我有一个这样的复选框翻转开关:
<form>
<input type="text" name="control_name[]"/>
<label for="flip-checkbox-4">Flip toggle switch checkbox:</label>
<input type="checkbox" data-role="flipswitch" name="control_name_switch[]" id="flip-checkbox-4" checked=""/>
</form>
使用此 php 代码读取表单:
if (!empty($_POST)) {
$controlname = $_POST['control_name'];
$controlstate = $_POST['control_name_switch'];
$program = $_POST['program'];
$i = 0;
$ctrlSize = sizeof($controlname);
for ($i = 0; $i < $ctrlSize; $i++) {
if (empty($_POST['control_name_switch'])) {
$controlstate[$i] = " ";
} else {
$controlstate[$i] = "checked";
}
}
当我在 php 中阅读帖子时,即使我点击它并关闭它,我也会让它始终“打开”。请帮忙。