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up vote -1 down vote favorite

Why second FOR loop don't work ?

toget = new Array("var18", "var4", "var43");

for (var i=0; i < toget.length; i++) {

    for (var x=0; x < result.toget[i].list.length; x++) {
    alert(x);
    }

}

If I do like this: 

for (var i=0; i < toget.length; i++) {

    for (var x=0; x < result.var18.list.length; x++) {
    alert(x);
    }

}

It works but this is not correct way.

I need that values in toget array become variables to process them in second FOR loop.

Any ideas ? Thanks.

share|improve this question
    
can you clarify what "result" is? –  shoebox639 Aug 8 '11 at 20:16
1  
"result.var18.list.length" is what? –  Diodeus Aug 8 '11 at 20:16
    
Your code is missing what result is. Only guessing is possible for an answer until you include all relevant code. –  jfriend00 Aug 8 '11 at 20:18
2  
what is "result"? and better yet, why is this tagged as jquery? –  Sinetheta Aug 8 '11 at 20:18
    
... where and how is jQuery coming into play here? –  vector Aug 8 '11 at 20:19

2 Answers 2

up vote 3 down vote accepted 
for (var i = 0; i < toget.length; i++) {
    for (var x = 0; x < result[toget[i]].list.length; x++) {
        alert(x);
    }
}
share|improve this answer
up vote 2 down vote

Try jquery each loop even better than for loop

toget = new Array("var18", "var4", "var43");

$.each(toget, function(i){
    $.each(result[toget[i]].list, function(x){
      alert(x);
    });
});
share|improve this answer
    
result.toget[i].list is incorrect here. It should be result[toget[i]].list. –  Rocket Hazmat Aug 8 '11 at 20:25
    
@Rocket - Missed it, anyways I fixed it thanks. –  ShankarSangoli Aug 8 '11 at 20:27
    
I didn't know about the $.each() function. It looks very useful to me. –  Jon Trauntvein Aug 8 '11 at 21:53
    
Yes, it is very useful take a look at the documentation here –  ShankarSangoli Aug 8 '11 at 21:55

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