Need put variable into object

Question

How i can put variable into json see example

$('.thermometer-noconfig').thermometer({

    percent: 10,   
    speed: 'slow'
})
/* thermometers with config */
$('.thermometer-config').thermometer({

    percent: 10, 
    speed: 'slow'
})

I need instead of a number "10", I need put variable test How I can do that, please help me

Also, this isn't json, these are javascript object literals. — Matt Burland
– Matt Burland, Commented Apr 3, 2014 at 20:46

Marc Vilalta · Accepted Answer · 2014-04-03 20:44:39Z

1

your_var = 'whatever'; 
$('.thermometer-noconfig').thermometer({

    percent: 10,   
      speed: 'slow'
    })
    /* thermometers with config */
    $('.thermometer-config').thermometer({

      percent: your_var , 
      speed: 'slow'
    })

answered Apr 3, 2014 at 20:44

Marc Vilalta

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Comments

Rob Sedgwick · Accepted Answer · 2014-04-03 20:54:18Z

0

var myThing = { percent: 10, speed: 'slow' };

mything.percent="test";

var myJsonThing = JSON.stringify(myThing);

Now I don't know what the 'thermometer' is / or what it needs - but you can pass either of these :

$('.thermometer-config').thermometer(myThing);

Or if it needs the JSON format :

$('.thermometer-config').thermometer(myJsonThing);

The output of the JSON.stringyfy() on myThing is

{ "percent": "test", "speed": "slow" }

, but as you are not passing this data to a service over http, I don't know why you are asking for the json format, have to tell us more if so.

answered Apr 3, 2014 at 20:54

Rob Sedgwick

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5 Comments

Rob Sedgwick Over a year ago

* I see you have removed the reference to json in the question now :P

user3127698 Over a year ago

Hi I need to display some of my calculations as thermometer

user3127698 Over a year ago

but when I insert my calculations in the function, then it stops working

user3127698 Over a year ago

here is it $("#start2").click(function(){ G = $("#test").val(); H = 10; D = 0.2; p_0 = 14.69; T_0 = 293.15; Pi= 3.1415926; S = 0.25*PiDD; V_0 = SH; p = p_0 + G/(144*S); pp0 = p_0/p; st_pp0 = Math.pow(pp0,5./7.); V = V_0 * st_pp0; h = (V_0 - V)/S; st_pp02 = Math.pow(pp0,-2./7.); T = 1.8*T_0*st_pp02-459.67; h2 = h * 156; T2 = 2*(t-40); p2 = 2.5(p-10); $('.thermometer-noconfig').thermometer({ percent: T2 , speed: 'slow' }) / thermometers with config */ $('.thermometer-config').thermometer({ percent: h2 , speed: 'slow' }) });

Rob Sedgwick Over a year ago

Hi @user3127698, probally best to ask a new question

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