我有具有由如下的函数生成20下拉字段一个html形式....验证在PHP为下拉菜单
<?php
$sub_1 = subject_show (subject_1, $scl_unique_id, $cls_sec_bat);
if ($sub_1)
{?>
<tr>
<td><?php echo $sub_1;?></td>
<td><select name="subject_1" id="subject_1" style="min-width:100px">
<?php
if(isset($subject_1)){?>
<option value="<?php echo $subject_1; ?>" selected="selected"><?php echo $subject_1; ?></option>
<?php }else{ ?>
<option value="<?php echo $sub_1;?>" disabled="disabled" selected="selected"><?php echo $sub_1;?></option>
<?php } ?>
<option value="<?php echo MAR_ABS;?>"><?php echo MAR_ABS;?></option>
<option value="<?php echo MAR_NA;?>"><?php echo MAR_NA;?></option>
<?php
for ($i=0; $i<=100; $i++)
{
echo '<option value="'.$i.'">'.$i.'</option>';
}
?>
</select></td>
</tr>
<?php
}
?>
在代码中,如果函数对$ sub_1的值,则打印在带有名称subject_1的html输出上的下拉字段。如果不是,它将不会打印。
相同的代码我有20次。如果函数的值为$ sub_2或$ sub_3或$ sub_4,直到$ sub_20,则该字段的名称为subject_2或subject_3或subject_4,直到subject_20。
根据用户输入和功能输出,可以打印任何下拉字段。
当阅读时间,我读所有的20个名字,如
if (isset($_POST['save_exit']))
{
$roll_number = strtoupper(trim(implode(' ', preg_split('/\s+/', $_POST["roll_number"]))));
$subject_1 = $_POST["subject_1"];
$subject_2 = $_POST["subject_2"];
$subject_3 = $_POST["subject_3"];
$subject_4 = $_POST["subject_4"];
$subject_5 = $_POST["subject_5"];
$subject_6 = $_POST["subject_6"];
$subject_7 = $_POST["subject_7"];
$subject_8 = $_POST["subject_8"];
$subject_9 = $_POST["subject_9"];
$subject_10 = $_POST["subject_10"];
$subject_11 = $_POST["subject_11"];
$subject_12 = $_POST["subject_12"];
$subject_13 = $_POST["subject_13"];
$subject_14 = $_POST["subject_14"];
$subject_15 = $_POST["subject_15"];
$subject_16 = $_POST["subject_16"];
$subject_17 = $_POST["subject_17"];
$subject_18 = $_POST["subject_18"];
$subject_19 = $_POST["subject_19"];
$subject_20 = $_POST["subject_20"];
我不希望用户提交空的下拉列表,所以我wrinting审定
if(empty(subject_1))
{
say error
}
if(empty(subject_2))
{
say error
}
我wrting这所有科目20次。
问题出在用户在论坛中只有三个主题,如subject_1和subject_2和subject_3。它会给所有剩余的17个主题带来错误。
我该如何阅读只验证表单上可用的主题?
请让我知道,如果它不明确。
请张贴什么样的HTML表单看起来像PHP使得它后(从页面源)为例。这是在表单元素'name ='属性中使用PHP的'[]'数组语法重新构建的。 –
我已经发布了表单代码。请让我知道你想要什么?我怎么才能得到它.. – user2642907
您需要检查表单元素是否已发布,但值为空 - 'isset($ _ POST [“subject _#”])''''subject_1 =='''。所以你的'if'可以是 - 'if(isset($ _ POST [“subject_1”] && empty($ subject_1)){error message}' – Sean