【发布时间】：2018-06-24 21:01:29
【问题描述】：
我正在开发应该具有将图像上传到服务器的功能的项目。在 MainActivity 我有 3 个选项卡。第一个选项卡有 2 个按钮来捕获图像并从图库中选择图像。选择或捕获图像后，将显示到名为 Upload 活动的新 Activity。此上传活动有 1 个图像视图和 1 个按钮上传图像。我可以显示图像，但上传图像有问题。点击上传按钮后，来自服务器的响应并且logcat中没有错误仍然我的图像没有上传到服务器。
上传活动
public class Upload extends AppCompatActivity{
public static String URL = "https://smilestechno.000webhostapp.com/upload.php";
String filepath;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_upload);
final Button btnUpload = (Button) findViewById(R.id.btnUpload);
ImageView imageview = (ImageView) findViewById(R.id.imageview);
Intent intent = getIntent();
if (intent == null){
return;
}
final Uri imageUri = Uri.parse(intent.getStringExtra("image"));
filepath = getPath(Upload.this, imageUri);
imageview.setImageURI(imageUri);
btnUpload.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
if (filepath != null){
imageUpload(filepath);
}else {
Toast.makeText(Upload.this, "Image not Selected", Toast.LENGTH_SHORT).show();
}
}
});
}
private void imageUpload(final String imagePath){
SimpleMultiPartRequest smr = new SimpleMultiPartRequest(Request.Method.POST, URL, new Response.Listener<String>() {
@Override
public void onResponse(String response) {
try {
JSONObject jsonObject = new JSONObject(response);
Toast.makeText(Upload.this, "Image Uploaded", Toast.LENGTH_SHORT).show();
} catch (JSONException e) {
e.printStackTrace();
}
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(com.android.volley.error.VolleyError error) {
Toast.makeText(getApplicationContext(), error.getMessage(), Toast.LENGTH_SHORT).show();
}
});
smr.addFile("img", imagePath);
MyApplication.getInstance().addToRequestQueue(smr);
}
public static String getPath(Context context, Uri contentUri) {
//copy file and send new file path
File TEMP_DIR_PATH = new File(Environment.getExternalStorageDirectory(), "/My Children/Temp");
TEMP_DIR_PATH.mkdir();
String fileName = getFileName(contentUri);
if (!TextUtils.isEmpty(fileName)) {
File copyFile = new File(TEMP_DIR_PATH + File.separator + fileName);
copy(context, contentUri, copyFile);
return copyFile.getAbsolutePath();
}
return null;
}
public static String getFileName(Uri uri) {
if (uri == null) return null;
String fileName = null;
String path = uri.getPath();
int cut = path.lastIndexOf('/');
if (cut != -1) {
fileName = path.substring(cut + 1);
}
return fileName;
}
public static void copy(Context context, Uri srcUri, File dstFile) {
try {
InputStream inputStream = context.getContentResolver().openInputStream(srcUri);
if (inputStream == null) return;
OutputStream outputStream = new FileOutputStream(dstFile);
org.apache.commons.io.IOUtils.copy(inputStream, outputStream);
inputStream.close();
outputStream.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
PHP 脚本
<?php
$uploaddir = 'ImagesUpload/';
$uploadfile = $uploaddir . basename($_FILES['uploadedfile']['name']);
echo "<p>";
if (copy($_FILES['uploadedfile']['tmp_name'], $uploadfile))
{
echo "File is valid, and was successfully uploaded.\n";
$servername = "localhost";
$username = "xyz";
$password = "xyz";
$dbname = "xyz";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error)
{
die("Connection failed: " . $conn->connect_error);
}
$filename="ImagesUpload/".$_FILES["uploadedfile"]["name"];
$sql = "INSERT INTO ImgInfo (ImgStatus, ImgLink,ImgUploadDate) VALUES ('0', '$filename',CURRENT_TIMESTAMP)";
if ($conn->query($sql) === TRUE)
{
echo "New record created successfully";
} else
{
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
}
else
{
echo "Upload failed";
}
echo "</p>";
echo '<pre>';
echo 'Here is some more debugging info:';
print_r($_FILES);
print "</pre>";
?>
我试过这个脚本。我创建了 html 页面，并且可以使用此脚本从 html 上传图像。
【问题讨论】：
-
这个
$_FILES['uploadedfile']['name']有什么信息吗？调试我的意思是 -
是的。你可以从这里上传smilestechno.000webhostapp.com/upload.html
标签： php android android-volley image-uploading