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Home/ Questions/Q 6085289
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T11:37:21+00:00 2026-05-23T11:37:21+00:00

I’m trying to use loop variable in executeSql function that contained by loop. But

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I’m trying to use loop variable in executeSql function that contained by loop. But loop variable gets the last value if i don’t use a closure. When i use a closure , i don’t get the result list from executeSql function. Examples:

for (var i = 0; i < count; i++) {
tx.executeSql('SELECT AColumn FROM ATable WHERE refID=' + i, [],
function(tx,results) //success function
{
//do something
}
,errorfunction);

}

In success function “i” is always “count+1”.

To solve this i changed my code like this:

for (var i = 0; i < count; i++) {
tx.executeSql('SELECT AColumn FROM ATable WHERE refID=' + i, [],
(function(tx,results) //success function
{
//do something
})(i)
,errorfunction);
}

With this, i get the “i” right. But “results” is undefined.

I tried to pass “tx” and “i” like this:

    (function(tx,results) //success function
    {
    //do something
    })(tx,null,i)

With this i understand why i get “results” as null. I want to learn how can i get the right results of executeSql.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T11:37:22+00:00Added an answer on May 23, 2026 at 11:37 am

    You’re looking for the following:

    for (var i = 0; i < count; i++) {
       tx.executeSql('SELECT AColumn FROM ATable WHERE refID=' + i, [],
          (function(i){
             return function(tx,results) //success function
             {
                //do something
             };
          })(i),
       errorfunction);
    }
    

    At the end of the day you need to pass a function of the signature function(tx,res) which is clearly what that whole (function(i){ return function(tx,res){ ... }; })(i) does, since the outer anonymous function executes immediately and returns a function of that signature.

    The value i in that inner function has the value of i when the outer function was called (i.e. the value of each iteration), since the value i is passed by value into the outer anonymous function, so references to i in the returned inner function will resolve correctly.

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