确保输入的名称不以空格结尾

时间:2013-05-01 23:33:28

标签: javascript

我试图得到它,如果我输入一个以空格结尾的名称,文本字段将变为红色。大多数代码都工作,它只有一种方法似乎不起作用。

问题必须在最后一个索引部分的某处?

var NamePass = true;

function ValidateName() {
    var BlankPass = true;
    var GreaterThan6Pass = true;
    var FirstBlankPass = true;
    var BlankMiddleName = true;

    if (document.getElementById('Name').value == "") {
        BlankPass = false;
    }

    var Size = document.getElementById('Name').value.length;
    console.log("Size = " + Size);

    if (Size < 7) {
        GreaterThan6Pass = false;
    }

    if (document.getElementById('Name').value.substring(0, 1) == " ") {
        FirstBlankPass = false;
    }

    var LastIndex = document.getElementById('Name').value.lastIndexOf();

    if (document.getElementById('Name').value.substring((LastIndex - 1), 1) == " ") {
        FirstBlankPass = false;
    }

    string = document.getElementById('Name').value;
    chars = string.split(' ');
    if (chars.length > 1) {} else
        BlankMiddleName = false;

    if (BlankPass == false || GreaterThan6Pass == false || FirstBlankPass == false || BlankMiddleName == false) {
        console.log("BlankPass = " + BlankPass);
        console.log("GreaterThan6Pass = " + GreaterThan6Pass);
        console.log("FirstBlankPass = " + FirstBlankPass);
        console.log("BlankMiddleName = " + BlankMiddleName);
        NamePass = false;
        document.getElementById('Name').style.background = "red";
    } else {
        document.getElementById('Name').style.background = "white";
    }
}

http://jsfiddle.net/UTtxA/10/

1 个答案:

答案 0 :(得分:3)

lastIndexOf获取字符的最后一个索引,而不是字符串中的最后一个索引。我认为您打算使用length代替:

var lastIndex = document.getElementById('Name').value.length;

然而,另一个问题是substring采用了开始和结束索引,而不是起始索引和子字符串长度。您可以改用substr,但charAt更容易:

if (document.getElementById('Name').value.charAt(lastIndex - 1) == " ") {
    FirstBlankPass = false;
}

现在,对于一些通用的代码改进。不是从true处的所有变量开始并有条件地将它们设置为false,而只需将它们设置为条件:

var NamePass = true;

function ValidateName() {
    var value = document.getElementById('Name').value;

    var BlankPass = value == "";
    var GreaterThan6Pass = value.length > 6;
    var FirstBlankPass = value.charAt(0) == " ";
    var LastBlankPass = value.charAt(value.length - 1) == " ";
    var BlankMiddleName = value.split(" ").length <= 1;

    if (BlankPass || GreaterThan6Pass || FirstBlankPass || LastBlankPass || BlankMiddleName) {
        console.log("BlankPass = " + BlankPass);
        console.log("GreaterThan6Pass = " + GreaterThan6Pass);
        console.log("FirstBlankPass = " + FirstBlankPass);
        console.log("BlankMiddleName = " + BlankMiddleName);
        NamePass = false;
        document.getElementById('Name').style.background = "red";
    } else {
        document.getElementById('Name').style.background = "white";
    }
}

还有几点需要注意:

  • 使用camelCase变量名而不是PascalCase变量名称可能是个好主意,后者通常是为构造函数保留的
  • blah == false应该写成!blah
  • if后跟else也可以替换为if (!someCondition)
  • 该函数看起来应该返回truefalse,而不是设置全局变量NamePass

倒数第二,你可以在一个正则表达式中总结这一点,但如果你打算根据实际上的错误向用户提供更具体的错误信息,那么我就不会这样做。

function validateName() {
    return /^(?=.{6})(\S+(\s|$)){2,}$/.test(document.getElementById('name').value);
}

最后 - 请记住,并非所有人都有中间名,甚至是超过6个字符的名字,正如@poke指出的那样。