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。
我尝试了很多方法，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，无论我谷歌多少，我都只能使用奇怪的 API 或一些 firebase 的东西得到结果。< /p>
我已经试过了：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事情。
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;
}
};
}
}
和 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";
}
?>
我还有一个带有图像字段的数据库，其类型为 MEDIUMBLOB，大小为 500 万。
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，因为谷歌一点帮助都没有。
小更新：
所以我环顾四周，找到了某个地方，但仍然没有上传图片。 我现在在 Android 中有这个代码：
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'字段是一个大小为300的varchar
答案 0 :(得分：0)
您应该使用 encode 而不是 decode 来存储到数据库，因为它尚未编码为 base64，您还没有这样做。
$image = base64_encode($_POST['image']);