Question

我没有运行查询2（提交）表单。只需刷新页面即可。我如何解决这个功能“diе（）;”

            <?php
            if (isset($_POST['query1'])) {
//working.. next
            ?>
            <?php
            if (isset($_POST['query2'])) {
            die("Go die"); //not working, how to fix?
            }
            ?>
            <form method="post">
                ...Query2 (not working)
                <p><input type="submit" value="Query2" name="query2"></p>
            </form>
            <?php
            }
            ?>
            <form method="post">
                ...Query1 (working)
                <p><input type="submit" value="Query1" name="query1"></p>

Answer 1

query2条件位于query1条件内，因此永远不会使用您当前的代码执行（因为处理query1时未设置query2）。它需要移到外面，或者您需要更新form，以便在query1中设置form2。所以：

<?php
if (isset($_POST['query1'])) {
     //working.. next
     if (isset($_POST['query2'])) {
          die("Go die"); //not working, how to fix?
     }
?>
<form method="post">
    ...Query2 (not working)
    <p><input type="submit" value="Query2" name="query2"></p>
    <input type="hidden" value="just so we are set" name="query1">
</form>
<?php
}
?>

（上述方法假定每次都应执行query1代码）或

<?php
if (isset($_POST['query1'])) {
     //working.. next
}
if (isset($_POST['query2'])) {
          die("Go die"); //not working, how to fix?
}
if (isset($_POST['query1'])) {
?>
<form method="post">
    ...Query2 (not working)
    <p><input type="submit" value="Query2" name="query2"></p>
    <input type="hidden" value="just so we are set" name="query1">
</form>
<?php
}
?>

我没有运行（提交）表格

1 个答案: