解析数据时格式错误 JSON 响应
Malformed JSON response when parsing data
使用 Postman 查看 php 的这一部分是否按预期工作,但是 Postman returns 错误 Malformed JSON: Unexpected 'U'
并且数据库未更新,我无法理解这个错误对我来说没问题?
代码:
function placeVoteForCandidate()
{
global $connect;
$username = $_POST["username"];
$query = "UPDATE User SET votesAttained=votesAttained+1 WHERE USER_NAME = $username";
mysqli_query($connect, $query) or die (mysqli_error($connect));
mysqli_close($connect);
$message['success'] = 'Vote added';
echo json_encode($message);
}
仅执行时
echo json_encode('Vote added');
应该没有问题。是否抛出任何异常导致此错误?
尝试将 SQL 语句更改为
$query = "UPDATE User SET votesAttained=votesAttained+1 WHERE USER_NAME = " . $username;
使用 Postman 查看 php 的这一部分是否按预期工作,但是 Postman returns 错误 Malformed JSON: Unexpected 'U'
并且数据库未更新,我无法理解这个错误对我来说没问题?
代码:
function placeVoteForCandidate()
{
global $connect;
$username = $_POST["username"];
$query = "UPDATE User SET votesAttained=votesAttained+1 WHERE USER_NAME = $username";
mysqli_query($connect, $query) or die (mysqli_error($connect));
mysqli_close($connect);
$message['success'] = 'Vote added';
echo json_encode($message);
}
仅执行时
echo json_encode('Vote added');
应该没有问题。是否抛出任何异常导致此错误? 尝试将 SQL 语句更改为
$query = "UPDATE User SET votesAttained=votesAttained+1 WHERE USER_NAME = " . $username;