文本字段的Javascript电话面具与正则表达式 [英] Javascript phone mask for text field with regex

问题描述

我正在使用此功能来触摸屏幕并且几乎完美地工作。

I'm using this function to phone mask and works almost perfectly.

function mask(o, f) 
{ 
    v_obj = o; 
    v_fun = f; 
    setTimeout("execmask()", 1) 
};

function execmask() 
{ 
    v_obj.value = v_fun(v_obj.value) 
};

function mphone(v){
    v=v.replace(/\D/g,"");           
    v=v.substring(0, 11);
    v=v.replace(/^(\d{2})(\d)/g,"(OXX$1) $2"); 
    v=v.replace(/(\d)(\d{4})$/,"$1-$2"); 
    return v;
}

这里我在文本字段中运行掩码：

Here I run the mask in the text field:

<input type="text" id="phone" name="phone" onkeypress="mask(this, mphone);" onblur="mask(this, mphone);" />

问题是我需要更改代码的这部分（OXX $ 1）适用于（0XX $ 1）。

The problem is that I need to change this part of the code (OXX$1) for (0XX$1).

9位数的当前情况：（OXX99）99999-9999 。

Current situation with 9 digit: (OXX99) 99999-9999.

8位数的当前情况：（OXX99）9999-9999 。

Current situation with 8 digit: (OXX99) 9999-9999.

我需要9位数的正确格式：（0XX99）99999-9999

The correct formatting that I need with 9 digit: (0XX99) 99999-9999

8位数我需要的格式正确：（0XX99）99999-9999

The correct formatting that I need with 8 digit: (0XX99) 99999-9999

8位或9位数是用户的选择。

The amount of 8 or 9 digit is the choice of the user.

如果我将字符O更改为0，则会导致掩码出错。

If I change the character 'O' for '0', causes an error in the mask.

请帮助！

推荐答案

function mask(o, f) {
    setTimeout(function () {
        var v = f(o.value);
        if (v != o.value) {
            o.value = v;
        }
    }, 1);
}

function mphone(v) {
    var r = v.replace(/\D/g,"");
    r = r.replace(/^0/,"");
    if (r.length > 10) {
        // 11+ digits. Format as 5+4.
        r = r.replace(/^(\d\d)(\d{5})(\d{4}).*/,"(0XX$1) $2-$3");
    }
    else if (r.length > 5) {
        // 6..10 digits. Format as 4+4
        r = r.replace(/^(\d\d)(\d{4})(\d{0,4}).*/,"(0XX$1) $2-$3");
    }
    else if (r.length > 2) {
        // 3..5 digits. Add (0XX..)
        r = r.replace(/^(\d\d)(\d{0,5})/,"(0XX$1) $2");
    }
    else {
        // 0..2 digits. Just add (0XX
        r = r.replace(/^(\d*)/, "(0XX$1");
    }
    return r;
}

http： //jsfiddle.net/BBeWN/

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