Params를 사용하여 이미지를 업로드하려고합니다.
Android 코드 :
서버에 POST 데이터
RequestParams params = new RequestParams();
params.put("item_name", "Name of item");
params.put("item_image", encodedImage);
MyRestClient.post(MainActivity.this, "item",params, new JsonHttpResponseHandler(){
@Override
public void onSuccess(int statusCode, Header[] headers, JSONArray response) {
JSONArray jArr = response;
super.onSuccess(statusCode, headers, response);
}
@Override
public void onFailure(int statusCode, Header[] headers, String responseString, Throwable throwable) {
String responseFromAPI = responseString;
super.onFailure(statusCode, headers, responseString, throwable);
}
@Override
public void onSuccess(int statusCode, Header[] headers, String responseString) {
String responseStr = responseString;
super.onSuccess(statusCode, headers, responseString);
}
@Override
public void onSuccess(int statusCode, Header[] headers, JSONObject response) {
JSONObject jObj = response;
super.onSuccess(statusCode, headers, response);
}
@Override
public void onFailure(int statusCode, Header[] headers, Throwable throwable, JSONObject errorResponse) {
JSONObject jOBj = errorResponse;
super.onFailure(statusCode, headers, throwable, errorResponse);
}
@Override
public void onFailure(int statusCode, Header[] headers, Throwable throwable, JSONArray errorResponse) {
JSONArray jArr = errorResponse;
super.onFailure(statusCode, headers, throwable, errorResponse);
}
});
비트 맵 이미지 인코딩
public String getStringImage(Bitmap bmp){
ByteArrayOutputStream baos = new ByteArrayOutputStream();
bmp.compress(Bitmap.CompressFormat.JPEG, 100, baos);
byte[] imageBytes = baos.toByteArray();
String encodedImage = Base64.encodeToString(imageBytes, Base64.DEFAULT);
return encodedImage;
}
슬림 프레임 워크 코드 :
$app->post('/image', function ($request, $response) {
$input = $request->getParsedBody();
$uploaded_image = $input['image_image'];
$path = "/..../uploads/"."img-".date("Y-m-d-H-m-s").".jpg";
if (file_put_contents($path, base64_decode($uploaded_image)) != false)
{
$sql = "INSERT INTO item (item_name, item_image) VALUES (:restaurant_name, :restaurant_image)";
$sth = $this->db->prepare($sql);
$sth->bindParam("item_name", $input['item_name']);
$sth->bindParam("item_image", $input['item_image']);
$sth->execute();
$input['id'] = $this->db->lastInsertId();
}
return $this->response->withJson($input);
});
문제:
사진은 코드와 나의 이해에 따라 업로드되어야합니다. 원하는 폴더에 이미지를 업로드하지 않습니다.
나는 일을 올바르게하고 있습니까, 아니면 뭔가를 놓쳤습니까?
<?php
$app->post('/image', function ($request, $response) {
$files = $request->getUploadedFiles();
$file = $files['image_image']; // uploaded file
$parameters = $request->getParams(); // Other POST params
$path = "/..../uploads/"."img-".date("Y-m-d-H-m-s").".jpg";
if ($file->getError() === UPLOAD_ERR_OK) {
$file->moveTo($path); // Save file
// DB interactions here...
$sql = "INSERT INTO item (item_name, item_image) VALUES (:restaurant_name, :restaurant_image)";
$sth = $this->db->prepare($sql);
$sth->bindParam("item_name", $input['item_name']);
$sth->bindParam("item_image", $input['item_image']);
// if statement is executed successfully, return id of the last inserted restaraunt
if ($sth->execute()) {
return $response->withJson($this->db->lastInsertId());
} else {
// else throw exception - Slim will return 500 error
throw new \Exception('Failed to persist restaraunt');
}
} else {
throw new \Exception('File upload error');
}
});
이 기사는 인터넷에서 수집됩니다. 재 인쇄 할 때 출처를 알려주십시오.
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