【发布时间】:2016-01-03 02:10:34
【问题描述】:
<?php
include("dbFunctions.php");
$query ="SELECT * FROM `physical_examination` WHERE `PE_Opt_ans`= 0";//select form options name
$result = mysqli_query($link,$query);
?>
<div id="tabs-3">
<table>
<form action="<?php echo $_SERVER['PHP_SELF']?>" method="post" name="tab0">
<?php while ($arrayResult = mysqli_fetch_array($result)){ ?>
<tr>
<td><label for="input"><b><?php echo $arrayResult['PE_Opt_name']?></label></td>
<?php if ($arrayResult['PE_Opt_type'] == "textarea") { ?>
<td><textarea rows="8" cols="45"name = "other1"></textarea></td>
<?php } else { ?>
<td><input type="<?php echo $arrayResult['PE_Opt_type']?>" name="input<?php echo $arrayResult['id']?> " ></td>
</tr>
<?php } ?>
<?php } ?>
<br> <input value="Submit" type="submit" name="submit1">
</form>
</table>
<?php
include "dbFunctions.php";
if(isset($_POST['submit1'])) {
$number = $_POST['other1'];
我被困在这里了。
点击提交按钮后,我如何$_POST 基于while循环的表单?输入<?php echo $arrayResult['id']?>的名称值是否需要另一个while循环?
【问题讨论】:
-
HTML 中不能有
<table><form>,应该是<form><table>。