Home > Software design >  For loop breaks an unrelated array
For loop breaks an unrelated array

Time:11-06

Hey so I am using puppeteer and when I run this code:

console.log(dealsId[i]);

for (var i = 0; i < sizes.length; i  ) {
    var refIdClasses = await sizes[i].$eval('input', a => a.getAttribute('class'));
    if (refIdClasses != 'disabled') {
        var refId = await sizes[i].$eval('input', a => a.getAttribute('value'));
        var size = await sizes[i].$eval('input', a => a.getAttribute('data-size'));
        refIds.push({ size: size, refId: refId });
    }
}
console.log(dealsId[i]);

The Deals Id works before the for loop and after the loop it says undefined and I am so confused on why.

CodePudding user response:

Var i is globally-scoped, so by using var i=0 you are definitely creating scope conflict. Use let or another variable name instead.

CodePudding user response:

This is because your for loop is also using i to iterate, possibly conflicting with the i in dealsId[i].

Possible solution might be to change the variable in for loop from i to j.

The code becomes:

console.log(dealsId[i]);

for (var j = 0; j < sizes.length; j  ) {
    var refIdClasses = await sizes[i].$eval('input', a => a.getAttribute('class'));
    if (refIdClasses != 'disabled') {
        var refId = await sizes[i].$eval('input', a => a.getAttribute('value'));
        var size = await sizes[i].$eval('input', a => a.getAttribute('data-size'));
        refIds.push({ size: size, refId: refId });
    }
}
console.log(dealsId[i]);

CodePudding user response:

That's the problem with vars. You can declare and re-declare them in the same scope without warning.

I assume you declared a var i in the code before what you showed, but you re-declare it in the for loop. i < sizes.length being the condition to break the loop, the loop break and i is now out-of-bounds of your array, hence the undefined (if dealsId has a size that's inferior to sizes array).

It isn't "unrelated" since you used the same variable to loop through 2 different arrays. Since we don't know if the two arrays have the exact same size, we can't pinpoint with 100% accuracy what's going on, but I'll go with that.

  • Related