jshint

This is the error I get when using const: <error line="2" column="1" severity="warning" message="&apos;const&apos; is available in ES6 (use esnext option) or Mozilla JS extensions (use moz)." source="jshint.W104" /> My code looks like this: const Suites = { Spade: 1, Heart: 2, Diamond: 3, Club: 4 }; The code works fine only JSHint is warning me every time. When relying upon ECMAScript 6 features such as const , you should set this option so JSHint doesn't raise unnecessary warnings. /*jshint esnext: true */ ( Edit 2015.12.29 : updated syntax to reflect @Olga's comments ) /*jshint esversion: 6

I am writing some node command line utilities. They all start with the line: #!/usr/bin/env node With Eclipse Juno and the Nodeclipse Node.js plugin, this line of code produces an error as shown: OK, so # is not a valid comment character in javascript, but it is a valid character in Linux/UNIX as the shebang of the first line in a file. But how can I set up Eclipse to ignore this error? This is a problem for me because code formatting does not work if you have errors. I have to delete the line. Hit CTRL-SHIFT-F and add the line back. I have tried a lot of things and researched, but I can't

jshint is throwing an error when defining an angular module (or directive, or factory) as recommended by the Angular style guides (by John Papa or Todd Motto ). For example, for a controller like this: (function () { 'use strict'; angular .module('myApp') .controller('myAppCtrl', theController); function theController() {...} })(); ... jshint throws this error: 'theController' was used before it was defined. The angular app works perfectly despite these errors. However I don't know why jshint protests... What am I missing? I wonder if jshint is a good evaluator of the quality of the angular

I want my Gulp build to fail, if there are errors in JSHint. According to the documentation of gulp-jshint I can use the "fail reporter". However the following does not work: gulp.task("lint", function() { return gulp.src(JS_SOURCES) .pipe(jshint()) .pipe(jshint.reporter("jshint-stylish")) .pipe(jshint.reporter("fail")); }); The task above always returns with exit code 0, even when there are errors in JSHint. I am using gulp 3.8.10 and gulp-jshint 1.9.0. There are discussions in the github issues of gulp-jshint here and here ... but according those discussions I gather that above code should

This question already has an answer here: What does “use strict” do in JavaScript, and what is the reasoning behind it? 27 answers Jshint.com is giving the error: Line 36: var signin_found; Missing "use strict" statement. Czarek Tomczak Add "use strict" at the top of your js file (at line 1 of your .js file): "use strict"; ... function initialize_page() { var signin_found; /*Used to determine which page is loaded / reloaded*/ signin_found=document.getElementById('signin_button'); if(signin_found) { More about "use strict" in another question here on stackoverflow: What does "use strict" do in