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嗨，我是网络开发的新手。我有一个提交 php 页面，它正在从以前的表单页面提交数据并显示提交确认信息。但是现在我要让它要求用户在提交到 SQL 之前再输入 2 个东西（姓名和房间号），所以......来自以前表单和名称的信息，房间号将一起插入到 SQL 中。我想在同一个页面中处理这个提交，但是对于新添加的 2 个数据（姓名和房间号）我不想做一个单独的 submit()，因为我不想刷新页面，

我该怎么做？如何将输入标签（Cname）中的值应用于 $Cname ？不刷新页面？这是我的代码：

  ...

<head>
<?php

$T1comment1 = $_POST['T1comment1'];//these are from previous form
$T1comment2 = $_POST['T1comment2'];//these are from previous form
$T1comment3 = $_POST['T1comment3'];//these are from previous form
$T1comment4 = $_POST['T1comment4'];//these are from previous form

$item_1 = $_POST['item_1'];//these are from previous form
$item_2 = $_POST['item_2'];//these are from previous form
$item_3 = $_POST['item_3'];//these are from previous form
$item_4 = $_POST['item_4'];//these are from previous form

$Cname = $_POST['Cname'];
$CRnumber = $_POST['CRnumber'];


?>
</head>
<body>

 <div>

   <form method="POST" name="namdAndRm" id="namdAndRm"  action="<?php echo $PHP_SELF;?>"  >


      <input type="text" id="Cname" name="Cname" value="your name here"></input>


      <input type="text" id="CRnumber" name="CRnumber" value="room no."></input>

   </form>

 </div>



<div>

    <div class="Back"></div>

    <div class="submit" onclick="goSubmit();"></div>

</div>

<script type="text/javascript">

function goSubmit(){
    <?php
$Cname = $_POST['Cname'];  //is this the right way to do it? 
$CRnumber = $_POST['CRnumber'];   // I tested a lot, seams direct "=" is not working...

$SqlStatement = "INSERT INTO T3survey (T1item_1, T1comment1, T1item_2, T1comment2, T1item_3, T1comment3, T1item_4, T1comment4, Cname, CRnumber, day) VALUES ('$item_1', '$T1comment1', '$item_2', '$T1comment2', '$item_3', '$T1comment3', '$item_4', '$T1comment4', '$Cname', '$CRnumber',  NOW())";
    $result = mysql_query($SqlStatement,$connection);
 if (!$result){ die("Error " . mysql_errno() . " : " . mysql_error());}
    ?>;
};

</script>
</body>
</html>
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1 回答 1

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您需要做的就是在表单中创建一堆隐藏字段，如下所示：

<input type='hidden' name ='T1comment1' value='<?php $T1comment1;?>'> 
<input type='hidden' name ='T1comment2' value='<?php $T1comment2;?>'> 
<input type='hidden' name ='T1comment3' value='<?php $T1comment3;?>'> 
<input type='hidden' name ='T1comment4' value='<?php $T1comment4;?>'>
于 2013-04-04T17:25:23.293 回答