Question

我正在尝试在函数中的mysqli变量上编译if语句，但是我收到一个致命错误，声明`在第117行不在对象上下文时使用$ this。我做错了什么以及如何我可以在函数中使用if语句吗？

代码：

    $qandaquery = "SELECT ReplyType
                    FROM Reply";

    $qandaqrystmt=$mysqli->prepare($qandaquery);
    // get result and assign variables (prefix with db)
    $qandaqrystmt->execute(); 
    $qandaqrystmt->bind_result($qandaReplyType);
    $this->shared_result[] = $qandaReplyType; //ERROR HERE

function ExpandOptionType($option) { 


     if('Single' == $this->shared_result){

...

 }else if('Multiple' == $this->shared_result){

...

}
}

Answer 1

您正在尝试在通用函数（不是方法）中使用$this。它应该写成以下内容：

class A {

    private $shared_result = array();

    public function __construct($mysqli) {

        $qandaquery = "SELECT ReplyType FROM Reply";
        $qandaqrystmt = $mysqli->prepare($qandaquery);
        $qandaqrystmt->execute(); 
        $qandaqrystmt->bind_result($qandaReplyType);
        $this->shared_result[] = $qandaReplyType;

    }

    public function ExpandOptionType($option) { 

        if('Single' == $this->shared_result){
            ...
        } else if('Multiple' == $this->shared_result) {
            ...
        }

    }

}

有意义。请记住，$this始终引用您正在使用的类的实例。在非类上下文中它只是没有意义。

使用$ this时出现致命错误

1 个答案: