I have a traditional conditional statement below, and I was wondering if I could translate this into an inline conditional(ternary?) statement.

var type;

if type === 1 {

    $('#form1').append(new mySuperObject({ model: myModelB}).render().$el);

} else {


    $('#form1').append(new mySuperObject({ model: new myModelA({ type: myType.Gas }) }).render().$el);

}

Would this be the proper format?

$('#form1').append(new mySuperObject({ model: ((type === 1) ? myModelB : new myModelA({ type: myType.Gas) }) }).render().$el);

When I try it, I get the error below in Firebug console:

TypeError: this.model.get(...) is undefined

Thanks

I have a traditional conditional statement below, and I was wondering if I could translate this into an inline conditional(ternary?) statement.

var type;

if type === 1 {

    $('#form1').append(new mySuperObject({ model: myModelB}).render().$el);

} else {


    $('#form1').append(new mySuperObject({ model: new myModelA({ type: myType.Gas }) }).render().$el);

}

Would this be the proper format?

$('#form1').append(new mySuperObject({ model: ((type === 1) ? myModelB : new myModelA({ type: myType.Gas) }) }).render().$el);

When I try it, I get the error below in Firebug console:

TypeError: this.model.get(...) is undefined

Thanks

• javascript
• jquery

Share Follow edited Jan 10, 2014 at 18:24 kei 20.5k22 gold badges3737 silver badges6464 bronze badges asked Jan 10, 2014 at 18:17 SkyeBoniwellSkyeBoniwell 7,1321515 gold badges100100 silver badges218218 bronze badges 5

• 4 Ternary operators are better used in one-line expressions; if..else looks better in this case. – elclanrs Commented Jan 10, 2014 at 18:18
• 1 The ternary is illegible. The only thing you need in the conditional is the creation of the new mySuperObject, too. – Dave Newton Commented Jan 10, 2014 at 18:19
• 3 Also, your syntax is wrong. Check here jshint. – elclanrs Commented Jan 10, 2014 at 18:20
• Where are the parentheses that go around the if conditional? – crush Commented Jan 10, 2014 at 18:21
• 1 new myModelA({ type: myType.Gas) }) should be new myModelA({ type: myType.Gas }) ) – Blazemonger Commented Jan 10, 2014 at 18:21

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4 Answers 4

Sorted by: Reset to default 4

I think you just have a parenthesis out of place. Compare the two lines below. The first is the orginal.

$('#form1').append(new mySuperObject({ model: ((type === 1) ? myModelB : new myModelA({ type: myType.Gas) }) }).render().$el);
$('#form1').append(new mySuperObject({ model: ((type === 1) ? myModelB : new myModelA({ type: myType.Gas })) }).render().$el);

To make it easier to read though, you could go with:

var model = (type === 1) ? myModelB : new myModelA({ type: myType.Gas });
$('#form1').append(new mySuperObject({ model: model }).render().$el);

You have some syntax errors in there..

for readability you could use

var objToUse = (type === 1) ? myModelB : new myModelA({ type: myType.Gas});
$('#form1').append(new mySuperObject({ model: objToUse }).render().$el);

Don't jam all of that stuff together like that. To avoid the duplication of code, you can do something like this:

var type;
var model;

if(type === 1) {

    model = myModelB;

} else {

    model = new myModelA({ type: myType.Gas });

}

$('#form1').append(new mySuperObject({ "model": model}).render().$el);

The following:

if (type === 1) {
    $('#form1').append(new mySuperObject({ model: myModelB}).render().$el);
} else {
    $('#form1').append(new mySuperObject({ model: new myModelA({ type: myType.Gas }) }).render().$el);
}

is the same as:

 var model;
 if (type === 1) {
     model = myModelB;
 }
 else {
     model = new myModelA({type: myType.Gas});
 }
 $("#form1").append(new mySuperObject({model: model})).render().$el;

Which is the same as:

 var model = (type === 1) 
             ? myModelB 
             : new myModelA({type: myType.Gas});

 $("#form1").append(new mySuperObject({model: model})).render().$el;

And you can jam it together if you really want:

 $("#form1").append(new mySuperObject({model: (type === 1) ? myModelB : new myModelA({type: myType.Gas})})).render().$el;

I do remend the second or third variant vs. jamming it all together in one line, though. The functional programmer in me likes the third over the second, actually, because you only have one assignment and don't have to declare the variable separately from assigning it.