MySQL query in php does not recognize variable sent from ajax

Question

MySQL query in php does not recognize variable sent from ajax

Navigation

#1 by (1 votes)
#2 by (0 votes)

0

To pass the variable from JavaScript to php I do it with Ajax in this way:

 <script type="text/javascript">
       function ejecutar(num_js){
        $.ajax({
            url: 'mi_php.php',
            type:'POST',
            data: {num:num_js},
               success : function(json){
               $('#contenedor-texto').html(json);
                       console.log("success");
                       console.log(json);
                       alert(json);
                       }
              });
          }
 </script>

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

<?php
     $num = $_POST["num"];
?>

This is my full index.php file:

<!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(num_js){
        $.ajax({
            url: 'mi_php.php',
            type:'POST',
            data: {num:num_js},
               success : function(json){
               $('#contenedor-texto').html(json);
                       console.log("success");
                       console.log(json);
                       alert(json);
                       }
              });
          }
    </script>
</head>

<body>
    <button type="button" onclick="ejecutar('12');">Ejecutar</button>
</body>
</html>

This is my full mi_php.php file:

   <?php
        $num = $_POST["num"];
        echo "Num = ".$num." ";

        $link = mysql_connect('localhost', 'root', 'passw');
        if (!$link){
            die('Could not connect: ' . mysql_error());
        }
        $sql = "SELECT * from tabla where num = '".$num."' ";
        echo "Consulta SQL = ".$sql;
        $result_q = mysql_query($sql);
        $array_return = array();
        while($result = mysql_fetch_array($result_q)){
            $array_return[$result['campo']]['campo1'] = $result['campo1'];
            $array_return[$result['campo']]['campo2'] = $result['campo2'];
            $array_return[$result['campo']]['campo3'] = $result['campo3'];
            $array_return[$result['campo']]['campo4'] = $result['campo4'];
        }
?>

The console.log(json); and the alert(json); of the file index.php show me the following:

Num = 12 Consulta SQL = SELECT * from tabla where num = '12'

But the echo "Consulta SQL = ".$sql; of line 10 of the file mi_php.php shows me the following:

Consulta SQL = SELECT * from tabla where num = ''

And the result of the query SQL on the web page is as if it had done this:

SELECT * from tabla where num = ''

javascript php json ajax

asked by El Cóndor 14.07.2017 в 23:52

source

2 answers

Delete or change Footer error when connecting to MSSQL 2008 without instance

score 1 · Answer 1

I recommend you change your Ajax function so you can receive a JSON-type response

<script type="text/javascript">
       function ejecutar(num_js){
        $.ajax({
            url: 'mi_php.php',
            type:'POST',
            dataType: 'JSON',
            data: {num:num_js},
               success : function(json){
               $('#contenedor-texto').html(json);
                       console.log("success");
                       console.log(json);
                       alert(json);
                       }
              });
          }
 </script>

And then send the result to the console to see what it gives back

   <?php
        $num = $_POST["num"];

        $link = mysql_connect('localhost', 'root', 'passw');
        if (!$link){
            die('Could not connect: ' . mysql_error());
        }
        $sql = "SELECT * from tabla where num = '".$num."' ";
        $result_q = mysql_query($sql);
        $array_return = array();
        while($result = mysql_fetch_array($result_q)){
            $array_return[$result['campo']]['campo1'] = $result['campo1'];
            $array_return[$result['campo']]['campo2'] = $result['campo2'];
            $array_return[$result['campo']]['campo3'] = $result['campo3'];
            $array_return[$result['campo']]['campo4'] = $result['campo4'];
        }

      echo json_encode(array('num' => $num, 'sql' => $sql,'data' => $array_return));
?>

Add to return the value of the variable $ sql , but you should never send your queries in your response. This is only for a development environment.

score 0 · Answer 2

You need to indicate the dataType:'json' in your Ajax

<script type="text/javascript">
       function ejecutar(num_js){
        $.ajax({
            url: 'mi_php.php',
            type:'POST',
            data: {num:num_js},
            dataType:'json',
               success : function(json){
               $('#contenedor-texto').html(json);
                       console.log("success");
                       console.log(json);
                       alert(json);
                       }
              });
          }
 </script>

According to the JQuery source: link

Alternate to this you can try sending it in the URL

<script type="text/javascript">
       function ejecutar(num_js){
        $.ajax({
            url: 'mi_php.php?num='+num_js,
            type:'POST',
            data: {},
               success : function(json){
               $('#contenedor-texto').html(json);
                       console.log("success");
                       console.log(json);
                       alert(json);
                       }
              });
          }
 </script>