dtqf81594 2014-09-16 05:23
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Jquery Php根据结果发布值并获得结果

Here is my Jquery and Php file, in which i will post the values in php file through jquery,

In Simple :

    <script type="text/javascript" src="../jquery.min.js"></script>
   <script type="text/javascript" language="javascript">
  $(document).ready(function() {
      $("#driver").click(function(event){
       var name=$("#txt").val();   
       $.post("test_post.php",
             {name: name}).done(function(data){
              $('#stage').html(data);
               }

          );
      });
   });
   </script>
</head>
<body>
<input type="text" id="txt" value="Load Data" />
<input type="button" id="driver" value="Load Data" />
</body>

Here is the .php file. If the insert was success i should get alert as success else i should get alert as failure.

    <?php

 include ('conn.php');
 if( $_REQUEST["name"] )
{
$q = "INSERT INTO text (name) VALUES ($name)";
if(mysql_query($q))
{
echo 'success'.$q;
}
else
{
echo 'fail'.$q;
}
}
?>

And the echo should be displayed in the main page.

How can i do this ?

  • 写回答

1条回答 默认 最新

  • doulu1867 2014-09-16 05:26
    关注
    if( $_REQUEST["name"] )
    {
    $q = "INSERT INTO text (name) VALUES ($name)";
    

    Here you are not assigning the variable $name. So the query will definitely fail.

    Change it to:

    if( $_REQUEST["name"] )
    {
      $name=$_REQUEST['name'];
      $q = "INSERT INTO text (name) VALUES ($name)";
      -------------------------
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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