Php 在symfony2中，如何将查询放入json数据？

我要求：

SELECT Bank_ID, Status, COUNT(Bank_ID) FROM int_client_bank WHERE status = 30 or status = 50 or status = 35 or status = 37 GROUP BY Bank_ID, Status;

见数据：

"Bank_ID"   "Status"    "COUNT(Bank_ID)"
"1"         "30"        "772"
"1"         "35"        "58"
"1"         "50"        "151"
"2"         "30"        "124"
"2"         "35"        "27"
"2"         "50"        "25"
"3"         "30"        "227"
"3"         "35"        "16"
"3"         "37"        "1"
"3"         "50"        "143"
"4"         "30"        "337"
"4"         "35"        "23"
"4"         "37"        "1"
"4"         "50"        "98"
"5"         "30"        "72"
"5"         "35"        "7"
"5"         "50"        "9"
"6"         "30"        "113"
"6"         "35"        "3"
"6"         "50"        "68"
"7"         "30"        "16"
"7"         "50"        "10"
"8"         "30"        "13"
"8"         "35"        "1"
"8"         "50"        "6"
"9"         "30"        "16"
"9"         "35"        "2"
"9"         "50"        "6"
"10"        "30"        "4"
"10"        "35"        "2"
"11"        "30"        "2"
"11"        "50"        "2"
"12"        "30"        "4"
"12"        "35"        "1"
"12"        "50"        "1"
"13"        "30"        "3"
"13"        "50"        "2"
"14"        "30"        "5"
"15"        "30"        "1"
"15"        "50"        "1"
"16"        "30"        "1"
"17"        "30"        "1"
"18"        "30"        "2"

我如何将其放入symfony中以做出JsonResponse？：

return new JsonResponse(array('data' => $result, 'success' => true));

我需要以下数据：

{
    "data":[
        {"Bank_Id":"1","Status":"30","Count":"772"},
        {"Bank_Id":"1","Status":"35","Count":"58"},
        ...
    ],
    "success":true
}

---

现在还不清楚你在问什么，但我想让symfony根据你的数据做一个a，如下所示：

use Symfony\Component\HttpFoundation\JsonResponse;
$em = $this->getDoctrine()->getManager();
$query = $em->createQuery('SELECT Bank_ID, Status, COUNT(Bank_ID) FROM int_client_bank WHERE status = 30 or status = 50 or status = 35 or status = 37 GROUP BY Bank_ID, Status');

$bankResult = $query->getResult();
$response = new JsonResponse();
$response->setData(array(
    'data'    => $bankResult,
    'success' => true
));

---

您需要对数组进行json编码，然后将其作为json响应发送

$jsonArray = array(
            'data' => $result,
       'success' => true,
        );

        $response = new Response(json_encode($jsonArray));
        $response->headers->set('Content-Type', 'application/json; charset=utf-8');

        return $response;

---

但我怎样才能提出质疑呢？要将其放入json$bankresult中？

$bankresult

不是

json``JsonResponse

（）`自动对数组数据进行json编码并添加内容类型头。因此，只需确保

$bankResult

包含您的数据集/查询响应。在示例中添加了一些原则。Tim Dev i have error:[语义错误]第0行，第44列，靠近“int_client_bank”：错误：未定义类“int_client_bank”。您可以更改此查询请求吗$em=$this->getDoctrine（）->getManager（）->getRepository（'OmniSoftIntegratBundle:IntClientBank'）->getQuery？如果我尝试更改查询以从OmniSoftIntegrabundle中选择银行ID、状态、计数（银行ID）：IntClientBank，其中状态=30或状态=50或状态=35或状态=37按银行ID分组，状态[2/2]查询异常：[语法错误]第0行，第84列：错误：字符串预期结束，得到“状态”，这是查询中的一个简单错误，我不知道您的数据库。因此，任何查询和创建简单数据数组的问题都与上述问题无关。请阅读使用JsonResponse对象的手册，您不必手动设置标题。数字php数组在JSON中变成[]。在JSON中，关联数组变为{}。您没有向我们展示数据是如何呈现的。此外，为什么不使用实体呢？