Question

我正试图从表单中获取值但是在没有isset()的情况下调用时它给了我未定义的索引错误。

现在，当我使用isset()时，它会给我未定义的变量错误。

我该如何解决这个问题？

<html>
  <head></head>
  <body>
    <form action="conn.php" enctype="multipart/form-data" method="post">
        <p>Enter Database Username:  <input type="text" name="username"></p>
        <p>Enter Password:     <input type="password" name="pass"></p>
        <p><input type="submit" value="submit"></p>
    </form> 

<?php
if (isset($_POST['username'])) {
  $username=$_POST['username'];
}

if (isset($_POST['pass'])) {
  $pass=$_POST['pass'];
}

echo($username);
echo($pass);

?>
     </body>
</html>

Answer 1

因为您的$username和$pass变量都未定义且未经检查或isset()阻止

echo($username);
echo($pass);

首先使用空白值初始化这些变量

<?php  $username='';$pass='';
    if (isset($_POST['username'])) {
        $username=$_POST['username'];}
    if (isset($_POST['pass'])) {
        $pass=$_POST['pass'];}
    echo($username);
    echo($pass);
?>

或通过在check block中回显来简化： -

if (isset($_POST['username'])) {
  echo $username=$_POST['username'];}
if (isset($_POST['pass'])) {
  echo $pass=$_POST['pass'];}

Answer 2

是的，它是未定义的，因为你在if块的范围之外使用它，它首先检查存在。

仔细观察：

// if this fails
if (isset($_POST['username'])) {
    $username=$_POST['username'];
}

// if this fails
if (isset($_POST['pass'])) {
    $pass=$_POST['pass'];
}

// your echoing an undefined variable
echo($username);
echo($pass);

或者，您可以这样做：

if(isset($_POST['username'], $_POST['password'])) {
    // isset() can handle multiple parameters to check for its existence, if one of them is undefined, then its false
    $username = $_POST['username'];
    $password = $_POST['password'];

    echo $username . '<br/>' . $password;
}

Answer 3

因为当未发布表单时，变量将不会被设置。这个 -

if (isset($_POST['username']) && isset($_POST['pass'])) {
    $username=$_POST['username'];
    $pass=$_POST['pass']
    echo($username);
    echo($pass);
}

或

$username = $pass = '';
if (isset($_POST['username'])) {
    $username=$_POST['username'];}

if (isset($_POST['pass'])) {
    $pass=$_POST['pass'];}

echo($username);
echo($pass);

Answer 4

正如您所看到的，有很多方法可以设置初始变量......但您需要将它们设置为在没有警告的情况下回显它们：

// Set variables to post if set, else empty if not
$username = (isset($_POST['username']))? $_POST['username']: "";
$pass     = (isset($_POST['pass']))? $_POST['pass']: "";

echo($username);
echo($pass);

从PHP中的表单获取值时未定义的变量错误

4 个答案: