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使用邮递员查看这段php是否按预期工作,但邮递员返回错误Malformed JSON: Unexpected 'U'
并且数据库未更新,无法绕过此错误,因为它对我来说看起来很好?解析数据时出现格式错误的JSON响应
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);
}
“$ username”周围的某些引号如何? –
是的,忘记了我需要在PHP中这样做,欢呼! – GTucker
votesAttained = votesAttained + 1这是什么? votesAttained应该是可变的权利? – Senanayaka