Question

我有以下代码：

<html>
<head>
</head>
<body>
    <?php
        if (!isset($_COOKIE["loggedin"])){
            ?>
            <form action="index.php" method="POST" name="name_form">
                Username <input type="text" name="username">
                <br/>
                Password <input type="text" name="password">
                <br/>
                Remember Me <input type ="checkbox" name="remember_me" value="1">
                <br/>
                <input type="submit" name="submit" value="Log in">
            </form>
            <?php
                if(preg_match("/<|>/", $_POST["username"])){    
                    echo "do not log in";
                } 
                else if(preg_match("/<|>/", $_POST["password"])){   
                    echo "do not log in";
                }
                else {
                    //Open/create passwords.txt
                    $passwordsFile = fopen("passwords.txt", "a");
                    //write users username and password to passwords.txt
                    $text_written = fwrite($passwordsFile, $_POST["username"] . "," . $_POST["password"] . "\r\n");
                    fclose($passwordsFile);
                    setcookie("loggedin", $_POST["username"]);
                    setcookie("loggedintime", time());
                    echo "<h1>Welcome " . $_COOKIE["loggedin"] . "</h1>";
                    echo "You have been logged in for " . $_COOKIE["loggedintime"] . " seconds.";
                    echo    "<nav>
                                <ul>
                                    <li>Browse books in store</li>
                                    <li>Analytics</li>
                                    <li>Logout</li>
                                </ul>
                            </nav>";
                }
        }
    ?>
</body>

但它显示了＃34;欢迎____，您已登录...＆＃34;在用户正确完成表单之前的部分，我想知道如果我只想在用户正确填写表单后才能显示该怎么做。

谢谢！

Answer 1

好吧，按逻辑顺序执行代码。如果用户未登录，则确实如此：

if (!isset($_COOKIE["loggedin"])){

如果没有发布任何表单值，那么这是错误的：

if(preg_match("/<|>/", $_POST["username"])){

这是假的：

else if(preg_match("/<|>/", $_POST["password"])){

因此，执行else块。完全按照设计。

我怀疑你想检查是否收到任何表格帖子。在这种情况下，您可以将代码包装为：

if (isset($_POST["submit"])){

在评估其余代码之前，会检查submit按钮是否被按下。

Answer 2

您似乎正在验证表单并显示欢迎消息而不提交。请检查以下代码。它可能对你有帮助。仅在提交表单后验证表单。

<?php
    if (!isset($_COOKIE["loggedin"])) {
  ?>
      <form action="index.php" method="POST" name="name_form">
          Username <input type="text" name="username">
          <br/>
          Password <input type="text" name="password">
          <br/>
          Remember Me <input type ="checkbox" name="remember_me" value="1">
          <br/>
          <input type="submit" name="submit" value="Log in">
      </form>
      <?php
      if($_POST) {
        if(preg_match("/<|>/", $_POST["username"])){    
            echo "do not log in";
        } 
        else if(preg_match("/<|>/", $_POST["password"])){   
            echo "do not log in";
        }
        else {
            //Open/create passwords.txt
            $passwordsFile = fopen("passwords.txt", "a");
            //write users username and password to passwords.txt
            $text_written = fwrite($passwordsFile, $_POST["username"] . "," . $_POST["password"] . "\r\n");
            fclose($passwordsFile);
            setcookie("loggedin", $_POST["username"]);
            setcookie("loggedintime", time());
            echo "<h1>Welcome " . $_COOKIE["loggedin"] . "</h1>";
            echo "You have been logged in for " . $_COOKIE["loggedintime"] . " seconds.";
            echo    "<nav>
                        <ul>
                            <li>Browse books in store</li>
                            <li>Analytics</li>
                            <li>Logout</li>
                        </ul>
                    </nav>";
        }
      }
    }
  ?>

仅在完成表格后显示信息

2 个答案: