Question

我有一个基本的php验证的简单联系表格，这是代码的相关部分：

    $err = '';

    if(!preg_match("/^[a-zA-Z ]*$/",$name)) {
        $err .= "Name : Only letters and white space allowed.\n";
    }
    if(strlen($name)<3){
        $err .= "Name : Please enter your full name.\n";
    }
    if (!filter_var($email, FILTER_VALIDATE_EMAIL)) {
        $err .= "Email : Invalid email format.\n";
    }
    if(strlen($email)<8){
        $err .= "Email : Okay, how about using a correct email address?\n";
    }
    if(strlen($message)<30){
        $err .= "Message : Message too short.\n";
    }

    if($err != ''){
        $err = 'There were few errors : \n '.$err;

        //THE ALERT BELOW DOESN'T WORKS, gets added into body code, but does not popup
        echo '<script type="text/javascript">alert("'.$err.'");</script>';

        // While this one works well
        //echo '<script type="text/javascript">alert("This alert works. \nThe one above does not.");</script>';
    }
    else{
        //whatever
    }

问题是最后一个if条件中的alert()不起作用，它会添加到正文代码中但不会弹出。如果我使用普通文本而不是$err变量（就像注释掉的那样） - 这一切都很好。

过去我曾经多次遇到这个问题，我从来不明白为什么会发生这种情况。什么是解决方案，我总是宁愿删除警报并显示响应消息，通常在div或其他东西，但这次我需要知道 - 为什么??

Answer 1

将新行改为此

$err .= "Name : Please enter your full name.\\n";

参考：Cant add new lines in Javascript Alert Box

Answer 2

if($err != ''){
        $err = 'There were few errors : \n '.$err;


        echo '<script type="text/javascript">alert('.$err.');</script>';
    }
    else{
        //whatever
    }

使用PHP变量不会弹出Javascript警报

2 个答案: