我正试图开始编码，但遇到了一个问题——以下代码运行良好：
$acid=50;
$cocaine=0;
$hashish=0;
$heroin=0;
$ecstasy=0;
$smack=0;
$opium=0;
$crack=0;
$peyote=0;
$shrooms=0;
$speed=0;
$weed=0;
$drugs_current = array("acid" => $acid, "cocaine" => $cocaine, "hashish" => $hashish, "heroin" => $heroin, "ecstasy" => $ecstasy, "smack" => $smack, "opium" => $opium, "crack" => $crack, "peyote" => $peyote, "shrooms" => $shrooms, "speed" => $speed, "weed" => $weed);
但当我尝试使用这个时：
$durgs_value_current = array("acid" => 1000, "cocaine" => 15000, "hashish" => 450, "heroin" => 5000, "ecstasy" => 10, "smack" => 1500, "opium" => 500, "crack" => 1000, "peyote" => 100, "shrooms" => 600, "speed" => 70, "weed" => 300);
var_dump()返回NULL。你有什么建议吗？
这个https://3v4l.org/a5IRp运行那个确切的代码和var_dump似乎不一致。
也许你的$durgs_value_current在var_dump()输入时有错别字？
调用var_dump($undefined_variable)将导致NULL，如果您已打开错误报告或正在从日志中抑制E_NOTICE，则不会收到未定义的变量错误。