As my question states how would I determine the index of the last item in JavaScript array with a value > 0
So if myarray = [12, 35, 56, 0, 42, 0]
How would I find the index of the last positive integer in this array? i.e in this instance the answer would be 4 (no. 42).
There's no shortcut, just loop backward through the array from length - 1 to 0 (inclusive) and stop at the first >0 entry.
this should do it:
var myarray = [12, 35, 56, 0, 42, 0];
var found = {};
found.value = 0;
found.index = 0;
var j = myarray.length;
for (j; j>0; j--) {
if (myarray[j] > 0) {
found.value = myarray[j];
found.index = j;
break;
}
}
alert(found.index);
EDIT: note the above would not clarify the 0 index or none so:
found.value = undefined;
found.index = undefined;
var j = myarray.length;
for (j; j>0; j--) {
if (myarray[j] > 0) {
found.value = myarray[j];
found.index = j;
break;
}
}
alert(found.index);
Try this:
var array = [12, 35, 56, 0, 42, 0];
function getLastNotZero(array){
var t = 0, i = array.length-1;
while(t <= 0 && i >= 0){
t = array[--i];
}
return i-1;
}
getLastNotZero(array);
//Returns 4
= vs. ==, which is good. Mind you, this is still an infinite loop. Edit Ah, now you've fixed that too.
= was a typo, seemingly cost me this answer :/. This wasn't a infinite loop, though.i. I suppose the loop would have broken eventually, as undefined <= 0 is false.Hope this helps:
for(var i=myarray.length-1; i>=0;i--){
if(myarray[i] > 0){
alert(i);
break;
}
}
if(i<0)
alert(i);
Try this solution:
var myarray = [12, 35, 56, 0, 42, 0], index;
$.each(myarray,function(idx,el){
if(el > 0){
index = idx;
}
});
alert('Index:'+index+', value:'+myarray[index]);
You can get the reverse index with .reverse() and .findIndex, then do some basic arithmetic to flip it:
> myarray = [12, 35, 56, 0, 42, 0]
> var ridx = myarray.slice().reverse().findIndex(n => 0 < n)
1
> var idx = myarray.length - 1 - ridx
4
