Question

我使用Doctrine作为ORM。我对此非常陌生。所以我希望发送实体的Json响应。实体如下：

医学
泛型
内容

Medicine and Generics有很多关系，Generic和Content也有很多关系。我需要展示药物的细节以及药物的仿制药和仿制药的内容。

这是我的代码

public function jsonSerialize()
{
    $newArray= array();
    $generics=$this->getGenerics();
    foreach ($generics as $generic){
        $array = json_decode(json_encode($generic), true);
        var_dump($array);
        array_push($newArray, $array);
    }
    var_dump($newArray);
    $xyz=array();
    $xyz1=array();
    $arratJosn = array(
            'medicineId' => $this->id,
            'medicineName' => $this->medicineName,
            'medicineQuantity' => $this->medicineQuantity,
            'medicinePrice' =>$this->medicinePrice,
            'manufactureId' =>$this->manufactureId->getmanufatureName(),
            'medicineApproved' => $this->medicineApproved,
            'medicineVersion' =>$this->medicineVersion,
            'genericDetails'=> array('genericName'=>$xyz,
                                     'genericQuantity' => $xyz1,),

    );

    foreach ($generics as $generic){
        $genericName=$generic->getGenericName();
        array_push($arratJosn['genericDetails']['genericName'], $genericName);
        $contents=$generic->getContent();
        foreach ($contents as $content){
            $genericQuantity=$content->getGenericQuantity();
            array_push($arratJosn['genericDetails']['genericQuantity'], $genericQuantity);
            $genericPrice=$content->getGenericUnit();
        }
    }

    return $this;

}

在实体上使用jsonEncode我能够检索json数据，如下所示：

{"medicineId":27,
"medicineName":"MedTest11",
"medicineQuantity":20
"medicinePrice":70,
"manufactureId":"Company2",
"medicineApproved":1,
"medicineVersion":1,
"genericDetails":{"genericName":["paracetamol","newgeneric"],
                  "genericQuantity":["30","40"]}
}

虽然我希望结果显示为

{"medicineId":27,
"medicineName":"MedTest11",
"medicineQuantity":20, 
"medicinePrice":70,
"manufactureId":"Company2", 
"medicineApproved":1, 
"medicineVersion":1,
"genericDetails":[{"genericName":"paracetamol", "genericQuantity":"30"},{"genericName":"newgeneric", "genericQuantity":"40"}] 
}

我怎样才能实现这个目标？

Answer 1

要实现这一点，您需要将数组推送到像这样的通用详细信息

$arratJosn = array(
        'medicineId' => $this->id,
        'medicineName' => $this->medicineName,
        'medicineQuantity' => $this->medicineQuantity,
        'medicinePrice' =>$this->medicinePrice,
        'manufactureId' =>$this->manufactureId->getmanufatureName(),
        'medicineApproved' => $this->medicineApproved,
        'medicineVersion' =>$this->medicineVersion,
        'genericDetails'=> array()            
);

foreach ($generics as $generic){
    $genericName=$generic->getGenericName();
    $contents=$generic->getContent();
    foreach ($contents as $content){
        $genericQuantity=$content->getGenericQuantity();
        $genericPrice=$content->getGenericUnit();
        $genericArray = array (
            'genericName' => $genericName,
            'genericQuantity' => $genericQuantity
        );            
        array_push($arratJosn['genericDetails'],$genericArray);                
    }
}

很多很多关系Json

1 个答案: