【发布时间】：2013-09-21 10:43:19
【问题描述】：
我是 android 新手，我正在尝试将图像上传到 mysql 服务器。在我的应用程序中，当用户单击按钮时，我有一个名为上传图像的按钮，它将在 android 设备中打开一个摄像头。我在 imageview 中使用相机成功获取了捕获的图像，我想使用 php 将此图像保存在 mysql 数据库中，我已使用 php 成功连接到数据库并上传了在 edittext 中输入的几个字段，但我无法上传图像。我用谷歌搜索了它，但都显示了带有 url 的上传图片。但在我的情况下，我的代码略有不同，请帮助我。
FormActivity.java
public class FormActivity extends Activity {
final Context context = this;
private ProgressDialog pDialog;
static final String TAG_SUCCESS = "success";
JSONParser jsonParser = new JSONParser();
private static String url_submit_hourly = "http://www.example.com/FacebookApp/submit.php";
EditText tasktitle;
String title;
Button upload, submit;
private static final int CAMERA_REQUEST = 1888;
private ImageView imageView;
byte[] byteArray;
protected static final int TAKE_PHOTO_CODE = 0;
byte[] imgbyte;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.submit_hourly);
tasktitle = (EditText) findViewById(R.id.etfirst);
submit = (Button) findViewById(R.id.submit);
submit.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
new BookSlot().execute();
}
});
upload = (Button) findViewById(R.id.upload);
this.imageView = (ImageView) this.findViewById(R.id.image);
upload.setOnClickListener(new OnClickListener() {
@Override
public void onClick(View v) {
Intent intent = new Intent("android.media.action.IMAGE_CAPTURE");
startActivityForResult(intent, CAMERA_REQUEST);
}
});
}
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
if (requestCode == CAMERA_REQUEST && resultCode == RESULT_OK) {
Bitmap photo = (Bitmap) data.getExtras().get("data");
imageView.setImageBitmap(photo);
}
}
class BookSlot extends AsyncTask<String, String, String> {
@Override
protected void onPreExecute() {
super.onPreExecute();
pDialog = new ProgressDialog(FormActivity.this);
pDialog.setMessage("Creating Hourly..");
pDialog.setIndeterminate(false);
pDialog.setCancelable(true);
pDialog.show();
}
@Override
protected String doInBackground(String... args) {
title = tasktitle.getText().toString();
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("task_tilte", title));
JSONObject json = jsonParser.makeHttpRequest(url_submit_hourly,
"POST", params);
// check log cat fro response
Log.d("Create Response", json.toString());
// check for success tag
try {
int success = json.getInt(TAG_SUCCESS);
if (success == 1) {
// successfully created product
Intent i = new Intent(getApplicationContext(),
FacebookLoginActivity.class);
startActivity(i);
// closing this screen
finish();
} else {
// failed to create product
}
} catch (JSONException e) {
e.printStackTrace();
}
return null;
}
/**
* After completing background task Dismiss the progress dialog
* **/
protected void onPostExecute(String file_url) {
// dismiss the dialog once done
pDialog.dismiss();
}
}
}
提交.php
<?php
// array for JSON response
$response = array();
// check for required fields
if (isset($_POST['task_tilte'])) {
$title = $_POST['task_tilte'];
// include db connect class
define('__ROOT__', dirname(dirname(__FILE__)));
require_once(__ROOT__.'/FacebookApp/db_connect.php');
// connecting to db
$db = new DB_CONNECT();
// mysql inserting a new row
$result = mysql_query("INSERT INTO task_table(task_tilte) VALUES('$title')");
// check if row inserted or not
if ($result) {
// successfully inserted into database
$response["success"] = 1;
$response["message"] = "Product successfully created.";
// echoing JSON response
echo json_encode($response);
} else {
// failed to insert row
$response["success"] = 0;
$response["message"] = "Oops! An error occurred.";
// echoing JSON response
echo json_encode($response);
}
} else {
// required field is missing
$response["success"] = 0;
$response["message"] = "Required field(s) is missing";
// echoing JSON response
echo json_encode($response);
}
?>
我不知道如何将 imageview 中的图像保存到 mysql 数据库。
【问题讨论】：