Question

我正在使用PHP验证HTML表单。我的所有代码都很好，除了消息字段。如果我在消息字段中放置了某些内容并且其他字段显示错误，则它仍会提交。但是，如果我将某些内容放入其他字段并且其他字段有错误，则不会提交，这是正确的。我怀疑它与我的代码的最后一个if-else有关。先感谢您。

contact.php
<?php

include 'includes/config.php';
$errors = FALSE;
$displayErrors = NULL;

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

$first = $_POST['first'];
$last = $_POST['last'];
$email = $_POST['email'];
$subject = $_POST['subject'];
$message = $_POST['message'];

//Connect to MYSQL Database server
$connect = mysql_connect(DB_HOST, DB_USER, DB_PASS) or die("Could not connect to MYSQL Database.");
$result = mysql_select_db(DB_NAME, $connect) or die("Could not connect to MYSQL table.");

//Clean Data to prevent malicous injections
mysql_real_escape_string(strip_tags(stripcslashes(trim($first))));
mysql_real_escape_string(strip_tags(stripcslashes(trim($last))));
mysql_real_escape_string(strip_tags(stripcslashes(trim($email))));
mysql_real_escape_string(strip_tags(stripcslashes(trim($subject))));
mysql_real_escape_string(strip_tags(stripcslashes(trim($message))));

if (empty($first)) {
    $errors = TRUE;
    $displayErrors .= 'First name is invalid.<br/>';
}
if (empty($last)) {
    $errors = TRUE;
    $displayErrors .= 'Last name is invalid.<br/>';
}
if (empty($email) || !filter_var($email, FILTER_VALIDATE_EMAIL)) {
    $errors = TRUE;
    $displayErrors .= 'Email is invalid.<br/>';
}
if (empty($subject)) {
    $errors = TRUE;
    $displayErrors .= 'Subject is invalid.<br/>';
}
if (empty($message)) {
    $errors = TRUE;
    $displayErrors .= 'Message is invalid.<br/>';
} else {
    $errors = FALSE;
          //Database insertion goes here
    echo 'Form submission successful. Thank you ' . $first . '.';

}

}
?>

Answer 1

基本上你是说，如果邮件不是空的那么$errors=FALSE，但你没有考虑所有其他字段。

我建议将错误放在如下的数组中：

$errors['email'] = true;

然后使用foreach

在最后检查该数组

Answer 2

尝试将else转换为

if($errors == FALSE) {//just corrected the equality operator
    //Database insertion goes here
    echo 'Form submission successful. Thank you ' . $first . '.'
}

Answer 3

制作你所有的ifs else ifs 这样你的最后一个不会搞砸你的验证。

Answer 4

在所有这些if之前，你应该把$ errors = FALSE。然后代替代码末尾的else，你可以把if（！$ errors）{//数据库插入到这里}

Answer 5

首先，Don't use mysql_* functions in new code。它们不再被维护and are officially deprecated。请参阅red box？转而了解prepared statements，并使用PDO或MySQLi - this article将帮助您确定哪个。如果您选择PDO here is a good tutorial。

我已经整理了你的逻辑，而不是使用$error=true等你应该构建一个错误数组（包含你的错误文本），然后检查那里是否为空，我还清理了魔术引号问题吧好像你有，而且可怕的多个mysql_real_escape_string，希望它有帮助

contact.php

<?php
include 'includes/config.php';
//Connect to MYSQL Database server
$connect = mysql_connect(DB_HOST, DB_USER, DB_PASS) or die("Could not connect to MYSQL Database.");
$result = mysql_select_db(DB_NAME, $connect) or die("Could not connect to MYSQL table.");

$errors = array();

if ($_SERVER['REQUEST_METHOD']=='POST') {

    function clean(&$value){
        if (get_magic_quotes_gpc()){
            $value = mysql_real_escape_string(strip_tags(trim($value)));
        }else{
            $value = mysql_real_escape_string(trim($value));
        }
    }
    array_walk($_POST,'clean');


    if (empty($_POST['first'])) {
        $errors['first']= 'First name is invalid.<br/>';
    }else{
        $first = $_POST['first'];
    }

    if (empty($_POST['last'])) {
        $errors['last']= 'Last name is invalid<br/>';
    }else{
        $last = $_POST['last'];
    }

    if (empty($_POST['email']) || !filter_var($_POST['email'], FILTER_VALIDATE_EMAIL)) {
        $errors['email'] = 'Email is invalid.<br/>';
    }else{
        $email = $_POST['email'];
    }

    if (empty($_POST['subject'])) {
        $errors['subject']= 'Subject is invalid.<br/>';
    }else{
        $subject = $_POST['subject'];
    }

    if (empty($_POST['message'])) {
        $errors['message']= 'Message is invalid.<br/>';
    }else{
        $message = $_POST['messsage'];
    }

    if(empty($errors)){
        //Database insertion goes here
        echo 'Form submission successful. Thank you ' . htmlentities($first) . '.';
    }else{
        //Errors, so your have $errors['name'] ect to output
        //so somthing like:
        echo (isset($errors['name'])?$errors['name']:null);

    }

}
?>

您也应该添加一个条件，该值是最小特定长度，strlen($_POST['first']) > 2等。

PHP表单验证逻辑有点偏

5 个答案: