Question

以下代码提供了一个简单的注册表单，并使用JavaScript来验证用户的输入。在chrome中运行时，会出现预期的警报。但是在IE和Firefox中，页面只会在没有警报的情况下转到adduser.php，即使表格中没有输入任何内容。

CSS

.signup {
    font: normal 14px helvetica;
    color: #000000;
    border: solid 6px #555555;
    }

JS

function validate() {
    var fail = "";
    fail += validateForename(document.getElementById("forename").value);
    fail += validateSurname(document.getElementById("surname").value);
    fail += validateUsername(document.getElementById("username").value);
    fail += validatePassword(document.getElementById("password").value);
    fail += validateAge(document.getElementById("age").value);
    fail += validateEmail(document.getElementById("email").value);

    if (fail == "") return true;
    else alert(fail);
    return false;
}


function validateForename(str) {
    if (str == "") return "No forename has been found\n";
    return "";
}

function validateSurname(str) {
    if (str == "") return "No surname has benn found\n";
    return "";
}

function validateUsername(str) {
    if (str == "") return "No username has been found\n";
    if (str.length < 5) return "Username must be at least 5 characters\n";
    if (/[^a-zA-Z0-9_-]/.test(str)) return "Only a-z, A-Z, 0-9, - and _ are allowed in username\n";
    return "";
}

function validatePassword(str) {
    if (str == "") return "Password can not be empty\n";
    if (str.length < 6) return "Password must be at least 6 characters\n";
    if (!/[0-9]/.test(str) || !/[a-z]/.test(str) || !/[A-Z]/.test(str)) return "Password must have at least one each of a-z, A-Z, 0-9\n";
    return "";
}

function validateAge(str) {
    if (isNaN(str)) return "No age has been found\n";
    if (str < 18 || str > 110) return "Age must be between 18 and 110\n";
    return "";
}

function validateEmail(str) {
    if (str == "") return "No email address has been found\n";
    if (!(str.indexOf('.') > 0 && str.indexOf('@') > 0) || /[^a-zA-Z0-9_-.@]/.test(str)) return "The email address is invalid\n";
    return "";
}

HTML

<table class="signup" border="0" cellpadding="4" bgcolor="#eeeeee">
    <th colspan="2" align="center">Sign Up</th>
    <form method="post" action="adduser.php" onSubmit="return validate()">
        <tr>
            <td>Forname:</td>
            <td>
                <input type="text" name="forename" id="forename" maxlength="32" />
            </td>
        </tr>
        <tr>
            <td>Surname:</td>
            <td>
                <input type="text" name="surname" id="surname" maxlength="32" />
            </td>
        </tr>
        <tr>
            <td>Username:</td>
            <td>
                <input type="text" name="username" id="username" maxlength="16" />
            </td>
        </tr>
        <tr>
            <td>Password:</td>
            <td>
                <input type="password" name="password" id="password" maxlength="12" />
            </td>
        </tr>
        <tr>
            <td>Age:</td>
            <td>
                <input type="text" name="age" id="age" maxlength="3" />
            </td>
        </tr>
        <tr>
            <td>Email:</td>
            <td>
                <input type="text" name="email" id="email" maxlength="64" />
            </td>
        </tr>
        <tr>
            <td colspan="2" align="center">
                <input type="submit" value="Signup" />
            </td>
        </tr>
    </form>
</table>

Answer 1

问题在于这一行（55）：

if (!(str.indexOf('.') > 0 && str.indexOf('@') > 0) || /[^a-zA-Z0-9_-.@]/.test(str)) return "The email address is invalid\n";

您声明的字符类无效：

/[^a-zA-Z0-9_-.@]/  //--> Wrong
/[^a-zA-Z0-9_.@-]/  //--> Right

浏览器尝试将_-.解释为_和.之间的一系列字符，并且严重失败。

Answer 2

以下正则表达式（第55行）在Firebug中产生错误（SyntaxError：字符类中的无效范围）：

/ [^ A-ZA-Z0-9 _-。@] /

试试这个：

/ [^ \ W - ] + / GI

现在在FF和IE 11中为我工作。

Answer 3

这个正则表达式对象不太可靠：

/[^a-zA-Z0-9_-.@]/

Chrome可以容忍这一点，但Firefox并不了解_-.不是字符范围，例如a-z。您可以将-移到最后，以避免歧义。最终的正则表达式变为（i表示不区分大小写的匹配）

/[^a-z0-9_.@-]/i

JS在Chrome中工作但不在Firefox或IE中工作

3 个答案: