在android中将图像作为BLOB上传到SQL
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【中文标题】在android中将图像作为BLOB上传到SQL【英文标题】：Uploading an image as BLOB to SQL in android 【发布时间】：2021-09-07 02:25:17 【问题描述】：我正在制作一个 uni 项目，我选择制作一个也需要上传和检索图像的 android 应用。
我不明白的部分是上传和检索部分。
我尝试了很多事情，从尝试将位图转换为字节并以这种方式上传或创建路径但没有运气，无论我用多少谷歌搜索，我只能使用奇怪的 API 或一些 firebase 的东西获得结果。
我试过这个：https://www.maxester.com/blog/2019/10/04/upload-file-image-to-the-server-using-volley-in-android/
还有这个：https://www.youtube.com/watch?v=ULHyRwep3EU
还有其他一些不起作用的东西。
我有一个使用此代码的测试应用：
public void openGallery()
Intent intent = new Intent();
intent.setType("image/*");
intent.setAction(Intent.ACTION_GET_CONTENT);
startActivityForResult(Intent.createChooser(intent, "Select Imgae"), PICK_IMAGE);
public void onActivityResult(int requestCode, int resultCode, Intent data)
super.onActivityResult(requestCode, resultCode, data);
if(requestCode == PICK_IMAGE && resultCode == RESULT_OK && data != null)
Uri pickedImage = data.getData();
String[] filePath = MediaStore.Images.Media.DATA;
Cursor cursor = getContentResolver().query(pickedImage, filePath, null, null, null);
cursor.moveToFirst();
String imagePath = cursor.getString(cursor.getColumnIndex(filePath[0]));
BitmapFactory.Options options = new BitmapFactory.Options();
options.inPreferredConfig = Bitmap.Config.ARGB_8888;
Bitmap bmp = BitmapFactory.decodeFile(imagePath, options);
ByteArrayOutputStream stream = new ByteArrayOutputStream();
bmp.compress(Bitmap.CompressFormat.PNG, 100, stream);
byte[] byteArray = stream.toByteArray();
bmp.recycle();
cursor.close();
String url = "https://192.168.108.2/test/upload.php";
StringRequest request = new StringRequest(Request.Method.POST, url, new Response.Listener<String>()
@Override
public void onResponse(String response)
Toast.makeText(MainActivity.this, response.toString(), Toast.LENGTH_LONG);
, new Response.ErrorListener()
@Override
public void onErrorResponse(VolleyError error)
Toast.makeText(MainActivity.this, error.getMessage().toString(), Toast.LENGTH_LONG);
)
protected Map<String, String> getParams()
Map<String, String> map = new HashMap<String, String>();
map.put("image", byteArray.toString());
return map;
;
而php是：
$username = "root";
$password = "";
$dbname = "test";
$servername = "localhost";
$image = base64_decode($_POST['image']);
$conn = new mysqli($servername, $username, $password, $dbname);
$conn->set_charset("utf8");
if ($conn->connect_error)
die("Connection failed: " . $conn->connect_error);
$sql = "INSERT INTO image(img) VALUES('".$image."');";
if($conn->query($sql))
echo "Success";
else
echo "Failed";
?>
我还有一个图像字段类型为 MEDIUMBLOB 的数据库，大小为 5mil。
这显然不起作用（选择图像时应用程序崩溃）。 我希望有人指导我如何通过 android 将图像上传到 sql 以及如何检索它们，因为 google 一点帮助都没有。
小更新：
所以我环顾四周，我已经到了某个地方，但仍然没有上传图片。 我现在在 Android 中有这段代码：
public class MainActivity extends AppCompatActivity
Button uploadBtn, retrieveBtn;
ImageView image;
private static final int PICK_IMAGE = 1;
@Override
protected void onCreate(Bundle savedInstanceState)
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
image = findViewById(R.id.image);
uploadBtn = findViewById(R.id.uploadBtn);
retrieveBtn = findViewById(R.id.retrieveBtn);
uploadBtn.setOnClickListener(new View.OnClickListener()
@Override
