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Php 不能超过一个选择选项_Php - Fatal编程技术网

Php 不能超过一个选择选项

Php 不能超过一个选择选项,php,Php,这是一段让我发疯的剧本 if(isset($_POST['submit'])) { $cnt = $_POST['cnt']; for ($i ='0'; $i <= $cnt; $i++){ $group[$i] = $_POST['group']; $databox[$i] = $_POST['newdata'][$i]; $cve[$i] = $_POST['cve'][$i]; $play[$i] = $_POST['play'][$i];

这是一段让我发疯的剧本

if(isset($_POST['submit']))  {
$cnt = $_POST['cnt'];

for ($i ='0'; $i <= $cnt; $i++){
    $group[$i] = $_POST['group'];
    $databox[$i] = $_POST['newdata'][$i];
    $cve[$i] = $_POST['cve'][$i];
    $play[$i] = $_POST['play'][$i];
    $cause[$i] = $_POST['cause'][$i];
    }
}
但我得到了“值”的拆分(一个字母) 我的意思是,假设你选择了我得到的选项4 $group[0]=v $group[1]=e $group[2]=n


希望您理解我的意思

您将选择元素作为单个元素而不是数组提交。PHP脚本使用单个(字符串)元素的索引来获取单个字符

试一试



我认为团队[$I]会让它更稳定谢谢你们,它工作得很好
<form method="POST" name="tt" action="#" >
<?php
for ($i = 0; $i <= $cnt; $i++){
[...]

echo "<input type='text' name ='newdata[$i]' value ='$newdata' />
<input type='text' name ='cause[$i]' value ='$newcause' />
<input type='text' name ='play[$i]' value ='$newplay' />
<input type='text' name ='cve[$i]' value = '$newcve' />

<select name='group'>
<option></option>
<option value = 'afb'>1</option>
<option value = 'alc'>2</option>
<option value = 'abd'>3</option>
[...]
<option value = 'ven'>4</option>
<option value = 'nct'>5</option>
</select>";
} 

echo "<input type='hidden' name ='cnt' value= '$cnt' />
<input type='submit' name='submit' value='Go' />";
?>
</form>
$group[$i] = $_POST['group'][$i];
<select name='group[]'>