PHP - not returning a count number for filled array...

Question

Morning, this is eating me alive so Im hoping it's not something stupid, lol.

$arrg = array();
if( str_word_count( $str ) > 1 ) {
    $input_arr = explode(' ', $str);
    die(print_r($input_arr));
    $count = count($input_arr);
    die($count);

above is part of a function. when i run i get;

> Array (
>     [0] => luke
>     [1] => snowden
>     [2] => create
>     [3] => develop
>     [4] => web
>     [5] => applications
>     [6] => sites
>     [7] => alse
>     [8] => dab
>     [9] => hand
>     [10] => design
>     [11] => love
>     [12] => helping
>     [13] => business
>     [14] => thrive
>     [15] => latest
>     [16] => industry
>     [17] => developer
>     [18] => act
>     [19] => designs
>     [20] => php
>     [21] => mysql
>     [22] => jquery
>     [23] => ajax
>     [24] => xhtml
>     [25] => css
>     [26] => de
>     [27] => montfont
>     [28] => award
>     [29] => advanced
>     [30] => programming
>     [31] => taught
>     [32] => development
>     [33] => years
>     [34] => experience
>     [35] => topic
>     [36] => fully
>     [37] => qualified
>     [38] => electrician
>     [39] => city
>     [40] => amp
>     [41] => guilds
>     [42] => level )

Which im expecting;

run this however and nothing is returned!?!?!

$arrg = array();
if( str_word_count( $str ) > 1 ) {
    $input_arr = explode(' ', $str);
    //die(print_r($input_arr));
    $count = count($input_arr);
    die($count);

can anyone see anything that my eyes cant??

regards, Phil

Answer 1

Hmm. Is it doing this because $count is an integer, I wonder? What happens if you die(strval($count))?

Answer 2

die($count);

Kills your script with $count (an integer) as an exit code.

You'll want:

die((string) $count);

(Or comparable.)

Answer 3

From http://www.php.net/manual/en/function.exit.php (same as die()):

If status is a string, this function prints the status just before exiting.

If status is an integer , that value will also be used as the exit status. Exit statuses should be in the range 0 to 254, the exit status 255 is reserved by PHP and shall not be used. The status 0 is used to terminate the program successfully.

Answer 4

die() will not print the argument if it is numeric, it'll use it as exit status code instead.

The problem is only your debugging technique... :o)

Answer 5

Integer parameter to die is used as exit code of the process (die is equivalent to exit). Just check the documentation.

Answer 6

have you tried to output $count? at the moment you kill your scrippt with $count as errorcode. simly do

die(print($count));

to get what you want.