public void onClick(View v)
openGallery();
);
retrieveBtn.setOnClickListener(new View.OnClickListener()
@Override
public void onClick(View v)
getImage();
);
public void getImage()
public void openGallery()
Intent photoPickerIntent = new Intent(Intent.ACTION_PICK);
photoPickerIntent.setType("image/*");
startActivityForResult(photoPickerIntent, PICK_IMAGE);
protected void onActivityResult(int reqCode, int resultCode, Intent data)
super.onActivityResult(reqCode, resultCode, data);
if (resultCode == RESULT_OK)
try
final Uri imageUri = data.getData();
final InputStream imageStream = getContentResolver().openInputStream(imageUri);
final Bitmap bmp = BitmapFactory.decodeStream(imageStream);
image.setImageBitmap(bmp);
ByteArrayOutputStream baos = new ByteArrayOutputStream();
bmp.compress(Bitmap.CompressFormat.JPEG, 100, baos);
byte[] imageBytes = baos.toByteArray();
String encodedImage = Base64.encodeToString(imageBytes, Base64.DEFAULT);
//Toast.makeText(getApplicationContext(),"ByteArray created..",Toast.LENGTH_SHORT).show();
String url = "http://192.168.108.2/test/upload.php";
StringRequest request = new StringRequest(Request.Method.POST, url, new Response.Listener<String>()
@Override
public void onResponse(String response)
Toast.makeText(getApplicationContext(), response.toString(), Toast.LENGTH_LONG);
, new Response.ErrorListener()
@Override
public void onErrorResponse(VolleyError error)
Toast.makeText(getApplicationContext(), error.getMessage().toString(), Toast.LENGTH_LONG);
)
protected Map<String, String> getParams()
Toast.makeText(getApplicationContext(), "I'm in stringrequest", Toast.LENGTH_LONG);
Map<String, String> map = new HashMap<String, String>();
map.put("image", encodedImage.toString());
return map;
;
catch (FileNotFoundException e)
e.printStackTrace();
Toast.makeText(MainActivity.this, "Something went wrong", Toast.LENGTH_LONG).show();
else
Toast.makeText(MainActivity.this, "You haven't picked Image",Toast.LENGTH_LONG).show();
还有这个 php 代码：
<?php
$username = "root";
$password = "";
$dbname = "test";
$servername = "localhost";
$path = './images/';
$uploadfile = $path . basename($_FILES['image']['name']);
$echo "I'm in php";
$conn = new mysqli($servername, $username, $password, $dbname);
$conn->set_charset("utf8");
if ($conn->connect_error)
die("Connection failed: " . $conn->connect_error);
if(file_exists($path))
if(move_uploaded_file($_FILES['image']['name'], $uploadfile))
$sql = "INSERT INTO image(img) VALUES('".$uploadfile."');";
echo "Success!";
else
echo "Failed";
else
mkdir($path);
if(move_uploaded_file($_FILES['image']['name'], $uploadfile))
$sql = "INSERT INTO image(img) VALUES('".$uploadfile."');";
echo "Success!";
else
echo "Failed";
?>
当我选择一个图像时，它会选择它并将其放入图像视图中，但它不会将任何内容上传到 SQL。 怎么了？
忘了说：SQL 表中的 'img' 字段是大小为 300 的 varchar
【问题讨论】：
警告：您对SQL Injections 持开放态度，应该使用参数化的prepared statements，而不是手动构建查询。它们由PDO 或MySQLi 提供。永远不要相信任何形式的输入！即使您的查询仅由受信任的用户执行，you are still in risk of corrupting your data。 Escaping is not enough! 【参考方案1】：你应该使用编码而不是解码来存储到数据库，因为它还没有被编码成base64，你还没有这样做。
$image = base64_encode($_POST['image']);
【讨论】：
好吧，应用程序仍然崩溃，但是感谢您让我注意到这一点。有点想念那个：D以上是关于在android中将图像作为BLOB上传到SQL的主要内容，如果未能解决你的问题，请参考以下文章
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