问题
Good Morning stackoverflow... I'm having a problem.... this is my sample code
var i:Number = new Number();
trace("showarray length" + showArray.length);
for(i=0;i<showArray.length;i++){
trace("equal daw" + showArray.getItemAt(i).id + "==" + num);
if(showArray.getItemAt(i).id == num){
showArray.removeItemAt(i);
}
}
trace('alerts');
myproblem here is...wherenever the if is satisfied it stops looping it immediately goes out of the loop
this is a sample output given that the length of showArray is 2 and num = 0
showarray length2
equal daw0==0
alerts
please help me
回答1:
If you want to remove items while iterating over array, iterate in reverse order. This way element removal does not affect cycle condition:
for (var i:int = showArray.length - 1; i >= 0; i--) {
if (someCondition) {
showArray.removeItemAt(i);
}
}
Another small bonus that this is slightly faster, as it doesn't call showArray.length on each step.
回答2:
An even better way might be to use the filter method of the Array class.
array = array.filter(function (e:*, i:int, a:Array):Boolean {
return e.id != num;
});
回答3:
When your if is satisfied for id == num (which is 0 so happening in the first loop) and the item is removed, your array length decreases to 1 so the loop won't run any more.
回答4:
That's because you are removing items at the time you are iterating throught them.
array = [1, 2]
^ // pointer in first iteration
eliminate 1
array = [2]
^ // the pointer remains in the same location
//ups! out of the loop. all items were visited.
You can copy the array before you iterate through it and iterate the copy or mark the indices to remove and remove them later or iterate the array backwards.
PS: Sorry for my poor English.
回答5:
After showArray.removeItemAt(i);, add i--;
Because you removed the item at index i from the array, the item that was at i + 1 got moved to i. By subtracting one, you ensure that the moved item doesn't get skipped.
alxx's answer is also a good solution.
来源:https://stackoverflow.com/questions/5409507/basics-in-for-loop-in-actionscript-3-in-flex