PHP associative arrays - how to treat integer as string

Question

I have a simple associative array.

$a = array("a"=>"b", "c"=>"d");

I want to check if the key "1" exists in the array, e.g.

isset($a["1"]);

This string is being treated as an integer, so that

echo $a["1"]; //prints "d"

How do I get it to treat it as a string?

I don't want to use array_key_exists or in_array because my benchmarking shows isset will be a lot faster.

Procedure

.. the easy answer is, don't use string integers for your associate arrays. Add a prefix, or completely change your naming convention. Why create a work-around to accommodate this preventable design choice? – Dragster

I'm parsing words from a text and checking their index, some of those will be numbers, so I have to do it this way. – Procedure

-1 bogus question, php doesn't behave this way. – Upcast

@chris, what do you mean? – Vitebsk

@Vitebsk echo $a["1"]; doesnt print "d". – Upcast

It prints Undefined index: 1 as expected – Isidore

Answer 1

It doesn't appear that you can do what you want to do. from http://us.php.net/manual/en/language.types.array.php:

A key may be either an integer or a string. If a key is the standard representation of an integer, it will be interpreted as such (i.e. "8" will be interpreted as 8, while "08" will be interpreted as "08").

You'll probably have to use Fosco's suggestion of prefixing all your keys with something. If you use the same prefix on every key, then it doesn't matter if you're parsing a text that might have words and numbers - put the same prefix on everything regardless.

Blase

Answer 2

isset($a["1"]) | isset($a[1]) ?

Or just isset($a[1])

Or even isset($a[intval(1)]) to be 1000% sure.

Explicit

Other way around, looking for string, not integer – Procedure

Answer 3

if echo $a['1'] prints d, then your array has more elements than you realize.

see var_dump($a) and print_r($a) functions to help you debug your code.

Upcast