我想检查用户输入是否是JavaScript的电子邮件地址,然后将其发送到服务器或试图发送电子邮件,以防止最基本的误解。


当前回答

<pre>
**The personal_info part contains the following ASCII characters.
1.Uppercase (A-Z) and lowercase (a-z) English letters.
2.Digits (0-9).
3.Characters ! # $ % & ' * + - / = ? ^ _ ` { | } ~
4.Character . ( period, dot or fullstop) provided that it is not the first or last character and it will not come one after the other.**
</pre>
*Example of valid email id*
<pre>
yoursite@ourearth.com
my.ownsite@ourearth.org
mysite@you.me.net
xxxx@gmail.com
xxxxxx@yahoo.com
</pre>
<pre>
xxxx.ourearth.com [@ is not present] 
xxxx@.com.my [ tld (Top Level domain) can not start with dot "." ]
@you.me.net [ No character before @ ]
xxxx123@gmail.b [ ".b" is not a valid tld ]
xxxx@.org.org [ tld can not start with dot "." ]
.xxxx@mysite.org [ an email should not be start with "." ]
xxxxx()*@gmail.com [ here the regular expression only allows character, digit, underscore and dash ]
xxxx..1234@yahoo.com [double dots are not allowed
</pre>
**javascript mail code**

    function ValidateEmail(inputText)
    {
    var mailformat = /^\w+([\.-]?\w+)*@\w+([\.-]?\w+)*(\.\w{2,3})+$/;
    if(inputText.value.match(mailformat))
    {
    document.form1.text1.focus();
    return true;
    }
    else
    {
    alert("You have entered an invalid email address!");
    document.form1.text1.focus();
    return false;
    }
    }

其他回答

以下是 Regex 认证:

没有 @(-) 和(.) 之前的空间字符不应该在 @ 之后聚集在一起 没有 @ 2 字符之后的特殊字符必须在 @ 电子邮件之前的长度不应该低于 128 字符

function validateEmail(email) {
var chrbeforAt = email.substr(0, email.indexOf('@'));
if (!($.trim(email).length > 127)) {
    if (chrbeforAt.length >= 2) {
        var re = /^(([^<>()[\]{}'^?\\.,!|//#%*-+=&;:\s@\"]+(\.[^<>()[\]\\.,;:\s@\"]+)*)|(\".+\"))@(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?/;
        //var re = /[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:\.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*@(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?/;
        return re.test(email);
    } else {
        return false;
    }
} else {
    return false;
}
}

這是從 http://codesnippets.joyent.com/posts/show/1917 偷走的

email = $('email');
filter = /^([a-zA-Z0-9_\.\-])+\@(([a-zA-Z0-9\-])+\.)+([a-zA-Z0-9]{2,4})+$/;
if (filter.test(email.value)) {
  // Yay! valid
  return true;
}
else
  {return false;}
<pre>
**The personal_info part contains the following ASCII characters.
1.Uppercase (A-Z) and lowercase (a-z) English letters.
2.Digits (0-9).
3.Characters ! # $ % & ' * + - / = ? ^ _ ` { | } ~
4.Character . ( period, dot or fullstop) provided that it is not the first or last character and it will not come one after the other.**
</pre>
*Example of valid email id*
<pre>
yoursite@ourearth.com
my.ownsite@ourearth.org
mysite@you.me.net
xxxx@gmail.com
xxxxxx@yahoo.com
</pre>
<pre>
xxxx.ourearth.com [@ is not present] 
xxxx@.com.my [ tld (Top Level domain) can not start with dot "." ]
@you.me.net [ No character before @ ]
xxxx123@gmail.b [ ".b" is not a valid tld ]
xxxx@.org.org [ tld can not start with dot "." ]
.xxxx@mysite.org [ an email should not be start with "." ]
xxxxx()*@gmail.com [ here the regular expression only allows character, digit, underscore and dash ]
xxxx..1234@yahoo.com [double dots are not allowed
</pre>
**javascript mail code**

    function ValidateEmail(inputText)
    {
    var mailformat = /^\w+([\.-]?\w+)*@\w+([\.-]?\w+)*(\.\w{2,3})+$/;
    if(inputText.value.match(mailformat))
    {
    document.form1.text1.focus();
    return true;
    }
    else
    {
    alert("You have entered an invalid email address!");
    document.form1.text1.focus();
    return false;
    }
    }

所有电子邮件地址都包含一个“at”(即 @)符号。

email.includes('@')

或者,如果您需要支持 IE/older 浏览器:

email.indexOf('@') > 0

即使你能完全确定一个电子邮件是否是RFC合成有效,这不会告诉你它是否属于提供它的人。

要做到这一点,请发送验证消息。

使用浏览器/运行时间来处理通过预付一个协议并将其转移到URL API,捕捉任何错误并检查结果的用户名和主名属性。

function validEmail(email=''){
    var $0, url, isValid = false, emailPatternInput = /^[^@]{1,64}@[^@]{4,253}$/, emailPatternUrl = /^[^@]{1,64}@[a-z][a-z0-9\.-]{3,252}$/i;
    email = email.trim();
    try{
        url = new URL('http://'+email);
        $0 = `${url.username}@${url.hostname}`;
        isValid = emailPatternInput.test( email );
        if(!isValid) throw 'invalid email pattern on input:' + email;
        isValid = emailPatternUrl.test( $0 );
        if(!isValid) throw 'invalid email pattern on url:' + $0;
        console.log(`email looks legit "${email}" checking url-parts: "${$0 === email ? '-SAME-':$0}"`);
    }catch(err){
        console.error(`probably not an email address: "${email}"`, err);
    };
    return isValid;
}

['user+this@はじめよう.みんな', 'stuff@things', 'user+that@host.com', 'Jean+François@anydomain.museum','هيا@יאללה', '试@例子.测试.مثال.آزمایشی', 'not@@really', 'no'].forEach(email=>console.log(validEmail(email), email));

这是我可以提出的最简单和最一般允许的例子,请在可以更准确的情况下编辑,同时保持其简单性和合理的一般允许有效性。

此外,请参见 MDN URL 文件 URL, window.URL 和 Nodejs for URL APIs。