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我需要返回從PHP JSON對象,這是我的代碼:額外字符
$resultFromDB = PerformTransaction($requestToDB);
if($resultFromDB['status'] == '0'){
$myReturn = array('status' => '0','message' => 'success');
header('Content-Type: application/json');
$json_result = json_encode($myReturn,JSON_UNESCAPED_UNICODE);
echo trim($json_result,'');
}else{
$myReturn = array(
'status' => $resultFromDB['status'],
'message' => $resultFromDB['errorMsg']
);
header('Content-Type: application/json');
echo json_encode($return,JSON_UNESCAPED_UNICODE);
}
當從功能= 0結果狀態,那麼導致這個代碼應該是:
{"status":"0","message":"success"}
當通過理念的IntelliJ測試RESTful Web服務發送的,我得到的結果是:

{"status":"0","message":"success"}
如何正確返回結果,我不會有額外的字符
在header('Content-Type:application/json')之前添加函數ob_clean(); –