Testing jQuery/JS output on Chrome developer tools

Question

I have a method that creates 2 arrays with data. I want to be able to include these arrays into JSON. Here is my code:

$(document).ready(function() {
    $("button").click(function(){
        var configs = [];
        var testPages = [];
        $(".test input:checked").each(function() {
            testPages.push($(this).attr('id'));
        });
        $(".config input:checked").each(function() {
            configs.push($(this).attr('id'));
        });
        //console.log(configs);
        //console.log(testPages);

        var testJson = new Object();
        testJson.testpages = testPages;
        testJson.configurations = configs;
        var runTestJson = JSON.stringify(testJson);
        return runTestJson;
    });
});

I want to be able to test this against my page but I am new to jQuery, Javascript and also using Chrome developer tools. I'm not sure how to test this in the Chrome console.

The commented portion (console.log()s) were displayed but when I type configs or testPages to try to view the array, I would get the error that they were undefined. How do I test to see if I am getting the correct output for my code on Chrome developer tools?

Scimonster · Accepted Answer · 2014-10-21 20:03:38Z

2

You'd need to make those two variables global. Just beware that unnecessarily polluting the global scope is not recommended. But if it's just for testing, then i suppose it's OK.

// global scope
var configs = [];
var testPages = [];
$(document).ready(function() {
    $("button").click(function(){
        $(".test input:checked").each(function() {
            testPages.push($(this).attr('id'));
        });
        $(".config input:checked").each(function() {
            configs.push($(this).attr('id'));
        });
        //console.log(configs);
        //console.log(testPages);

        var testJson = new Object();
        testJson.testpages = testPages;
        testJson.configurations = configs;
        var runTestJson = JSON.stringify(testJson);
        return runTestJson;
    });
});

answered Oct 21, 2014 at 20:03

Scimonster

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Comments

T J · Accepted Answer · 2014-10-21 20:15:43Z

1

You can debug your JavaScript without populating global space as follows (Instruction based on chrome, but similar in major browsers):

Open developer console F12
go to Sources tab
Open your script file (Ctrl + o and type the filename)
Put a breakpoint by clicking on the line number you wish to debug
Refresh your page an execution will be stopped at that line

. You can further use the available controls in the debug toolbar:

to proceed the execution as you wish, each controls have self-descriptive tooltips. You can see the values in variables by hovering over them or manually selecting them.

edited Oct 21, 2014 at 20:15

answered Oct 21, 2014 at 20:08

T J

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3 Comments

Liondancer Over a year ago

When you mean populating global space, do you mean putting variables in the global scope? I am also curious as to if all the the JS/jQ code is required to be within $(document).ready(function() {}). I read somewhere that code should be within that function so that the code is loaded before the rest of the page.

T J Over a year ago

@Liondancer Globalspace meaning the window object. If you simply create a variable outside a function or omit the var keyword, the variable will be a property of the window object which you can access like window.variablename.

T J Over a year ago

... .ready() method is prefered to make sure that the DOM is created before your script runs, ensuring all the events handlers, styles etc are properly attached... If part of your script doesn't interfere with DOM then it doesn't necessarily have to be in ready()

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