-1
我正在製作個人資料更新Android應用程序。我需要幫助來獲取JSON值,因爲我得到JSON結果爲空 - 任何人都可以發現錯誤?Android更新MySQL表(Android個人資料)
配置文件更新響應:
{"tag":"profile_update","error":false,"user":{"fname":null,"lname":null,"email":null,"mobile":null,"class":null,"school":null,"uid":null,"profile_pic":null,"created_at":null}}
我的PHP代碼:
public function profileUpdate($fname, $lname, $email, $mobile, $class, $school, $uid, $profile_pic){
$result = mysqli_query($this->con, "SELECT * FROM users WHERE unique_id = '$uid'")
or die(mysqli_error($this->con));
$path = "userImages/$uid.png";
$actual_path = "http://192.168.1.101/cedu/login/$path";
$no_of_rows = mysqli_num_rows($result);
if ($no_of_rows > 0) {
$result = mysqli_fetch_array($result);
$old_email = $result['email'];
$old_profile_pic = $result['profile_pic'];
$status = 0;
$otp = rand(100000, 999999); // otp code
if ($old_email == $email) {
if ($old_profile_pic == $profile_pic){
$result = mysqli_query($this->con, "UPDATE `users` SET `firstname` = '$fname',`lastname` = '$lname', `mobile` = '$mobile',`class` = '$class',`school` = '$school'
WHERE `unique_id` = '$uid'") or die(mysqli_error($this->con));
} else {
$result = mysqli_query($this->con, "UPDATE `users` SET `firstname` = '$fname',`lastname` = '$lname', `mobile` = '$mobile',`class` = '$class',`school` = '$school' , `profile_pic` = '$actual_path'
WHERE `unique_id` = '$uid'") or die(mysqli_error($this->con));
file_put_contents($path, base64_decode($profile_pic));
}
} else {
if ($old_profile_pic == $profile_pic){
$result = mysqli_query($this->con, "UPDATE `users` SET `firstname` = '$fname',`lastname` = '$lname', `email` = '$email', `mobile` = '$mobile',`class` = '$class',`school` = '$school' , `otp` = '$otp', `verified` = '$status'
WHERE `unique_id` = '$uid'") or die(mysqli_error($this->con));
} else {
$result = mysqli_query($this->con, "UPDATE `users` SET `firstname` = '$fname',`lastname` = '$lname', `email` = '$email', `mobile` = '$mobile',`class` = '$class',`school` = '$school' , `profile_pic` = '$actual_path', `otp` = '$otp', `verified` = '$status'
WHERE `unique_id` = '$uid'") or die(mysqli_error($this->con));
file_put_contents($path."user".$uid.".jpg", base64_decode($profile_pic));
}
}
} else {
//
return false;
}
}
我們不會爲您調試。調試你的代碼,看看有什麼不對。如果您遇到特定代碼行的問題,請返回。 –
我有一個返回值沒有更新的問題 –
我在這裏要做的第一件事就是使用參數綁定 - 這看起來很容易受到SQL注入的影響。 – halfer