更新数据库

我尝试写的代码，你可以通过一个的fancybox iframe中更新数据库

我的代码犯规里面的表单更新数据库的内容似乎即使显示没有错误工作。数据库没有更新

这里是我的代码

editschool.php（这是我的fancybox iframe的内容）

<?php 
    $temp = mysql_query("SELECT * from tertiary_school where tschool_id = $_GET[tschool_id]"); 
    $temp = mysql_fetch_array($temp); 
    ?> 
    <center> 
    <form class="form-inline" method = 'post' enctype="multipart/form-data"> 
    <table> 
    <tr> 
    <td width="40%"> 
    Edit School Name: 
    </td> 
    <td> 
    <input name = "tschool_name" type="text" class="input-xlarge" id = "tschool_name" value="<?php echo $temp[tschool_name]; ?>"><button type="submit" value = "submit" name = "submit" class="btn btn-primary" onclick="event.preventDefault(); parent.$.fancybox.close();">Save Changes</button> 
    </td> 
    </tr> 
    </table> 
    </form> 
    </center> 
    <?php 
    if(isset($_POST['submit'])) 
     { 
      $tschool_name = $_POST['tschool_name']; 
      $tschool_id = $_GET['tschool_id']; 

      mysql_query("UPDATE tertiary_school SET tschool_name=$tschool_name WHERE tschool_id=$tschool_id") or die(mysql_error()); 
     } 
    ?>

在此先感谢

来源

2012-11-19 itsover9000

尝试呼应'$ tschool_name'和'$ tschool_id' – Ravi

我没有和正确的数据就出来了。现在我的问题是为什么它不更新我的数据库？我是否需要在我的代码中添加某种父标签？ – itsover9000

如果一切越来越细见下文 – Ravi

没有$ _GET [ 'schoolid']当表单发布其数据时。

您可以更新表单动作属性，以便沿着

<form action="editschool.php?schoolid=<?php echo $_GET['schoolid'];?>" ...

来源

2012-11-19 12:53:08 Dale

我获取变量来自我的父页面，所以我的变量存在 – itsover9000

它存在于加载时，但不是当表单被发送时 – Dale

我认为这就是发生什么.odd ... – itsover9000

尝试下面的代码和后期输出。

<?php 
    $temp = mysql_query("SELECT * from tertiary_school where tschool_id = $_GET[tschool_id]"); 
    $temp = mysql_fetch_array($temp); 
    ?> 
    <center> 
    <form class="form-inline" method = 'post' enctype="multipart/form-data"> 
    <table> 
    <tr> 
    <td width="40%"> 
    Edit School Name: 
    </td> 
    <td> 
<input type="hidden" name="tschool_id" value="<?php echo $_GET[tschool_id];?>" /> 
    <input name = "tschool_name" type="text" class="input-xlarge" id = "tschool_name" value="<?php echo $temp[tschool_name]; ?>"><button type="submit" value = "submit" name = "submit" class="btn btn-primary" onclick="event.preventDefault(); parent.$.fancybox.close();">Save Changes</button> 
    </td> 
    </tr> 
    </table> 
    </form> 
    </center>

来源

2012-11-19 12:58:15 VibhaJ

我试过但数据库仍然不更新 – itsover9000

$query = "UPDATE `tertiary_school` SET `tschool_name`='$tschool_name' WHERE `tschool_id`='$tschool_id'"; 
mysql_query($query) or die(mysql_error());

来源

2012-11-19 12:59:11 Ravi

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