javascript - jshint可能严格违规错误

Question

I am developing a plugin of table export at client-side. Plugin works fine. But when I validated my code in jshint, it throws me an error saying possible strict violation. Below is the function:

function disableExport(){
        if(_(this).exportPlugin !== undefined){
            _(this).exportPlugin.setStyle('pointer-events', 'none');
            _(this).exportPlugin.find('.ebIcon').removeModifier('interactive','ebIcon');
            _(this).exportPlugin.find('.ebIcon').setModifier('disabled','','ebIcon');
        }else{
            console.log('ERROR!');
        }
    }

And it says: "If a strict mode function is executed using function invocation, its 'this' value will be undefined."

Complete code of plugin is available on https://jsfiddle.net/t47L8yyr/

How do I resolve this ? any other solution than using /*jshint validthis:true*/

Answer 1

Inside your disableExport() function, you make references to this . If you invoke the function normally...

disableExport()

... in strict Javascript mode , this will be undefined. Outside strict mode, this will usually be the window object. So:

disableExport() // `this` will be `window`

"use strict"
disableExport() // `this` will be undefined

This is not good. If you want to target the window object, use it explicitly:

_(window).exportPlugin(...) // reference `window`, not `this`

If you're attempting to use this as a parameter to the function, invoking it with Function.call() or Function.apply() , it's much better to take an actual parameter than to use this :

function disableExport(target) {
  if(_(target).exportPlugin !== undefined) {
    // ...
  }
}

You can then call disableExport(window) or any other target . It's usually better to use this only when dealing with methods of an object, defined in the prototype of a function, or through the ES6 class syntax .