How to find an object comprising multiple key/value pairs in a javascript array?

Question

Given:

var objects_found = [];
var entry = {any_number:77, any_text:"P1"}; // note: both primitive types
objects_found.push(entry);

.. why -with console logging showing such key/value pairs to be correctly stored- the following:

if (entry in objects_found) { ... }; // never fires
if (!(entry in objects_found)) { ... }; // always fires

Other than looping through the array as discussed here), is there a simple fix?

the keys in your 'entry' object should be strings like this: 'any_number': 77 — Tarun Dugar
– Tarun Dugar, Commented Jul 15, 2015 at 5:50
the in operator finds indexes and properties, e.g. 'any_number' in entry returns true as the object entry has the property any_number. It's not used for finding objects in arrays. AFAIK there is no way to magically pull a value out of an array without iterating over the array - even indexOf listed below will iterate over your array to find the index. — Kyle Muir
– Kyle Muir, Commented Jul 15, 2015 at 5:53
@Tarun - The property names any_number and any_text are strings whether you quote them or not. — Michael Geary
– Michael Geary, Commented Jul 15, 2015 at 6:00
Google "javascript in operator". That will lead you to this page: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…. There you will find it says "The in operator returns true if the specified property is in the specified object". Not the value of the property, but the property itself. — user663031
– user663031, Commented Jul 15, 2015 at 6:02
Can you be more specific about the goal you have in mind here? It's a fairly rare situation where you would need to search for an object in an array when you have just pushed the object in the array yourself. Perhaps you were writing some simplified code as an example, which is a good thing, but I'm afraid there isn't enough detail about the actual goal of the code to give much relevant advice. If you can elaborate on what you're trying to do, it will help people help you. — Michael Geary
– Michael Geary, Commented Jul 15, 2015 at 6:05

royhowie · Accepted Answer · 2015-07-15 06:16:25Z

2

Use Array.prototype.indexOf

if (objects_found.indexOf(entry) !== -1) {
    // `entry` is in `objects_found`
} else {
    // not in the array
}

Fiddle

edited Jul 15, 2015 at 6:16

royhowie

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answered Jul 15, 2015 at 5:50

Bobby Tables

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