Not returning a count number for filled array

Question

$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);

pinkgothic · Accepted Answer · 2010-04-23 08:46:45Z

11

die($count);

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

You'll want:

die((string) $count);

(Or comparable.)

answered Apr 23, 2010 at 8:46

pinkgothic

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Comments

zaf · Accepted Answer · 2010-04-23 08:47:18Z

5

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.

answered Apr 23, 2010 at 8:47

zaf

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2 Comments

deceze Over a year ago

Arguably this is worded wrong, isn't it? "Will also be" should be "will be". :)

zaf Over a year ago

You should pass that on to the php documentation team, eh?

Matt Gibson · Accepted Answer · 2010-04-23 08:45:19Z

2

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

answered Apr 23, 2010 at 8:45

Matt Gibson

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Comments

deceze · Accepted Answer · 2010-04-23 08:47:18Z

2

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)

answered Apr 23, 2010 at 8:47

deceze♦

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Comments

oezi · Accepted Answer · 2010-04-23 08:49:17Z

1

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.

answered Apr 23, 2010 at 8:49

oezi

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1 Comment

pinkgothic Over a year ago

This is actually a pretty neat solution, since print returns 1, so after printing, die(); still gets an error code to work with. Of course, this is debugging code, so it's not as useful as it could be, but the construct in general is worth keeping in mind. +1

Michal Čihař · Accepted Answer · 2010-04-23 08:48:52Z

0

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

answered Apr 23, 2010 at 8:48

Michal Čihař

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