Question

我试图让我的编码开始，但我遇到了一个问题，下面的代码工作正常：

$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);

但是当我尝试使用它时：

$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？

Answer 1

这个https://3v4l.org/a5IRp运行的确切代码和var_dump'似乎不同意。

也许你在$durgs_value_current时遇到var_dump()的错字？

调用var_dump($undefined_variable)将导致NULL，如果您关闭了错误报告或正在从日志中抑制E_NOTICE，那么您将不会收到未定义的变量错误。

PHP var_dump（）为assoc数组返回NULL

1 个答案: