Question

我正在使用下面的脚本验证表单数据。当我提交表单时，如果有任何错误，错误消息正确显示，但如果没有错误和验证成功，我尝试回显变量来测试脚本，但脚本只显示这个：[] 请检查代码并帮我解决这个问题。

    <?php

//included files
include("./includes/connect.php");
include("./includes/functions.php");
$errors = array();

//checking if user have submitted the form
if(isset($_POST['submitted'])) {
    //validating and cleaning submitted form data ... 
    if (isset($_POST['name']) && !empty($_POST['name'])) {
        if(preg_match("/^[a-zA-Z ]{2,20}$/", strip_trim($_POST['name']))) {
            $cln_name = clean_data($_POST['name']);
        } else {
            $_POST['name'] = FALSE;
            $errors[] = "The name you entered is not valid";
        }

    } else {
        $errors[] = "You have not enter your name!";
    }

    if(isset($_POST['email']) && !empty($_POST['email'])) {
        $cln_email = filter_var($_POST['email'] , FILTER_SANITIZE_EMAIL);
        if(filter_var($cln_email, FILTER_VALIDATE_EMAIL)) {
            $cln_email = clean_data($cln_email);
        } else {
            $_POST['email'] = FALSE;
            $errors[] = "The email you entered is not valid";
        }
    } else {
        $errors[] = "You have not provide you email!";
    }

    if(isset($_POST['plate_num']) && !empty($_POST['plate_num'])) {
        if(ctype_alnum($_POST['plate_num']) && strlen($_POST['plate_num']) >= 5) {
            $cln_plate_num = clean_data($_POST['plate_num']);
        } else {
            $_POST['plate_num'] = FALSE;
            $errors[] = "The plate number you provided is not a valid plate number";
        }
    } else {
        $errors[]= "You have not provide a plate number";
    }
    //checking for errors and printing errors.. 
    if (count($errors > 0)) {

        $errors_to_json = json_encode($errors);
        echo $errors_to_json;
        //foreach ($errors as $error) {
            //echo $error . "<br />";
        //}
    } else {
        echo $cln_name . "<br />";
        echo $cln_email . "<br />";
        echo $cln_plate_num;
    }
} else {
    echo "You did not submit the form!";
}

?>

此脚本仅返回此内容： []

有什么想法吗？

functions.php：

 <?php

function clean_data($data) {
    if(function_exists('mysql_real_escape_string')) {
        global $dbc;
        $data = mysql_real_escape_string(trim($data), $dbc);
        $data = strip_tags($data);
    } else {
        $data = mysql_escape_string(trim($data));
        $data = strip_tags($data);
    }
    return $data;
}


function strip_trim($data) {
    $data = stripslashes(trim($data));
    return $data;
}



?>

Answer 1

你的if条件有问题：

//checking for errors and printing errors.. 
if (count($errors > 0)) {
...

这将始终返回TRUE，因为$error = []和count([] > 0)会导致TRUE

这就是为什么你总是最终：

$errors_to_json = json_encode($errors);
echo $errors_to_json;
// Will indeed display '[]' because json_encode([]) is '[]'

我相信你的意思是：

if (count($errors) > 0) {
...

php脚本显示了一个weired结果

1 个答案: