一段时间以来,我一直在使用下面的代码返回JSON,并且它可以正常工作,但是知道我已经升级到Xcode 11,它告诉我无效的JSON。
if($response) {
$resultArray = array();
$tempArray = array();
// Loop through each row in the result set
while($row = mysqli_fetch_object($response))
{
// Add each row into our results array
$tempArray = $row;
array_push($resultArray, $tempArray);
}
// Finally, encode the array to JSON and output the results
echo json_encode($resultArray);
}else{
echo("Error: something went wrong: ");
echo mysqli_error($dbc);
}
mysqli_close($dbc);
我在JSON验证器中测试了输出,它也返回了无效的JSON。我需要输出为有效的JSON。
当前输出:
[
{
"key":"2025",
"financial_institution":"",
"variable_rate":"0",
"six_months":"0",
"one_year":"0",
"two_year":"0",
"three_year":"0",
"four_year":"0",
"five_year":"0",
"date":"2019-06-17"
},
]
答案 0 :(得分:-1)
有效的JSON如下所示,
[
{
"key":"2025",
"financial_institution":"",
"variable_rate":"0",
"six_months":"0",
"one_year":"0",
"two_year":"0",
"three_year":"0",
"four_year":"0",
"five_year":"0",
"date":"2019-06-17"
}
]