ID: 39811 Updated by: [EMAIL PROTECTED] Reported By: wishm at bk dot ru -Status: Closed +Status: Bogus Bug Type: Arrays related Operating System: win32 PHP Version: 4.4.4 New Comment:
. Previous Comments: ------------------------------------------------------------------------ [2006-12-13 11:29:18] wishm at bk dot ru ok the problem was just at this function's "feature" - multiple occurrences in first array are all treated the same way. I was wondered that applies to the associative arrays, too. So indexes are simply ignored. Screwed up a bit... array_diff_assoc fixes that :) Sorry for the bothering. ------------------------------------------------------------------------ [2006-12-13 11:29:15] [EMAIL PROTECTED] There is even Russian translation of this particular page: http://ru.php.net/manual/ru/function.array-diff.php ------------------------------------------------------------------------ [2006-12-13 11:27:09] [EMAIL PROTECTED] >I think you dont understand the purpose of the array_diff() function. And I think you didn't read what I said and keep ignoring the docs. >The element is considered present if and only if its' >VALUE is equal to the same-named element's VALUE in the >second array! No. Read the documentation first. http://php.net/array_diff ------------------------------------------------------------------------ [2006-12-13 11:10:14] wishm at bk dot ru I think you dont understand the purpose of the array_diff() function. The element is considered present if and only if its' VALUE is equal to the same-named element's VALUE in the second array! Try array_diff(array("a"=>"1"),array("a"=>"2")). The element a = 1 does not present at second array so it will be returned. You didnt even tried to understand the essence of the bug! It is 100% PHP error when 2 is considered equal to ANY value under conditions I've described. Please fix that! ------------------------------------------------------------------------ [2006-12-13 10:06:39] [EMAIL PROTECTED] http://php.net/array_diff array_diff() returns an array containing **all the values of array1 that are not present in any of the other arguments*. In your case all values of $a are present in $b. ------------------------------------------------------------------------ The remainder of the comments for this report are too long. To view the rest of the comments, please view the bug report online at http://bugs.php.net/39811 -- Edit this bug report at http://bugs.php.net/?id=39811&edit=1
