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[英]Return MySQL Error Code and Custom Response using Prepared Statement PHP
[英]Insert into table using PHP and prepared statement getting response code 500
我正在嘗試使用以下PHP代碼向表中插入一行:
$sql = "INSERT INTO `user_sessions`(`email`, `ip_address`, `location`, `lat`, `lng`, `device_type`, `device_imei`, `login_date`, `login_time`, `active`) VALUES (?,?,?,?,?,?,?,?,?,?);";
$stmt2 = mysqli_stmt_init($con);
if(!mysqli_stmt_prepare($stmt2, $sql)){
$data['result'] = "SQL error";
echo json_encode($data);
exit();
} else {
mysqli_stmt_bind_param($stmt2, "sssddsssss", $email, $ip, $city, $lat, $lng, $user_device, $user_device_imei, $user_date, $user_time, "1");
mysqli_stmt_execute($stmt2);
$data['result'] = "login success";
$data['user_id'] = $row['user_id'];
echo json_encode($data);
exit();
}
我不斷收到響應代碼500,這是根據我的研究得出的服務器內部錯誤
有什么建議么?
問題在第11行:
mysqli_stmt_bind_param($stmt2, "sssddsssss", $email, $ip, $city, $lat, $lng, $user_device, $user_device_imei, $user_date, $user_time, "1");
最后一個參數是“ 1”,但是mysqli_stmt_bimd_param()僅接受變量作為參數。
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