在登录表单上测试错误消息时 Foreach 无效

Question

我又回来了。我已经为表单制作了这个 php 代码-process.php;

    <?php
//prevent access if they haven't submitted the form.
if (!isset($_POST['submit'])) {
die(header("Location: form.php"));
}

session_start();

$_SESSION['formAttempt'] = true;

if (isset($_SESSION['error'])){
    unset($_SESSION['error']);
}

$required = array ("name", "email", "password1", "password2");

$_SESSION['error'] = array();

//Check required fields
foreach ($required as $requiredField) {
    if(!isset($_POST[$requiredField])  || $_POST[$requiredField] == "") {
        $_SESSION['error'] [] = $requiredField . "is required." ;
    }
}

//Validating Text in name
if (!preg_match('/^[/w.]+$/',$_POST['name'])) {
    $_SESSION['error'] [] = "Name must be letters and numbers only.";
}

//Validating Drop Down Selection
$validStates = array("Alabama","California", "Colorado", "Florida", "Illinois", "New York");
if (isset($_POST['state']) && $_POST['state'] != "") {
    if(!in_array($_POST['state'], $validStates)) {
        $_SESSION['error'] []="Please choose a valid state.";
    }
}

//Validating an Email
if(!filter_var($_POST['email'], FILTER_VALIDATE_EMAIL)) {
    $_SESSION['error'] = "Invalid email address";
}

//Ensure that the Passwords Match
if ($_POST['password1'] != $_POST['password2']) {
    $_SESSION['error'] [] = "Passwords do not match.";
}

//Final Disposition
if (isset($_SESSION['error']) && count ($_SESSION['error']) > 0) {
    die(header("Location: form.php")); 
} else {
    unset($_SESSION['formAttempt']);
    die(header("Location:success.php"));
}
?>

登录表单Div错误php代码；

<?php 
       if (isset($_SESSION['error']) && isset($_SESSION['formAttempt'])) {
           unset ($_SESSION['formAttempt']);
           print "Errors encountered <br />\n";
           foreach ($_SESSION['error'] as $error) {
           print $error . "<br />\n"; } 
   } 
 ?>

但是，在实际的登录表单上测试时，我得到了这个错误：

Warning: Invalid argument supplied for foreach() in E:\XAMPP\htdocs\form.php on line 20

我是 PhP 的新手，所以我可能遗漏了一些东西，但我不太清楚...抱歉

Answer 1

$_SESSION['error'] = "Invalid email address";

这将使它成为字符串而不是先前初始化的数组，从而导致 foreach 出错。尝试：

$_SESSION['error'][] = "Invalid email address";

还有

$_SESSION['error'] [] = "Passwords do not match.";

$_SESSION['error'] 和 [] 之间不应有 space，不确定这是否会导致问题，有多个地方有额外的 space 在你的验证中。

在登录表单上测试错误消息时 Foreach 无效

Invalid Foreach when testing error messages on login form

php