Question

我正在使用JavaScript在html中验证我的表单，但它似乎不起作用。我觉得我的代码是正确的。我不确定会出现什么问题。没有显示任何警告。任何人都知道这里有什么问题吗？提前感谢。

<script>
    function validate() {
        var password2 = document.getElementById("user_password").value;
        var repassword1 = document.getElementById("user_repassword").value;
        var email1 = document.getElementById("user_email").value;
        var atpos = email1.indexOf("@");
        var dotpos = email1.lastIndexOf(".");
        if (password2.lenght < 7) {
            alert("Password must be at least 8 characters long.");
            return false;
        } else if (password2 !== repassword) {
            alert("* Password Does Not Match");
            return false;
        } else if (atpos < 1 || dotpos < atpos + 2 || dotpos + 2 >= email.length) {
            alert("Not a valid e-mail address");
            return false;
        } else {
            return true;
        }
    }
</script>
<form name="submitform" method="post" action="signupsuccessfull.php"      onsubmit="return validate(this)" >
   <p id="error_para">* fields are required</p>
   <span id="error_para">*</span> Name:<input type="text" name="name" required  /><br /><br />
   <span id="error_para">*</span> Username:<input type="text" name="username"   required id="field_username"  /><br /><br />
   <span id="error_para">*</span> Password:<input type="password" name="password" id="user_password" required />
   <br /><br />
   <span id="error_para">*</span> Re-Enter Password:<input type="password" name="repassword" id="user_repassword" required /><br /><br />
   <span id="error_para">*</span> Email:<input type="text" name="email" id="user_email" required />
   <br /><br />
   <span id="error_para">*</span> Gender:<input type="radio" name="gender" value="male" required />Male
   &nbsp;<input type="radio" name="gender" value="female" />Female<br />
   About Urself:
   <textarea rows="5" cols="18" name="about" ></textarea>
   <br /><br />
   <input type="submit" value="Create Account" name="submit" />
   &nbsp;<input type="button"  value="Cancle" onClick="parent.location='../example/try2.html'"><br /><br /> 
</form>

Answer 1

您的代码中存在错误。可能会有更多..

if (password2.lenght < 7) {   // length is the right usage
...

} else if (password2 !== repassword) { // it is repassword1 as per you defined.

使用浏览器控制台（开发人员工具）找到并解决所有这些问题。

除非调用代码，否则函数内部的javascript错误不会显示，因此其余代码仍然有效。

Answer 2

有2个拼写错误。

应为password2.length

它也应该是repassword1，而不仅仅是repassword。

Answer 3

要修复，请更改行

 if (password2.lenght < 7) {

到

 if (password2.length < 7) {

将repassword更改为repassword1

Answer 4

我使用下面的代码

来完成这项工作

function validate() {

    var password2 = document.getElementById("user_password").value;
    var repassword1 = document.getElementById("user_repassword").value;
    var email1 = document.getElementById("user_email").value;
    var atpos = email1.indexOf("@");
    var dotpos = email1.lastIndexOf(".");

    var emilPat = /( *?[0-9a-zA-Z] *?){8,}/; 
    var m;
    m = emilPat.exec(password2) 
    if(m== null){
        alert("Password must be at least 8 characters long.");
        return false;
    }
    if (password2 !== repassword) {
        alert("* Password Does Not Match");
        return false;
    } else if (atpos < 1 || dotpos < atpos + 2 || dotpos + 2 >= email.length) {
        alert("Not a valid e-mail address");
        return false;
    } else {
        return true;
    }
}

Answer 5

像这样使用：

<script>
    function validate() {
        var password2 = document.getElementById("user_password").value;
        var repassword1 = document.getElementById("user_repassword").value;
        var email1 = document.getElementById("user_email").value;
        var atpos = email1.indexOf("@");
        var dotpos = email1.lastIndexOf(".");
        if (password2.lenght < 7) {
            alert("Password must be at least 8 characters long.");
            return false;
        } else if (password2 != repassword1) {
            alert("* Password Does Not Match");
            return false;
        } else if (atpos < 1 || dotpos < atpos + 2 || dotpos + 2 >= email1.length) {
            alert("Not a valid e-mail address");
            return false;
        } else {
            return true;
        }
    }
</script>

javascript表单验证无法正常工作且未显示任何警报

5 个答案: