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PlayerWet
Asked: 2020-06-05 01:18:10 +0800 CST 2020-06-05 01:18:10 +0800 CST

ajax request gives error and does not open php file or send data

  • 772

I have a problem with an Ajax request . I don't know if I'm doing it correctly, the fact is that I've always done it with the Jquery library , but now I want to do it only with Javascript .

Code

var enviarObjeto={'titulo': 'hola','usuario': 'Pier'};
var objeto = "json_name=" + JSON.stringify(JSON.stringify(enviarObjeto));
var xhr = new XMLHttpRequest();
xhr.open('POST', "guardar.php", true);
xhr.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
if (xhr.status == 200) {
   xhr.send(objeto);
   xhr.onload = function () {
   alert(this.responseText);
 };

alert('enviado ok');
} 
else {
    alert('error');
}

Can you tell me what I have done wrong? It is the first time that I do it with Javascript and I would be grateful if you tell me where my errors are.

php

2 Answers

  1. If you want to send an object Json, try adding the following:

    xhr.setRequestHeader('Content-Type', 'application/json');

    On the other hand, you are not sending the values, for example withxhr.send({});

    At this URL is an example similar to what you want to do.

    Cheers

    • 1
  2. Best Answer

    I recommend that you use FormData to send data. Here I leave you an example.

      envioAjax();
        function envioAjax() {
            formData = new FormData();
            formData.append("nombre", "Jose");
            formData.append('apellido', 'Sotteccani');
            formData.append('edad', '26');
            var xmlHttp = new XMLHttpRequest();
            xmlHttp.onreadystatechange = function () {
                if (xmlHttp.readyState == 4 && xmlHttp.status == 200) {
                    alert(xmlHttp.responseText);
                }
            }
            xmlHttp.open("post", "test.php");
            xmlHttp.send(formData);
        }

    This is the file in php that receives the data.

    <?php
    $datos = $_POST['nombre'];
    $apellido = $_POST['apellido'];
    $edad = $_POST['edad'];
    echo "Los datos de usuario son: ".$datos." ".$apellido." ".$edad." ";
?>
    • 1