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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T00:53:13+00:00 2026-05-22T00:53:13+00:00

I’ve got a fairly large site, with a lot of jQuery code for lots

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I’ve got a fairly large site, with a lot of jQuery code for lots of different pages. We’re talking about 1000 lines of fairly well optimized code (excluding plugins).

I know jQuery is fairly good at ignoring listeners for page elements that don’t exist, but it still has to test their existence when the page loads. I’m also creating a load of vars (including decent sized arrays and objects), but only a few of them are used on each page.

My Question is: What’s the best method of cutting down the amount of work each page has to do?

However, I do NOT want to split up the code into separate files. I want to keep all my code in 1 place and for the sake of efficiency I only want to call the server to download JS once (it’s only 30kb, smaller than most images).

I’ve thought of several ways so far:

  1. Put all my code chunks into named functions, and have each page call the functions it needs from inline <script> tags.
  2. Have each page output a variable as the pageID, and for each chunk of have an if statement: if (pageID = 'about' || pageID = 'contact') {code...}
  3. Give each page (maybe the body tag) a class or ID that can be used to identify the chunks that need executing: if ($('.about').length || $('.contact').length) {code...}
  4. Combine 1 and 2 (or 1 and 3), so that each page outputs a variable, and the if statements are all together and call the functions: if (pageID = 'about') {function calls...}

Any other ideas? Or which is the best/most efficient of those?

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  1. Editorial Team
    Editorial Team
    2026-05-22T00:53:14+00:00Added an answer on May 22, 2026 at 12:53 am

    Your first option will be fastest (by a minute margin).
    You’ll need to remember to call the functions from the correct pages.

    However, don’t bother.
    Unless you’ve measured a performance impact in a profiler, there is no need to optimize this much.

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