Question

PHP代码：

<?php
    if (isset($_POST['pwsubmitted']))  {
        $pwsub=$_POST['pass'];
        if ($pwsub != "TEST"){
                $s=1;
                $msg = "Incorrect Password";
                $msg2 = "Try Again";
            }
        else if ($pwsub == "TEST"){
                $s=2;
                $msg = "Password Accepted";
                $msg2 = "Your Download Is Below";
                $msg3 = "";
            }
            // so I can see what's going on when form submit happens
            echo "s="; var_dump($s); echo "</br>";
            echo "msg="; var_dump($msg); echo "</br>";
            echo "msg2="; var_dump($msg2); echo "</br>";
            echo "msg3="; var_dump($msg3); echo "</br>";
    }

    ?>

表格，展示位置如下：

<div class="passform">
            <form id="pwform" method="post" action="">
            <input type="hidden" name="submitted" value="pwsubmitted" />
              <center>
                <span class="titleblue">Enter The Password</span>
              </center>

           <input name="pass" id="pass" type="password" class="password"  />

        <input name="submit" type="submit" class="submit" style="cursor: pointer;" value="" /></div>
        </form>

        </div>

其他代码：

<?php 
if (isset($_POST['pwsubmitted'])) {
if ($s == 1) { Do This };
if ($s == 2) { Do This };
}
<?php if(!isset($POST['pwsubmitted'])) {
?>
<HTML FORM FROM ABOVE HERE>
<?php } ?>

当我提交表格时......没有任何反应。原始表单保持不变，就像未设置pwsubmitted post变量一样。发生了两件不同的事情，要么是说再次尝试，要么是显示内容。都没有发生。

我做错了什么？

Answer 1

在您的HTML中，您有此输入字段

<input type="hidden" name="submitted" value="pwsubmitted" />

在您的PHP中，您正在使用

if (isset($_POST['pwsubmitted']))  {

但该字段的名称为submitted。尝试

if (isset($_POST['submitted']))  {

Answer 2

您应该检查$ _POST ['submitted']（名称，而不是输入值）。

这个PHP密码表单有什么问题？我找不到它

2 个答案: