嘿即時嘗試從我的服務器使用角度POST獲取一些數據我得到的參數我發送,我得到了服務器的響應。我只是無法處理我的反應,並實際得到我想要的參數。如果我在控制檯看我如何處理POST響應
我得到這樣的響應:
data from server Response {_body: " {"data":[{"temperature":"20","dispenses":5,"lates…08:36:15","latest_reset":"2017-10-15 08:42:47"}]}", status: 200, ok: true, statusText: "OK", headers: Headers, …}
我怎麼會去grapping的溫度是多少?或者我應該改變我的返回JSON?請指導我在正確的方向
我的角度代碼:
getCustomerData()
{
var headers = new Headers();
headers.append('Content-Type', 'application/x-www-form-urlencoded');
let urlSearchParams = new URLSearchParams();
urlSearchParams.append('customerID', this.customerID);
//urlSearchParams.append('password', 'wtf');
let body = urlSearchParams.toString()
this.http.post('HIDDEN BUT WORKS',body,{headers: headers}).subscribe(data => {
// Read the result field from the JSON response.
console.log('data from server', data);
let jsonResponse = data.json();
//console.log('nextstep',data.temperature);
console.log('hmm',jsonResponse._body.data.temperature);
//console.log('size',data.toString);
},(error) => {
console.log('error', error);
});
}
我響應代碼:
while ($stmt->fetch()) {
$json[] = array(
'temperature' => $temperature,
'dispenses' => $dispenses,
'latest_cleaning' => $latest_cleaning,
'latest_reset' => $latest_reset
);
}
$finalresult['data'] = $json;
//logToFile('data.log',json_encode($finalresult));
echo json_encode($finalresult);
刪除我的$ finalresult ['data'] = $ json;你的答案都有效! TY –