PHP parse_ini_file

Question

I try to parse something from a file, and can't get it to work..

If i do:

$idd = 11;
$aInifile = parse_ini_file('status.ini', 'reply');
echo 'test : ' . $aInifile['reply']['11'];

Then i get the correct reply. But if i use:

$idd = 11;
$aInifile = parse_ini_file('status.ini', 'reply');
echo 'test : ' . $aInifile['reply'][$idd];

Then it do not work... So how can i get it working, using the $idd variable?

My .ini file

[reply]
0 = "In progress (a normal message submission, with no error encountered so far)"
10 = "Delivered upstream"
11 = "Delivered to mobile"

Have tried a lot of things, also searched on Google, so betting on help from you guys here!

Show us ini file. Also the second argument must be a bool value, not string. — sectus, May 16 '13 at 02:22
Try `var_dump($aInifile);`. What does that output look like? — Travesty3, May 16 '13 at 02:23
@LukeMills: http://stackoverflow.com/a/4100765/259457 - In an array, if a key is the standard representation of an integer, it will be interpreted as such (i.e. "8" will be interpreted as 8, while "08" will be interpreted as "08"). — Travesty3, May 16 '13 at 02:26
Please var_dump($aInifile). It will show exactly what the type of the key is. As @Travesty3 points out, it shouldn't matter whether the key is a string or an int, but there's probably something going on that var_dump should show. — Luke Mills, May 16 '13 at 02:47
'Then it do not work...' - what does it mean? Error message? — sectus, May 16 '13 at 03:30
@sectus If i use "echo 'test : ' . $aInifile['reply'][$idd];" i do not get any Error message, but just the "test " text from the echo, and nothing more.. — KjetilP, May 16 '13 at 03:38

score 0 · Answer 1 · answered May 16 '13 at 02:28

0

The index is a string, so you can do this:

$idd = "11";
...
echo 'test : ' . $aInifile['reply'][$idd];

Note that 11 !== "11"

answered May 16 '13 at 02:28

Ibu

According to [the manual](http://php.net/manual/en/language.types.array.php), "Strings containing valid integers will be cast to the integer type. E.g. the key "8" will actually be stored under 8. On the other hand "08" will not be cast, as it isn't a valid decimal integer." It's not a strict comparison. http://codepad.viper-7.com/UtLvnw – Travesty3 May 16 '13 at 02:30
Yes, $idd = "11"; did the trick.. But the "clue" is that i get $idd out of an page that i "save" to the variable $idd using curl, so is there a way to add "" to the number (ex the number 11)inside the $idd variable? – KjetilP May 16 '13 at 02:38

1 Answers1