php – 不返回填充数组的计数
发布时间:2020-12-13 22:18:31 所属栏目:PHP教程 来源:网络整理
导读:$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] = deve
$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. 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($count); 解决方法die($count); 用$count(整数)作为exit code杀死你的脚本. 你想要: die((string) $count); (或可比较.) (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |