I'm trying to upload zip file to the url https://anypoint.mulesoft.com/designcenter/api-designer/projects/{projectId}/branches/master/import. Content-Type must be application/zip, cant change to multipart/form-data. In Mule 3, a java transform class is used (com.test.FileReader) with the FileReader.class is stored in lib. It worked in Mule 3.
I tried to use ReadFile component to read test.zip and set as payload but it's not working. Any suggestion how to upload zip file in Mule 4?
package com.test;
import org.mule.transformer.*;
import org.mule.api.*;
import org.mule.api.transformer.*;
import java.io.*;
public class PayloadFileReader extends AbstractMessageTransformer
{
public Object transformMessage(final MuleMessage message, final String outputEncoding) throws TransformerException {
byte[] result = null;
try {
result = this.readZipFile("test.zip");
}
catch (Exception e) {
e.printStackTrace();
}
message.setPayload((Object)result);
return message;
}
public String readFileTest(final String path) throws FileNotFoundException, IOException, Exception {
final ClassLoader classLoader = this.getClass().getClassLoader();
final File file = new File(classLoader.getResource(path).getFile());
final FileReader fileReader = new FileReader(file);
BufferedReader bufferReader = null;
final StringBuilder stringBuffer = new StringBuilder();
try {
bufferReader = new BufferedReader(fileReader);
String line;
while ((line = bufferReader.readLine()) != null) {
stringBuffer.append(line);
}
}
catch (IOException e) {
e.printStackTrace();
if (bufferReader != null) {
try {
bufferReader.close();
}
catch (IOException e) {
e.printStackTrace();
}
}
}
finally {
if (bufferReader != null) {
try {
bufferReader.close();
}
catch (IOException e2) {
e2.printStackTrace();
}
}
}
return stringBuffer.toString();
}
public byte[] readZipFile(final String path) {
final ClassLoader classLoader = this.getClass().getClassLoader();
final File file = new File(classLoader.getResource(path).getFile());
final byte[] b = new byte[(int)file.length()];
try {
final FileInputStream fileInputStream = new FileInputStream(file);
fileInputStream.read(b);
fileInputStream.close();
}
catch (FileNotFoundException e) {
System.out.println("Not Found.");
e.printStackTrace();
}
catch (IOException e2) {
System.out.println("Error");
e2.printStackTrace();
}
return b;
}
}
'
Assuming that your zip file corresponds to a valid API spec, in Mule 4, you don't need to use a custom java code to achieve what you want: you can read the file content using the File connector Read operation, and use an HTTP Request to upload it to Design Center using Design Center API. Your flow should look like:
For the Read operation, you only need to set the file location, in the File Path operation property.
No need to set content type in the HTTP Request (Mule 4 will configure the content type automatically based on the file content loaded by the Read operation).
You can't use Java code that depends on Mule 3 classes in Mule 4. Don't bother trying to adapt the code, it is not meant to work. Their architecture are just different.
While in Mule 4 you can use plain Java code or create a module with the SDK, there is no reason to do so for this problem and it would be counterproductive. My advise it to forget the Java code and resolve the problem with pure Mule 4 components.
In this case there doesn't seem a need to actually use Java code. The File connector read operation should read the file just fine as it doesn't appear the Java code is doing anything else than reading the file into the payload.
Sending through the HTTP Request connector should be straightforward. You didn't provide any details of the error, (where is it happening, complete error message, HTTP status error code, complete flow with the HTTP request in both versions, etc) and the API Designer REST API doesn't document an import endpoint so it is difficult to say if the request is correctly constructed.
Related
I am using Netbeans on OS X and cannot seem to write text to a text file that I have in a package named "assets".
Below is the way I tried to accomplish writing to the text file and so far my method of doing this is not working.
The way I tried to approach this problem was converting a string to url, then converting the url to a uri. Then I used the uri for the new file parameter. After I tried to write a string using the class print writer.
public class Experiment {
File createFile(String path) {
java.net.URL url = getClass().getResource(path);
URI uri;
try {
uri = url.toURI();
}
catch (URISyntaxException e) {
uri = null;
}
if ((url != null) && (uri != null)) {
System.out.println("file loading sucess");
return new File(uri);
}
else {
System.out.println("Error file has not been loaded");
return null;
}
}
File file = createFile("/assets/myfile.txt");
public static void main(String[] args) {
Experiment testrun = new Experiment();
try {
PrintWriter writer = new PrintWriter(new FileWriter(testrun.file));
writer.println("it works");
writer.flush();
writer.close();
System.out.println("string was written");
}
catch (IOException e) {
System.out.println("there was an error while writing");
}
}
}
The output given from my try catch statements say that the file write code was executed.
file loading sucess
string was written
BUILD SUCCESSFUL (total time: 2 seconds)
I have also tried using absolute string paths for making a new file, but with null results. I am running out of ideas and hoping for some guidance or solution from somebody.
I have put a file "template.html" inside RAW folder and I want to read it into a InputStream. But it is returning me null. Can't understand what is wrong in the below code
e.g. fileName passed as parameter is "res/raw/testtemplate.html"
public String getFile(String fileName) {
InputStream input = this.getClass().getClassLoader().getResourceAsStream(fileName);
return getStringFromInputStream(input);
}
Also, there might be a better solution by putting these files in a particular subfolder and putting it inside Asset folder but then I believe I would need to pass context in AssetManager. I don't understand that solution, sorry I am new to android development. Can someone shed some light regarding how this approach can be achieved.
EDIT
I have started implementing this solution with Assets. Below method is supposed to return a string containing the entire text of the file stored as template.html.
getFile("template.html") // I am sending extension this time
Problem getting error getAssets() is undefined.
public String getFile(String fileName) {
BufferedReader reader = null;
StringBuilder sb = new StringBuilder();
String line;
try {
reader = new BufferedReader(new InputStreamReader(getAssets().open(fileName)));
while ((line = reader.readLine()) != null) {
sb.append(line);
}
}
catch (IOException e) {
e.printStackTrace();
} finally {
if (reader != null) {
try {
reader.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
return sb.toString();
}
use this
new BufferedInputStream(getResources().openRawResource(filepath));
this will return a buffered input stream
The file name should be without extension :
InputStream ins = getResources().openRawResource(
getResources().getIdentifier("raw/FILENAME_WITHOUT_EXTENSION",
"raw", getPackageName()));
For this purposes uses assets folder:
assets/
This is empty. You can use it to store raw asset files. Files that you save here are compiled into an .apk file as-is, and the original filename is preserved. You can navigate this directory in the same way as a typical file system using URIs and read files as a stream of bytes using the AssetManager. For example, this is a good location for textures and game data.
So, you could easy get access at assets with context: context.getAssets()
BufferedReader reader = null;
try {
reader = new BufferedReader(
new InputStreamReader(context.getAssets().open("filename.txt")));
}
} catch (IOException e) {
//log the exception
} finally {
if (reader != null) {
try {
reader.close();
} catch (IOException e) {
//log the exception
}
}
}
I am passing a file as input stream to parser.parse() method while using apache tika library to convert file to text.The method throws an exception (displayed below) but the input stream is closed in the finally block successfully. Then while renaming the file, the File.renameTo method from java.io returns false. I am not able to rename/move the file despite successfully closing the inputStream. I am afraid another instance of file is created, while parser.parse() method processess the file, which doesn't get closed till the time exception is throw. Is that possible? If so what should I do to rename the file.
The Exception thrown while checking the content type is
java.lang.NoClassDefFoundError: Could not initialize class com.adobe.xmp.impl.XMPMetaParser
at com.adobe.xmp.XMPMetaFactory.parseFromBuffer(XMPMetaFactory.java:160)
at com.adobe.xmp.XMPMetaFactory.parseFromBuffer(XMPMetaFactory.java:144)
at com.drew.metadata.xmp.XmpReader.extract(XmpReader.java:106)
at com.drew.imaging.jpeg.JpegMetadataReader.extractMetadataFromJpegSegmentReader(JpegMetadataReader.java:112)
at com.drew.imaging.jpeg.JpegMetadataReader.readMetadata(JpegMetadataReader.java:71)
at org.apache.tika.parser.image.ImageMetadataExtractor.parseJpeg(ImageMetadataExtractor.java:91)
at org.apache.tika.parser.jpeg.JpegParser.parse(JpegParser.java:56)
at org.apache.tika.parser.CompositeParser.parse(CompositeParser.java:244)
at org.apache.tika.parser.CompositeParser.parse(CompositeParser.java:244)
at org.apache.tika.parser.AutoDetectParser.parse(AutoDetectParser.java:121)
Please suggest any solution. Thanks in advance.
public static void main(String args[])
{
InputStream is = null;
StringWriter writer = new StringWriter();
Metadata metadata = new Metadata();
Parser parser = new AutoDetectParser();
File file = null;
File destination = null;
try
{
file = new File("E:\\New folder\\testFile.pdf");
boolean a = file.exists();
destination = new File("E:\\New folder\\test\\testOutput.pdf");
is = new FileInputStream(file);
parser.parse(is, new WriteOutContentHandler(writer), metadata, new ParseContext()); //EXCEPTION IS THROWN HERE.
