时间戳

首页 > 解决方案 > 验证电话号码的字符串操作

javascript - 验证电话号码的字符串操作

问题描述

编辑：我不能使用正则表达式，因为这是一个类，我的教授希望我们学习字符串操作的价值！

我需要使用字符串操作来验证表单中的电话号码。我需要使用的逻辑（对于课程要求）是将电话号码与 2 个字符串进行比较并输出一个新字符串。然后将新字符串与 4 种掩码电话号码类型进行比较以验证为真。

出于某种原因，它只打印“dddddddddd”。我是否使用了错误类型的循环？我关闭它太早了吗？请看下面：

Javascript：

    <script type="text/javascript" language="JavaScript">
    function validatePhone()
    {
        // kb- this is the phone number the end user will input into the form
        phoneString = document.PizzaForm.phone.value;
        // this is a number string I will use to compare the phone number against to confirm if it is a number
        var numberString = "1234567890";
        // kb- this is an operator string I will use to compare the phone number against to confirm if it has symbols in it such as (,), or -
        var operatorString= "()-";
        //kb- this is a dummy string I will be creating through this function
        var dummyString= "";
        // kb- these are the 4 mask ID phone numbers. If the phone number does not fit one of these masks, it will not be a valid phone number
        var mask1= "dddddddddd";
        var mask2= "(ddd)ddddddd";
        var mask3= "(ddd)ddd-dddd";
        var mask4= "ddd-ddd-dddd";
        console.log(phoneString);
    // kb- this is my for loop to go through the entire phone number string character by character
        for (var i=0; i <phoneString.length; i++)
        {
            var c = phoneString.charAt(i);

        //kb- if the character in the phone string is true (a #), please add a D to the dummy string
         if (numberString.indexOf(phoneString) != -1)
            {
                dummyString +="d";
            }

            // kb- if the character is not a number, if it is a '(', please add that to the dummy string
            else if (operatorString.indexOf(phoneString) ==0)
            {
                dummyString +="(";
            }

            //kb- if the character is not a number or a '(', if it is a ')',please add that to the dummy string
            else if (operatorString.indexOf(phoneString) ==1)
            {
                dummyString +=")";
            }

            //kb- if the character is not a number, a '(' or ')', but it is a '-', please add that to the dummy string
            else if (operatorString.indexOf(phoneString) ==2)
            {
                dummyString += "-";
            } 
        } 
        // kb- please print this to the console 
        console.log("dummyString is" + dummyString);
        // if the final dummy string matches 1 of the 4 mask IDs, alert as a true phone number

        if (dummyString == mask1 || dummyString== mask2 || dummyString ==mask3 || dummyString == mask4)
        {
            alert (dummyString);
        }
        return;
    }

HTML：

form name= "PizzaForm"> <input type= "text" name="phone"> <input type = "button" value = "Submit Order" onClick = "validatePhone()"/> </form>

标签： javascriptstring

解决方案

我的印象只是查看代码而不运行它，您应该phoneString在第一次if检查时替换为c（如下所示：if (numberString.indexOf(c) != -1)...）。毕竟，您要测试的是 中的每个字符（又名c） ，而不是整个字符，对吗？phoneStringphoneString

推荐阅读