Android應用程序用PHP連接到mysql

Question

我不知道php，但代碼應該是這樣的

$con = mysqli_connect($host_name, $user_name, $user_pass, $db_name);


if($con)
{
$image = $_POST ["image"];
$name = $_POST ["name"];
$sql = "insert into imageinfo(name) values('$name')";
$upload_path = "uploads/$name.jpg";

if(mysqli_query($con,$sql))
{
file_put_contents($upload_path,base64_decode($image));
echo json_encode(array('response'=>'Image Upload Successfully'));
}
else
{
echo json_encode(array('response'=>'Image upload failed1'));
}
}
else
{
echo json_encode(array('response'=>'Image Upload Failed2'));
}
mysqli_close($con);
?>`

我收到了來自($image = $_POST ["image"]..)的未識別的objets圖像和名稱的錯誤。 如果我使用if(isset)我得到以下回復：

圖片上傳失敗2

Answer 1

刪除空間

$image = $_POST["image"];
$name = $_POST["name"];

Answer 2

我猜你沒有從api請求發送“圖像”和“名稱”，這就是為什么它沒有設置。

Answer 3

當數據庫連接失敗時，您會收到“Image upload failed2”響應。

請通過添加以下代碼來檢查您的連接：

if (mysqli_connect_errno())
{
   die("Failed to connect to MySQL: " . mysqli_connect_error());
}

您的代碼看起來像：

$con = mysqli_connect($host_name, $user_name, $user_pass, $db_name);

if (mysqli_connect_errno())
{
   die("Failed to connect to MySQL: " . mysqli_connect_error());
}
if ($con)
{
    $image = $_POST ["image"];
    $name = $_POST ["name"];
    $sql = "insert into imageinfo(name) values('$name')";
    $upload_path = "uploads/$name.jpg";

    if (mysqli_query($con, $sql))
    {
        file_put_contents($upload_path, base64_decode($image));
        echo json_encode(array('response' => 'Image Upload Successfully'));
    } else
    {
        echo json_encode(array('response' => 'Image upload failed1'));
    }
} else
{
    echo json_encode(array('response' => 'Image Upload Failed2'));
}
mysqli_close($con);