php 返回无效的 json

以下 php 代码返回无效的 json 错误，不知道为什么

<?php
include("dbConnect.php");

$sql = "SELECT QID, Question, Answer,CatID FROM Questions";


$res = mysqli_query($con,$sql);

$result = array();

while($row = mysqli_fetch_array($res)){
array_push($result,
array('qid'=>$row[0],
'question'=>$row[1],
'answer'=>$row[2],
'catid'=>$row[3]
));
}

echo json_encode(array("result"=>$result));

mysqli_close($con);

当我运行 php 的链接时，它返回以下数据：

{"result":[{"qid":"1","question":"Question 1","answer":"Answer 1","catid":"1"},{"qid":"2","question":"Question 2","answer":"Answer 2","catid":"2"},{"qid":"3","question":"Question 3","answer":"Answer 3","catid":"3"},{"qid":"4","question":"Question 4","answer":"Answer 4","catid":"1"},{"qid":"5","question":"Question 5","answer":"Answer 5","catid":"3"},{"qid":"6","question":"Question 6","answer":"Answer 6","catid":"3"}]} 

我尝试运行 php 的链接，使用它返回 json 数据json 格式化程序网站 https://jsonformatter.curiousconcept.com/

我收到这些错误：

Error:Invalid media type, expecting application/json.[Code 28, Structure 0]

Error:Invalid encoding, expecting UTF-8, UTF-16 or UTF-32.[Code 29, Structure 0]

Error:Strings should be wrapped in double quotes.[Code 17, Structure 114]

Error:Invalid characters found.[Code 18, Structure 114]

Error:Strings should be wrapped in double quotes.[Code 17, Structure 116]

Error:Invalid characters found.[Code 18, Structure 116]

如果我尝试复制生成的 json 数据，我会得到有效的 JSON 格式，但是当我尝试从服务器上存储的 php 链接运行它时，我会收到上述错误

UPDATE:

链接到 PHP http://aamaa.comli.com/retrieve.php

问题不在于 JSON 字符串，而在于标头数据。 您必须指定Content-Type手动标头，否则它将以 text/html 形式发送输出：

<?php
include("dbConnect.php");

$sql = "SELECT QID, Question, Answer,CatID FROM Questions";


$res = mysqli_query($con,$sql);

$result = array();

while($row = mysqli_fetch_array($res)){
array_push($result,
array('qid'=>$row[0],
'question'=>$row[1],
'answer'=>$row[2],
'catid'=>$row[3]
));
}

header("Content-Type: application/json; charset=utf-8");
echo json_encode(array("result"=>$result));

mysqli_close($con);