我在PHP中有一个函数返回数组而不是值 你能告诉我我做错了吗
function get_current_call_count() {
//Variable declearation
$line1="";
$line2="";
$total_call_count="";
$current_call_count="";
$var=array();
// Executing shell command to get total number.
$shell_command = ("/usr/sbin/asterisk -rx 'core show calls'");
exec($shell_command,$result,$status);
//print_r($result);
$line1=explode(" ",$result['0']);
$current_call_count=$line1['0'];
$line2=explode(" ",$result['1']);
$total_call_count=$line2['0'];
$var=array("$current_call_count","$total_call_count");
return($var);
//echo("Current call count is $current_call_count and Total system call count is $total_call_count");
}
答案 0 :(得分:1)
你的函数正在返回数组,因为你这样做了:
$var=array("$current_call_count","$total_call_count");
return($var);
所以一切都很好。我怀疑稍后您尝试将此函数返回的值用作非数组,因此进行强制转换并以“Array”字符串结束。但是在以后的代码中你的错。如果你想使用数组中的某个值,你必须从那里得到它,这很可能在你的其他代码中丢失了。