如何使用JSLint全局设置'use strict'

Question

我是javascript的新手,我正在尝试通过JSLint进行验证.我应该在哪里使用"use strict"在全球范围内使用它并验证？

这给了我错误"在语句位置使用严格'的意外表达式.":

    "use strict";
    console.log('doing js in head-section');

    function helloWorld() {
        console.log('called function helloWorld()');
        alert('Hello World from a JS function showing an alert!');
    }

    function helloMyNumber() {
        console.log('called function helloMyNumber()');
        var max = 42;
        var yourLuckyNumber = prompt('Enter your lucky number (between 1 and '+ max +')');
        var myLuckyNumber = Math.floor(Math.random()*(max+1));
        var paragraph = document.getElementById('luckynumber');
        paragraph.innerHTML = paragraph.innerHTML + ' Your lucky number is: ' + yourLuckyNumber + '. Mine is: ' + myLuckyNumber + '. They ' + (yourLuckyNumber == myLuckyNumber ? 'DID ' : 'did NOT ') + 'match!';
    }



    console.log('doing JS in body-section');
    document.writeln('<p class="green">Hello World from JS within a body-section in HTML!</p>');

Answer 1

根据文档,browserJSLint 的选项会自动禁用"use strict";全局级别的使用.据我所知,没有办法把它重新打开.

您可以关闭浏览器选项,并browser使用以下选项获取与选项相同的预先声明的全局变量:

/*global
Audio, clearInterval, clearTimeout, document, event, history, Image, location, name, navigator, Option, screen, setInterval, setTimeout, XMLHttpRequest
*/

或者,您可以将所有代码包装在IIFE中并"use strict";在其顶部使用.

或者,您可以切换到JSHint(有更多选项),并使用其strict: global选项允许"use strict";在全局范围内.