String contentType = metadata.get(Metadata.CONTENT_TYPE);
System.out.println(contentType);
}
catch(Exception e1)
{
e1.printStackTrace();
}
catch(Throwable t)
{
t.printStackTrace();
}
finally
{
try
{
if(is!=null)
{
is.close(); //CLOSES THE INPUT STREAM
}
writer.close();
}
catch(Exception e2)
{
e2.printStackTrace();
}
}
boolean x = file.renameTo(destination); //RETURNS FALSE
System.out.println(x);
}
This might be due to other processes are still using the file, like anti-virus program and also it may be a case that any other processes in your application may possessing a lock.
please check that and deal with that, it may solve your problem.
I have coded a AJAX file upload feature in my application. It works perfectly when running it from my laptop. When I try the exact same file using the same app, but deployed on a jBoss server, I get the following exception:
2013-02-18 11:30:02,796 ERROR [STDERR] java.io.FileNotFoundException: C:\Users\MyUser\Desktop\TestFile.pdf (The system cannot find the file specified).
getFileData method:
private byte[] getFileData(File file) {
FileInputStream fileInputStream = null;
byte[] bytFileData = null;
try {
fileInputStream = new FileInputStream(file);
} catch (FileNotFoundException e1) {
e1.printStackTrace();
}
if (fileInputStream != null) {
ByteArrayOutputStream byteArrayOutputStream = new ByteArrayOutputStream();
byte[] bytBuffer = new byte[1024];
try {
for (int readNum; (readNum = fileInputStream.read(bytBuffer)) != -1;) {
byteArrayOutputStream.write(bytBuffer, 0, readNum);
}
} catch (IOException e) {
e.printStackTrace();
}
bytFileData = byteArrayOutputStream.toByteArray();
}
return bytFileData;
}
Getting the file content in a variable (from the method above):
byte[] bytFileData = this.getFileData(file);
Making the file:
private boolean makeFile(File folderToMake, File fileToMake, byte[] bytFileData) {
Boolean booSuccess = false;
FileOutputStream fileOutputStream = null;
try {
if (!folderToMake.exists()) {
folderToMake.mkdirs();
}
if (!fileToMake.exists()) {
if (fileToMake.createNewFile() == true) {
booSuccess = true;
fileOutputStream = new FileOutputStream(fileToMake);
fileOutputStream.write(bytFileData);
fileOutputStream.flush();
fileOutputStream.close();
}
}
} catch (Exception e) {
e.printStackTrace();
booSuccess = false;
}
return booSuccess;
}
Any idea?
Thank you
Charles
It seems you're just passing the file path as part of the request to the server, not actually uploading the file, then attempting to use that file path to access the file.
That will work on your laptop because the code, when running locally, has access to your file system and will be able to locate the file. It won't work deployed on a server because it's an entirely separate machine, and as a result won't have access to your file system.
You'll need to modify your client-side (AJAX) code to actually upload the file, then modify your server-side code to use that uploaded file. Note that AJAX file uploads aren't generally possible - there are plugins for frameworks such as jQuery that provide this functionality using workarounds.
I'm not 100%, but I think proper AJAX file uploads may be possible using HTML5 features, but browser support for that is likely going to be pretty poor right now.
I've hosted a text file which I would like to load into a string using java.
My code doesn't seem to work producing errors, any help?
try {
dictionaryUrl = new URL("http://pluginstudios.co.uk/resources/studios/games/hangman/dictionary.dic");
} catch (MalformedURLException catchMalformedURLException) {
System.err.println("Error 3: Malformed URL exception.\n"
+ " Dictionary failed to load.");
}
// 'Dictionary' scanner setting to file
// 'src/Main/Dictionary.dic'
DictionaryS = new Scanner(new File(dictionaryUrl));
System.out.println("Default dictionary loaded.");
UPDATE 1: The file doesn't seem to load going to the catch. But the file exists.
You could do something that this tutorial does
public class WebPageScanner {
public static void main(String[] args) {
try {
URLConnection connection =
new URL("http://java.net").openConnection();
String text = new Scanner(
connection.getInputStream()).
useDelimiter("\\Z").next();
} catch (IOException e) {
e.printStackTrace();
}
}
}
You need to use HttpClient and retrieve the data as a string or string buffer.
then use parse or read as file.
Something like this should work in your case:
DictionaryS = new Scanner(dictionaryUrl.openStream());
JavaDoc tells us:
File(URI uri)
Creates a new File instance by converting the given file: URI into an abstract pathname.
We can't create and use a File instance for any other resource type (like http).