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Asked: 2020-07-12 19:14:53 +0800 CST 2020-07-12 19:14:53 +0800 CST

Error passing variable from Javascript to PHP from the same file?

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First of all mention that I found questions similar to this one with possible solutions or closed for being duplicates but without a solution.

To pass the variable from JavaScriptto phpI decided to do it Ajaxthis way:

<script type="text/javascript">
        function ejecutar(){
		console.log("Inicia");
		var count = 5;
		$.ajax({
			url: "index.php",
			type: "POST",
			data: {num:count},

			success : function(json) {
		        console.log("success");
		    },

		    error : function(xhr, status) {
		        console.log("error "+" xhr: "+xhr+" Status: "+status);
		    },

		    complete : function(xhr, status) {
		        console.log("complete "+" xhr: "+xhr+" Status: "+status);
		        console.log("count = "+count);
		        console.log("num = "+num);
		    }
		});
		console.log("Fin");
            }
	</script>

And so I want to get the value of the variable in php:

<?php
if(isset($_POST['num'])){
	echo "num = ".$identificador;
}else{
	echo "variable vacia";
}
?>

This is my complete file index.php:

<!DOCTYPE html>
<html>
<head>
    <title>Ajax con php</title>
    <meta charset="utf-8">
    <script src="js/jquery-3.2.1.min"></script>

    <script type="text/javascript">

        function ejecutar(){
            console.log("Inicia");
            var count = 5;
            $.ajax({
                url: "index.php",
                type: "POST",
                data: {num:count},
                success : function(json) {
                    console.log("success");
                },

                error : function(xhr, status) {
                    console.log("error "+" xhr: "+xhr+" Status: "+status);
                },

                complete : function(xhr, status) {
                    console.log("complete "+" xhr: "+xhr+" Status: "+status);
                    console.log("count = "+count);
                    console.log("num = "+num);
                }
            });
            console.log("Fin");
        }
    </script>
</head>

<body>
<?php
    if(isset($_POST['num'])){
        echo "num = ".$identificador;
    }else{
        echo "variable vacia";
    }
?>
    <button type="button" onclick="ejecutar();">Ejecutar</button>
</body>
</html>

The result I get in the Google Chrome Developer Tools console is the following:

enter image description here and in the browser the following:

enter image description here

Finally mention that it is the 1st time that I do this so if I am omitting something basic or simple an apology

php

1 Answers

  1. Best Answer

    Specific errors can be seen at first sight, such as:

    • Trying to print variables without being declared or assigned with any value (PHP) specifically the variable $identificadorfor which it will show an undefined variable error.
    • I don't understand for what purpose you are trying to print the variable numsince you are using it as the clavevalue of the variable countyou access from PHP.
    • The error that it shows in the console is precisely because of this, it is not declared at the function level to be able to access it, it will only be accessible to ajax. Therefore, the first solution would be to declare it before and initialize it, but it would not make sense since it does not use it for anything except for the key of the sent value.
    • As you are making the call ajaxto the same file , it will PHPreturn the content HTMLas well, preferably create another file for only the data processing from PHPand it would change in the urlin ajaxfor example test.phpwhere only your code would go PHPor cut with exit finish the execution and place the code at the beginning of the html, I recommend the first option , create a new file

    Your code would be:

     <?php
  /* Para validar que la solicitud sea de tipo POST , (solo para ajax se ejecutará)*/
  if ($_SERVER['REQUEST_METHOD'] === 'POST') {
    if(isset($_POST['num'])){
        echo "num = ".$_POST['num'];
        exit(); /* Para que no siga imprimiendo el resto*/
    } else{
        echo "variable vacia";
        exit();
      }
   }
  ?>
    <!DOCTYPE html>
    <html>
    <head>
    <title>Ajax con php</title>
    <meta charset="utf-8">
    <script src="js/jquery-2.1.1.min.js"></script>
    <script type="text/javascript">

        function ejecutar(){
            console.log("Inicia");
            var count = 5;
            $.ajax({
                url: "index.php",
                type: "POST",
                data: {num:count},
                success : function(json) {
                    console.log("success");
                    console.log(json);
                },


            });

        }
    </script>
</head>

<body>

    <button type="button" onclick="ejecutar();">Ejecutar</button>
</body>
</html>
    • 1