stock => 4,
         nombre => "cosa2",
      ]  
      '3' => [
         stock => 2,
         nombre => "cosa1",
      ]
    ]

That is, it replaces my indexes, which are codes with letters and numbers, for a numerical index.

php
Martin Hope
Pavlo B.
Asked: 2020-01-31 00:59:28 +0800 CST
  • 9

How can you keep the footerbottom of the page always, but without using fixed or absolute, because several cases can occur:

In the event that the content is not enough to fill the page, the footer does not stay below unless you put an position: fixed; bottom: 0orposition: absolute; bottom: 0

But if I put one of those two, since the content is larger than the screen, the footer is always visible, in the case of fixed it scrolls but it always stays on the screen, but in the case of putting absolute, when it goes down it stays through the content medium.

<html>
  <body>
    <header style="background:grey">Header</header>
    <div style="background:yellow">Contenido</div>
    <footer style="background:green">tiene que estar abajo pero no se queda abajo siempre...</footer>
  </body>
</html>

css
Martin Hope
Pavlo B.
Asked: 2020-03-07 09:36:13 +0800 CST
  • 9

I am trying to place an icon on my tab, something like this:

enter image description here

The favicon.ico file is in the same folder as the file where I put this code:

<link rel="shortcut icon" href="favicon.ico" />

But it doesn't do anything, it leaves me the xamp icon...

html
Martin Hope
Pavlo B.
Asked: 2020-02-28 03:06:00 +0800 CST
  • 9

I am making a graph and I need to make it put a horizontal line at X point of it

this is the "script" of my graph;

var ctx = document.getElementById('myChart').getContext('2d');
  var myChart = new Chart(ctx, {
    type: 'line',
    data: {
      labels: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, ],
      datasets: [{
        label: 'Reparaciones x dia',
        data: ['0', '4', '1', '3', '1', '0', '0', '1', '11', '4', '8', '7', '5', '1', '2', '5', '3', '8', '1', ],
        backgroundColor: "rgba(0,255,51,0.5)"
      }, {
        label: 'Totales',
        data: [0, 4, 5, 8, 9, 9, 9, 10, 21, 25, 33, 40, 45, 50, 51, 53, 58, 61, 69, 73],
        backgroundColor: "rgba(0,153,255,0.5)"
      }]
    }
  });
<script src="https://cdnjs.cloudflare.com/ajax/libs/Chart.js/2.5.0/Chart.min.js"></script>

<canvas id="myChart"></canvas>

I need a line for example at number 15 of the vertical axis...

jquery
Martin Hope
Pavlo B.
Asked: 2020-02-01 00:57:10 +0800 CST
  • 5

I have a PHP file in which I send the actions with forms in which there are type hidden inputs with the variables that I need. But there comes a time when I need to reload the page and keep the variables of the $_POST. Reloading is necessary to avoid form resubmission.

How could I do something like that? JavaScript maybe? or is there any way to do it with PHP?

php
Martin Hope
Pavlo B.
Asked: 2020-01-24 03:47:53 +0800 CST
  • 6

If I have a text string like so:

var cadena1 = "cadena_1";

And I only need to get the 1 from the previous string to put it in a variable, how can I do it?

Doing this:

var res = cadena1.split("_");

the result iscadena,1

javascript
Martin Hope
Pavlo B.
Asked: 2020-01-21 08:22:29 +0800 CST
  • 14

If I have, for example, several <span>and I click on one, how can I, with jquery, know which one I clicked on? Being the IDs of the elements automatically numbered with item_<?php echo $i; ?>?

That is to say, I want that when doing in the span with id=item_2it makes me, for example, a alert.()"you have clicked on item_2"

<div id="items_en_uso">
  <span style="background:#c9c9c9;padding:3px;" id="item_1">Hola</span><br><br>
  <span style="background:#c9c9c9;padding:3px;" id="item_2">adios</span><br><br>
  <span style="background:#c9c9c9;padding:3px;" id="item_3">buenas</span><br><br>
  <span style="background:#c9c9c9;padding:3px;" id="item_4">malas</span><br><br>
</div>

EDIT:

Spans are added when clicking on another element with

$("#clikeable").click(function(){
   $("#items_en_uso").append("<span id='item"+count+"'>Hola</span>");
});

If the spans are set by hand (not by jquery) it works, but if I add them with jquery when clicking on them, it doesn't work for me... How can I fix this?

jquery
Martin Hope
Pavlo B.
Asked: 2020-01-13 08:13:43 +0800 CST
  • 3

If I have two tables that have a column called direccion, and I want to get only one of them, how do I do it?

Client table: ID, address, level, email, password, name

Company table: EID, address, name

What I want to do is:

SELECT ID, direccion, nivel FROM cliente, empresa WHERE ID = 1;

The result it gives me is that the column direccionis ambiguous.

mysql
Martin Hope
Pavlo B.
Asked: 2020-01-12 00:51:56 +0800 CST
  • 3

i have a div

function imprSelec(historial){
  var ficha=document.getElementById(historial);
  var ventimp=window.open(' ','popimpr');
  ventimp.document.write(ficha.innerHTML);
  ventimp.document.close();
  ventimp.print();
  ventimp.close();
}
    <div id="historial">
      <h3><center>Mi contenido</center></h3>
      ...
    </div>
<a href="javascript:imprSelec('historial')">Imprimir</a>

And I want to print it, I've tried like this but it doesn't work for me, in fact it doesn't do anything. But I want only the content of the div to be printed, historialpreserving the styles of the div.

I add images:

This is how it looks on the web enter image description here

And this is how it looks, it's more or less fine since I've put the styles directly in the attribute, although the background colors don't work well for me.

enter image description here

javascript
Martin Hope
Pavlo B.
Asked: 2020-12-30 07:12:59 +0800 CST
  • 4

I am trying to make a checkboxcustom in which the background color of my is changed labeldepending on whether it is selected or not, and in this way hide the checkboxitself. The code I have is this:

.milabel {
  background-color:#c9c9c9;
  border-radius:5px;
  padding:20px;
  cursor:pointer;
  }
input:checked{
  }
<input id="micb" type="checkbox"><br><br>
<label class="milabel" for="micb"> Hola mundo! Soy clickeable </label>

I know that with input:checkedit you can modify the checkboxitself to give it a certain style but I don't know.

html
Martin Hope
Pavlo B.
Asked: 2020-12-29 09:42:14 +0800 CST
  • 5

I am trying to make some type inputs fileon my website but I cannot see what it is when a file is selected, I have tried to follow the example here: https://tympanus.net/codrops/2015/09/15/styling-customizing -file-inputs-smart-way/

.inputfile {
	width: 0.1px;
	height: 0.1px;
	opacity: 0;
	overflow: hidden;
	position: absolute;
	z-index: -1;
}
.fileUpload {
    position: relative;
    overflow: hidden;
    margin: 10px;
}
.fileUpload input.upload {
    position: absolute;
    top: 0;
    right: 0;
    margin: 0;
    padding: 0;
    font-size: 20px;
    cursor: pointer;
    opacity: 0;
    filter: alpha(opacity=0);
}
<input id="image1" type="file" class="inputfile hidden-xs hidden-md" value="image1" name="image1" required />
<label for="image1" class="my-btn btn-danger"><center>Archivo 1</center></label><br>

What I want is that when I select a file instead of putting "File 1" put the name of the file and I don't know how to do it, I have tried to copy the js code that is on that website but when I put it it stops working at all .

javascript
Martin Hope
Pavlo B.
Asked: 2020-12-29 00:48:16 +0800 CST
  • 5

I am trying to put the cookie policies on my website but it does not work correctly. I have tried two variants and neither works correctly for me.

function GetCookie(name) {
    var arg=name+"=";
    var alen=arg.length;
    var clen=document.cookie.length;
    var i=0;
    while (i<clen) {
        var j=i+alen;
 
        if (document.cookie.substring(i,j)==arg)
            return "1";
        i=document.cookie.indexOf(" ",i)+1;
        if (i==0)
             break;
     }
    return null;
}
 
function aceptar_cookies(){
    var expire=new Date();
    expire=new Date(expire.getTime()+7776000000);
    document.cookie="cookies_surestao=aceptada; expires="+expire;
 
    var visit=GetCookie("cookies_surestao");
    if (visit==1){
        popbox3();
    }
}
 
jQuery(function() {
    var visit=GetCookie("cookies_surestao");
    if (visit==1){
        $('#overbox3').toggle();
    } else {
        var expire=new Date();
        expire=new Date(expire.getTime()+7776000000);
        document.cookie="cookies_surestao=aceptada; expires="+expire;
    }
});
 
function popbox3() {
    $('#overbox3').toggle();
}
#overbox3 {
    position: fixed;
    bottom: 0px;
    left: 0px;
    width: 100%;
    z-index: 999999;
    display: block;
}
#infobox3 {
    margin: auto;
    position: relative;
    top: 0px;
    height: 58px;
    width: 100%;
    text-align:center;
    background-color: #eeeeee;
}
#infobox3 p {
    line-height:58px;
    font-size:12px;
    text-align:center;
}
#infobox3 p a {
    margin-right:5px;
    text-decoration: underline;
}
<div id="overbox3">
    <div id="infobox3">
        <p>Esta web utiliza cookies para obtener datos estadísticos de la navegación de sus usuarios. Si continúas navegando consideramos que aceptas su uso.
        <a href="politica-privacidad.php">Más información</a>
        <a onclick="aceptar_cookies();" style="cursor:pointer;">X Cerrar</a></p>
    </div>
</div>
As you can see, it gives an error (which does not even appear on my local server) when accepting them, the div that contains it does not close. On the other hand, I have tried to do it like this:

function getCookie(c_name){
    var c_value = document.cookie;
    var c_start = c_value.indexOf(" " + c_name + "=");
    if (c_start == -1){
        c_start = c_value.indexOf(c_name + "=");
    }
    if (c_start == -1){
        c_value = null;
    }else{
        c_start = c_value.indexOf("=", c_start) + 1;
        var c_end = c_value.indexOf(";", c_start);
        if (c_end == -1){
            c_end = c_value.length;
        }
        c_value = unescape(c_value.substring(c_start,c_end));
    }
    return c_value;
}
 
function setCookie(c_name,value,exdays){
    var exdate=new Date();
    exdate.setDate(exdate.getDate() + exdays);
    var c_value=escape(value) + ((exdays==null) ? "" : "; expires="+exdate.toUTCString());
    document.cookie=c_name + "=" + c_value;
}
 
if(getCookie('tiendaaviso')!="1"){
    document.getElementById("barraaceptacion").style.display="block";
}
function PonerCookie(){
    setCookie('tiendaaviso','1',365);
    document.getElementById("barraaceptacion").style.display="none";
}
#barraaceptacion {
    display:none;
    position:fixed;
    left:0px;
    right:0px;
    bottom:0px;
    padding-bottom:20px;
    width:100%;
    text-align:center;
    min-height:40px;
    background-color: rgba(0, 0, 0, 0.5);
    color:#fff;
    z-index:99999;
}
 
.inner {
    width:100%;
    position:absolute;
    padding-left:5px;
    font-family:verdana;
    font-size:12px;
    top:30%;
}
 
.inner a.ok {
    padding:4px;
    color:#00ff2e;
    text-decoration:none;
}
 
.inner a.info {
    padding-left:5px;
    text-decoration:none;
    color:#faff00;
}
<div id="barraaceptacion" style="display: block;">
    <div class="inner">
        Solicitamos su permiso para obtener datos estadísticos de su navegación en esta web, en cumplimiento del Real 
        Decreto-ley 13/2012. Si continúa navegando consideramos que acepta el uso de cookies.
        <a href="javascript:void(0);" class="ok" onclick="PonerCookie();"><b>OK</b></a> | 
        <a href="http://politicadecookies.com" target="_blank" class="info">Más información</a>
    </div>
</div>

