没有返回填充数组的计数

Question

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

以上是函数的一部分.当我跑,我得到;

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 )

我期待的;

然而,运行此并没有返回任何内容:

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

Answer 1

die($count);

使用$count(整数)作为退出代码终止您的脚本.

你想要:

die((string) $count);

(或可比较.)

Answer 2

来自http://www.php.net/manual/en/function.exit.php(与die()相同):

如果status是字符串,则此函数在退出之前打印状态.

如果status是整数,则该值也将用作退出状态.退出状态应在0到254范围内,退出状态255由PHP保留,不得使用.状态0用于成功终止程序.