使用方法填充时未填充公共数组属性

Question

我正在创建一个泛型函数来在一个公共属性的数组中推送键值对。

当我调用pushDetailsToArray函数时，在getHospitalDetails函数内部似乎没有填充公共属性数组。虽然当您尝试在pushDetailsArray函数内打印数组时，它会打印出来。有谁知道我在这里做错了什么？提前谢谢。

public function pushDetailsToArray($row, $array){

        foreach($row as $key => $value){

            $array[$key] = $value;

        }

                    //print_r($this->hospDetails);

        return $array;

    }

    public function getHospDetails(){

        $row = $this->queryThis( "SELECT * from tblhospitals WHERE HospID = '$this->sessionId'" );

        /*foreach($row as $key => $value){

            $this->hospDetails[$key] = $value;

        }*/

        $this->pushDetailsToArray($row, $this->hospDetails);

        print_r($this->hospDetails);


    }

顺便说一句，getHospitalDetails方法中注释掉的foreach循环有效。我只是希望能够制作一个循环的通用方法。

Answer 1

如果您确定要引用用户，可以将pushDetailsToArray更改为：

 <?php
 public function pushDetailsToArray($row, &$array)
 {
     foreach($row as $key => $value)
     {
         $array[$key] = $value;
     }
 }

然后像这样调用它（phpversion＆gt; = 5.3）

 $this->pushDetailsToArray($row, $this->hospDetails);

或（phpversion＆lt; 5.3）：

 $this->pushDetailsToArray($row, &$this->hospDetails);

-

         ********** But, I suggest not to use `reference` **************

您只需将pushDetailsToArray更改为：

即可

 <?php
 public function pushDetailsToArray($row)
 {
     $array = array();
     foreach($row as $key => $value)
     {
         $array[$key] = $value;
     }
     return $array;
 }

然后：

 $this->hospDetails = $this->pushDetailsToArray($row);

Answer 2

public function pushDetailsToArray($row){
        $array=array();
        foreach($row as $key => $value){

            $array[$key] = $value;

        }

        //print_r($this->hospDetails);

        return $array;

    }

    public function getHospDetails(){

        $row = $this->queryThis( "SELECT * from tblhospitals WHERE HospID = '$this->sessionId'" );

        /*foreach($row as $key => $value){

            $this->hospDetails[$key] = $value;

        }*/

         $this->hospDetails=$this->pushDetailsToArray($row);

        print_r($this->hospDetails);


    }

Answer 3

也许我错过了什么，但你为什么不这样做：

选项1

public function getHospDetails(){

    $row = $this->queryThis( "SELECT * from tblhospitals WHERE HospID = '$this->sessionId'" );

    $this->hospDetails = $row;

}

除非$row不是数组，否则选项2。

选项2

public function pushDetailsToArray($row) {
    $output = array();
    foreach($row as $key => $value){
        $output[$key] = $value;
    }
    return $output;
}

public function getHospDetails() {
    $row = $this->queryThis( "SELECT * from tblhospitals WHERE HospID = '$this->sessionId'" );
    $this->hospDetails = $this->pushDetailsToArray($row);
}

另一种方式

public function pushRowToHospDetails($row) {
    foreach($row as $key => $value){
        $this->hospDetails[$key] = $value;
    }
    return $this
}

public function getHospDetails() {
    $row = $this->queryThis( "SELECT * from tblhospitals WHERE HospID = '$this->sessionId'" );
    $this->pushRowToHospDetails($row);
}