I also get the same error...

php
Martin Hope
Pavlo B.
Asked: 2020-11-12 03:49:03 +0800 CST
  • 7

Having this:

<footer>
  <div class="row no-margin no-padding">
    <div class="col-md-4  col-xs-6 oneplayer">
    <br>
      <a class="logo navbar-brand no-margin no-padding" href="index.php"><img class="logo-img" src="res/images/oneplayer_logo.png"></img></a>
    </div>
    <div class="col-md-8 col-xs-12">
      <div class="row no-margin no-padding">
        <div class="col-md-4 footer-links-container">
          <a href="aboutus.php" class="header-links"><?php echo $lang['footer_string_1']; ?></a><br><br>
          <a href="index.php#contact" class="header-links"><?php echo $lang['footer_string_4']; ?></a>
        </div>
        <div class="col-md-4 footer-links-container">
          <a href="politica_privacidad.php" class="header-links"><?php echo $lang['footer_string_2']; ?></a><br><br>
          <a href="condiciones.php" class="header-links"><?php echo $lang['footer_string_3']; ?></a>
        </div>
        <div class="col-md-4 footer-links-container">
          <form action="" method="post">
            <p class="footer-contact-info"><?php echo $lang['footer_string_5']; ?></p>
            <input type="email" class="my-form-control" placeholder="<?php echo $lang['footer_string_6']; ?>" name="newsemail" /> <button class="newsletter-button"><?php echo $lang['footer_string_7'];?></button><br> <br>
          </form>
          <?php
            if (isset ($_POST['newsemail'])){
              include("db_files/db.php");
              $newsemail = mysqli_real_escape_string($db, $_POST['newsemail']);
              $strSQL = "INSERT INTO newsletter (`email`) VALUES ( '$newsemail')";
              $query = mysqli_query($db, $strSQL);
              unset($_POST['newsemail']);
            }
          ?>
          <p class="footer-contact-info"> 999 999 999
        </div>
      </div>
    </div>
  </div>
</footer>

I want that after doing the insert, the variable $_POST['newsemail']is removed, since when reloading the page it tries to do the insert.

It would also be worth doing this in some other way but it doesn't occur to me, since this form is in the one footerof the page ( it footeris in a separate file "template/footer.php")

What options do I have to do something like that?

this is what i want to avoid: enter image description here

I have changed the title of the question

php
Martin Hope
Pavlo B.
Asked: 2020-11-10 01:42:14 +0800 CST
  • 8

This is the error:

Warning: Cannot modify header information - headers already sent by (output started at C:\xampp\htdocs\newop\template\header.php:80) in C:\xampp\htdocs\newop\mainindex.php on line 16

Line 80 of the header is this (it's a single line, only unpacked):

<div class="col-md-2 col-xs-2 device-select">
    <a href="budget.php?device=2">
        <img class="full-width" src="res/images/device2.jpg">
        </img>
    </a>
    <center class="popular-devices">
<?php echo $lang['header_dropdown_string_2'];?>
    </center>
</div>

... and this the code:

<?php
session_start();
if (!isset($_SESSION['user'])){ header("Location: index.php"); }
?>
<html>
  <body>
<?php
  include("template/head.php");
  include("template/header.php");
  include ("db_files/db.php");
  $usermail = mysqli_real_escape_string($db, $_SESSION['user']);
  $strSQL = "SELECT nivel FROM usuarios where email ='$usermail'";
  $query = mysqli_query($db, $strSQL);
  while ($result = mysqli_fetch_array($query)){
    if ($result['nivel'] == 1 || $result['nivel'] == 2){
    header("location: userlist:php");
  }
  
