javascript - iterate through array until first occurence of specified element is found?

Question

I have an array ["#page1", "yield", "yield", "yield", "#page2", "yield", "#page3" ]

I want to iterate through the array top-down, starting at the last yield and looking for the next occurence of an element not being yield (which is #page2).

This is what I have:

var longest = ["#page1", "yield", "yield", "yield", "#page2", "yield", "#page3" ],
    longestLen = longest.length;
for (i = longestLen-2; i>=0; i--) { 
     if ( longest[i] != "yield") {
          var gotoPage = longest[i];
          }
    }

I'm starting at i=5 (6 iterations) and checking if the element is not yield. However, the current way runs through all 6 iterations, so I end up with the page#1 instead of the #page2. I don't know how I can stop the iterations.

Using return did not work, what other means are there? Is lastIndexOf something I could use?

Thanks for help!

w3schools.com/js/js_break.asp

Blazemonger
– Blazemonger

2011-10-24 14:04:10 +00:00
Commented Oct 24, 2011 at 14:04 — Blazemonger
– Blazemonger, Commented Oct 24, 2011 at 14:04

jAndy · Accepted Answer · 2011-10-24 15:41:07Z

6

Just because noone mentioned it so far, you can also go with ES5 .some() for that.

Looks like:

var longest = ["#page1", "yield", "yield", "yield", "#page2", "yield", "#page3" ],
    gotoPage;

longest.reverse().slice(1).some(function( elem ) {
    return (gotoPage = elem) !== "yield";
});

alert('goto page: ' + gotoPage);

edited Oct 24, 2011 at 15:41

answered Oct 24, 2011 at 14:09

jAndy

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6 Comments

pimvdb Over a year ago

Nice solution, but this does not start at the last yield does it?

jAndy Over a year ago

@pimvdb: since I'm calling .reverse() before .some() it will!

pimvdb Over a year ago

But if I run this code I obtain #page3 for gotoPage, because .some already stops at #page3. Unless I'm mistaking, the OP wants to start at the last yield.

frequent Over a year ago

Correct. I can't start at the last entry, because that is the page I'm currently on (on panel-x). "yields" are added to the array when page changes on a correspondig panel (y). In order to get a backward-transition to the proper page, I need to find the last non-"yield" entry on the panel with the currently active page (= #page3). break works fine for me, although I will keep this mind, too

jAndy Over a year ago

well, I didn't realize that. If its always 100% sure that you need to skip the very last entry, I extended the snippet with .slice(1) which pretty much does the job.

|

Jamiec · Accepted Answer · 2011-10-24 14:03:32Z

4

you're looking for the break^tutorial keyword, which breaks out of a loop.

if ( longest[i] != "yield") {
    var gotoPage = longest[i];
    break; // wont iterate any further in the for(...) loop
}

answered Oct 24, 2011 at 14:03

Jamiec

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Comments

ipr101 · Accepted Answer · 2011-10-24 14:05:31Z

0

Try 'break' -

var longest = ["#page1", "yield", "yield", "yield", "#page2", "yield", "#page3" ],
    longestLen = longest.length;
for (i = longestLen-2; i>=0; i--) { 
     if ( longest[i] != "yield") {
          var gotoPage = longest[i];
          break;
          }
    }

answered Oct 24, 2011 at 14:05

ipr101

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Comments

TeckniX · Accepted Answer · 2011-10-24 14:13:49Z

0

Have a look at the 'break' operation with labels, so that you can break out where ever you'd like.

Javascript Loop Control with Break and Continue

answered Oct 24, 2011 at 14:13

TeckniX

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