function CheckSourceInfoOne($source) {
foreach ($array as list($a, $b, $c)) {
list($latitude, $longitude) = explode(',', $c);
$La = mysqli_query($dbc, "SELECT * from Spotting_TB WHERE latitude = '".$latitude."'");
$Lo = mysqli_query($dbc, "SELECT * from Spotting_TB WHERE longitude = '".$longitude."'");
if ($checkLatitude == false && $checkLongitude == false) {
if ($insert = $dbc->query("INSERT INTO Spotting_TB (spotting_number,P,longitude,latitude,I,time_posted,hidden) VALUES (NULL,'{$b}','{$Lo}','{$La}','0',NOW(),'0')")) {
}
}
}
}
我一直在尝试一切,我无法在代码中发现我的错误:/它接受了解码的json并通过它并发布它,如果它不在数据库中,但它无法正常工作