}
  $state = array(
    1 => $lang['state_name_string_1'],
    2 => $lang['state_name_string_2'],
    3 => $lang['state_name_string_3'],
    4 => $lang['state_name_string_4'],
  );
?>
  <div class="content container">
    <div class="row no-margin no-padding">
      <div class="col-md-2 hidden-sm hidden-xs">
        <div class="mainuserimage">
          <img class="full-width" src="res/images/users/user1.png"/>
        </div>
        <div>
<?php
            $strSQL = "SELECT nombre, email, telefono FROM usuarios where email ='$usermail'";
            $query = mysqli_query($db, $strSQL);
            while ($result = mysqli_fetch_array($query)){
              echo "<i class='glyphicon glyphicon-user'></i> ".$result['nombre']."<br>";
              echo "<i class='glyphicon glyphicon-envelope'></i> ".$result['email']."<br>";
              echo "<i class='glyphicon glyphicon-earphone'></i> ".$result['telefono']."<br>";
            }
?>
        </div>
      </div>
      <div class="col-md-10 col-xs-12">
      <h3 class="main-title"><?php echo $lang['mainindex_string_1']; ?></h3><br>
        <div class="row">
          <div class="col-md-2 col-xs-3">
            <p><?php echo $lang['mainindex_string_2']; ?></p>
          </div>
          <div class="col-md-2 col-xs-3">
            <p><?php echo $lang['mainindex_string_3']; ?></p>
          </div>
          <div class="col-md-2 col-xs-3">
            <p><?php echo $lang['mainindex_string_4']; ?></p>
          </div>
          <div class="col-md-2 hidden-sm hidden-xs">
            <p><?php echo $lang['mainindex_string_5']; ?></p>
          </div>
          <div class="col-md-2 col-xs-3">
            <p><?php echo $lang['mainindex_string_6']; ?></p>
          </div>
          <div class="col-md-2 hidden-sm hidden-xs">
            <p><?php echo $lang['mainindex_string_7']; ?></p>
          </div>
<?php
            $strSQL = "SELECT id, marca, modelo, imei, descripcion_problema, problem, problem2, problem2_namees, estado, fecha_creacion FROM tickets, problem2 WHERE tickets.email = '$usermail' AND tickets.problem2 = problem2.problem2_id";
            $query = mysqli_query($db, $strSQL);
            while($result = mysqli_fetch_array($query)){ ?>
              <div class="col-md-2 col-xs-3">
                <p><a href="repairdetail.php?id=<?php echo $result['id']; ?>"><i class="glyphicon glyphicon-list-alt"></i></a>  <?php echo $result['marca']; ?></p>
              </div>
              <div class="col-md-2 col-xs-3">
                <p><?php echo $result['modelo']; ?></p>
              </div>
              <div class="col-md-2 col-xs-3">
                <p><?php echo $result['imei']; ?></p>
              </div>
              <div class="col-md-2 hidden-sm hidden-xs">
                <p><?php echo $result['problem2_namees']; ?></p>
              </div>
              <div class="col-md-2 col-xs-3">
                <p><?php echo $state[$result['estado']]; ?></p>
              </div>
              <div class="col-md-2 hidden-sm hidden-xs">
                <p><?php echo $result['fecha_creacion']; ?></p>
              </div>
<?php
            }
?>
          </div>
        </div>
    </div>
  </div>
<?php
  include("template/footer.php");
?>
  </body>
</html>

php
Martin Hope
Pavlo B.
Asked: 2020-11-08 08:07:42 +0800 CST
  • 4

I need to add a new column in one of my tables, but this column in turn is a foreign key of another table but I don't know how to do it...

I tried this but obviously it gave me an error:

ALTER TABLE ordenes ADD COLUMN ticket varchar(50) FOREIGN KEY (ticket) REFERENCES tickets(ticket)
mysql
Martin Hope
Pavlo B.
Asked: 2020-11-04 04:19:27 +0800 CST
  • 3

How can you create a random hexadecimal number, say 5 digits? That is to say something like: A31D4.

