What the title says. Specifically if I have
$array1['name'] = 'zoo';
$array2['name'] = 'fox';
How can I determine that alphabetically $array2
's name should come above $array1
's?
Use strcmp
. If the first argument to strcmp is lexicographically smaller to the second, then the value returned will be negative. If both are equal, then it will return 0. And if the first is lexicograpically greater than the second then a positive number will be returned.
nb. You probably want to use strcasecmp(string1,string2)
, which ignores case...
You can compare both strings with strcmp
:
Returns < 0 if str1 is less than str2; > 0 if str1 is greater than str2, and 0 if they are equal.
I'm a little late (then again I wasn't a programmer yet in 2009 :-) No one mentioned this yet, but you can simply use the operators which you use on number as well.
< > <= >= == !=
and more
For example:
'a' > 'b'
returns false
'a' < 'b'
returns true
http://php.net/manual/en/language.operators.comparison.php
IMPORTANT
There is a flaw, which you can find in the comments below.
'10' < '2'
evaluates to false even though the string '10'
should come before '2'
if you compare alphabetically (which is what the question is about). This is because whenever strings look like integers PHP automatically converts them to integers and compares them numerically instead of alphabetically. You should go with strcmp()
as suggested if your goal is to really compare alphabetically.
I often use natsort
(Natural Sort), since I usually just want to preserve the array for later use anyway.
Example:
natsort($unsorted_array);
var_dump($usorted_array); // will now be sorted.
EDIT just realised values from different arrays, could array_merge first but not sure thats what you want
s($array1['name'])->compareTo($array2['name'])
ors($array1['name'])->compareToIgnoreCase($array2['name'])
helpful, as found in this standalone library.