Doing the following:

$mirand = dechex(rand());

It generates random numbers for me but sometimes they are 3 digits and most of the time they are 4, never 5.

php
Martin Hope
Pavlo B.
Asked: 2020-10-21 03:15:10 +0800 CST
  • 4

I have a dropdown that opens when I hover over it made with HTML and CSS like so:

.dropdown {
    position: relative;
    display: inline-block;
}
.dropbtn {
    height:30px;
    width:160px;
    background-color: #da291c;
    color: #ffffff;
    font-size: 16px;
    border: none;
    cursor: pointer;
}
.dropdown-content {
    display: none;
    position: absolute;
    background-color: #653734;
    min-width: 100px;
    box-shadow: 0px 8px 16px 0px rgba(0,0,0,0.2);
    padding: 5px 7px;
}

.dropdown:hover .dropdown-content {
    display: block;
}
<div class="dropdown">
					<button class="dropbtn"><?php echo $lang['topbar_string_1']; ?></button>
	<div class="dropdown-content">
		<div class="row no-margin no-padding">
			<form>
				<div class="col-xs-12 no-padding ">
					<label class="white-text"><?php echo $lang['topbar_string_2']; ?></label>
					<input type="text" class="form-control"/>
				</div>
				<div class="col-xs-12 no-padding ">
					<label class="white-text"><?php echo $lang['topbar_string_3']; ?></label>
					<input type="password" class="form-control"/>
				</div>
				<div class="col-xs-12 no-padding ">
					<button class="my-btn btn-primary" type="submit"><?php echo $lang['topbar_string_4']; ?></button>
				</div>
				<div class="col-xs-12 no-padding ">
					<a class="white-text" href="#"><?php echo $lang['topbar_string_5']; ?></a>
				</div>
			</form>
		</div>
	</div>
</div>

But I have run into a problem and it is that even if I click on an input to write and move the mouse out of the content of the dropdown, it disappears, and I need that if I click on it it no longer disappears, that it only disappears if I have I hovered the mouse over it and didn't click anything.

html
Martin Hope
Pavlo B.
Asked: 2020-10-14 23:45:51 +0800 CST
  • 7

I have been looking at a lot of questions for this same topic and I have not found anything so I am forced to formulate a new one...

I have a form like this:

var value=$.trim($(".form-control").val());
if(value.length>0)
{
    $("#user-data-next-button").attr('disabled', true);
} else{
    $("#user-data-next-button").attr('disabled', false);
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<form action="userdevdata.php" method="post">
	<label>Email</label>
	<input type="email" class="form-control" name="email"/>
	<label>Full name</label>
	<input type="text" class="form-control" name="name"/>
	<label>Address</label>
	<input type="text" class="form-control" name="address"/>
	<label>Postal code</label>
	<input type="number" class="form-control" name="cp"/>
	<label>City</label>
	<input type="text" class="form-control" name="city"/>
	<label>Province</label>
	<input type="text" class="form-control" name="province"/>
</form>
<button id="user-data-next-button" class="btn my-button" disabled="true" ><i class="glyphicon glyphicon-chevron-right"></i><i class="glyphicon glyphicon-chevron-right"></i></button>

What I need is that, as long as all the fields of my form are not completed, the button is deactivated (disabled). Instead, when I complete all the form fields, the button must be activated ( disabled="false") to be clickable.

     var userdevdata = 6;
 if($("#usermail").val().length > 0){ userdevdata = userdevdata-1;}
 if($("#username").val().length > 0){ userdevdata = userdevdata-1;}
 if($("#useraddress").val().length > 0){ userdevdata = userdevdata-1;}
 if($("#usercp").val().length > 0){ userdevdata = userdevdata-1;}
 if($("#usercity").val().length > 0){ userdevdata = userdevdata-1;}
 if($("#userprovince").val().length > 0){ userdevdata = userdevdata-1;}
 if(userdevdata == 0){
     $('#user-data-next-button').attr('disabled', false);
 } else {
     $('#user-data-next-button').attr('disabled', true);
 }

I'm not very good with this comment thing. I've done this and it doesn't work for me I don't know if it's right or wrong.

Thank you.

jquery