Develop Reference

Getting current jar file's name

I have this code which on the dev-environment return the information.
But when I run from the jar the code doesn't follow how it should.
The name of the jar is hardcoded and would like to get it's name, because versions vary.
private static String getManifestUrlForClass(Class<?> cl) throws URISyntaxException, IOException {
URL url = cl.getResource(cl.getSimpleName() + ".class");
String s = url.toString();
System.out.println("URL Path: " + url.getPath());
System.out.println("URL File: " + url.getFile());
String path = MYCLASS.class.getProtectionDomain().getCodeSource().getLocation().getPath();
String revisionNumber = "";
String decodedPath = "";
JarFile jarfile = null;
try {
decodedPath = URLDecoder.decode(path, "UTF-8").replace("classes", "");
try {
jarfile = new JarFile(decodedPath + "MYJAR-ver.si.on.jar");
} catch (IOException e1) {
System.out.println("or Path to file cannot decode...");
e1.printStackTrace();
}
Manifest manifestFromJar = jarfile.getManifest(); //
System.out.println("Manifest from " + jarfile.getName().toString() + " = "
+ manifestFromJar.getMainAttributes().getValue("Revision-Number").toString());
revisionNumber = manifestFromJar.getMainAttributes().getValue("Revision-Number").toString();
} catch (IOException e) {
System.out.println(url.getFile().toString() + "is not jar");// TODO Auto-generated catch block
System.out.println("or Path to file cannot decode...");
e.printStackTrace();
}
return revisionNumber;
}
MYJAR will always be the same but the |ver.si.on| will most likely vary and hardcoding the name isn't a best practice.
What I want to do?
1. Get the MYJAR-ver.si.on.jar's location no matter where it is located
2. Use the location to access it's Manifest
3. Use the Manifest to extract revision number
4. Show the revision number in the ui
I'm new yet to java and don't understand it pretty well. I've read something about using "rsrc:" to get to the jar, or something similar to this https://stackoverflow.com/a/40680501/6756124 .

Encoding url in java avoiding special characters

I need to pass the encoded url but it needs to avoid special characters as well. So how do I encode it? None of the answers on stackoverflow worked for me. Can any one help? I want to do it in java
public String tweet_quote_func(Map attrs, FilterParam param) {
String url = param.getShorturi();
String text = attrs.get("display");
if (url != null && text != null) {
try {
text = StringEscapeUtils.unescapeHtml(text); // for double quotes
String encodedurl = "https://twitter.com/intent/tweet?url="+URLEncoder.encode(url, "UTF-8");
encodedurl = encodedurl + "&text=" + StringUtil.escapeUrl(text);
return "<span class=\"tweet_quote\"> " + text.trim() + "<span></span></span>";
} catch (UnsupportedEncodingException e) {
System.err.println(e);
return null;
}
} else
return null;
}

You should use URLEncoder for all URL query argument names and their values. Not StringUtils.escapeURL(), whatever that is.
I don't know what you think is different about 'the special characters ".", "-", "*", and "_"', but URLEncoder is defined to do the right thing.
For the URL paths themselves you should use new URI(null, path, null).toASCIIString().

Merge two URL in JAVA

I merge two url with the following code.
String strUrl1 = "http://www.domainname.com/path1/2012/04/25/file.php";
String arg = "?page=2";
URL url1;
try {
url1 = new URL(strUrl1);
URL reconUrl1 = new URL(url1,arg);
System.out.println(" url : " + reconUrl1.toString());
} catch (MalformedURLException ex) {
ex.printStackTrace();
}
I'm surprise by the result : http://www.domainname.com/path1/2012/04/25/?page=2
I expect it to be (what browser do) : http://www.domainname.com/path1/2012/04/25/file.php?page=2
Tha javadoc about the constructor URL(URL context, String spec) explain it should respect the RFC.
I'm doing something wrong ?
Thanks
UPDATE :
This is the only problem I encountered with the fonction.
The code already works in all others cases, like browser do
"domain.com/folder/sub" + "/test" -> "domain.com/test"
"domain.com/folder/sub/" + "test" -> "domain.com/folder/sub/test"
"domain.com/folder/sub/" + "../test" -> "domain.com/folder/test"
...

You can always merge the String first and then created the URL based on the merged String.
StringBuffer buf = new StringBuffer();
buf.append(strURL1);
buf.append(arg);
URL url1 = new URL(buf.toString());

try
String k = url1+arg;
URL url1;
try {
url1 = new URL(k);
//URL reconUrl1 = new URL(url1,arg);
System.out.println(" url : " + url1.toString());
} catch (MalformedURLException ex) {
ex.printStackTrace();
}

I haven't read through the RFC, but the context (as mentioned in the Java Doc for URL) is presumably the directory of a URL, which means that the context of
"http://www.domainname.com/path1/2012/04/25/file.php"
is
"http://www.domainname.com/path1/2012/04/25/"
which is why
new URL(url1,arg);
yields
"http://www.domainname.com/path1/2012/04/25/?page=2"
The "workaround" is obviously to concatenate the parts yourself, using +.

you are using the constructor of URL here which takes paramter as URL(URL context, String spec). So you dont pass the php page with the URL but instead with the string. context needs to be the directory. the proper way to do this would be
String strUrl1 = "http://www.domainname.com/path1/2012/04/25";
String arg = "/file.php?page=2";
URL url1;
try {
url1 = new URL(strUrl1);
URL reconUrl1 = new URL(url1,arg);
System.out.println(" url : " + reconUrl1.toString());
} catch (MalformedURLException ex) {
ex.printStackTrace();
}

Try this
String strUrl1 = "http://www.domainname.com/path1/2012/04/25/";
String arg = "file.php?page=2";
URL url1;
try {
url1 = new URL(strUrl1);
URL reconUrl1 = new URL(url1,arg);
System.out.println(" url : " + reconUrl1.toString());
} catch (MalformedURLException ex) {
ex.printStackTrace();
}

When you read the java doc it mentions about the context of the specified URL
Which is the domain and the path:
"http://www.domainname.com" + "/path1/2012/04/25/"
Where "file.php" is considered the text where it belongs to the context mentioned above.
This two parameter overloaded constructor uses the context of a URL as base and adds the second param to create a complete URL, which is not what you need.
So it's better to String add the two parts and then create URL from them:
String contextURL = "http://www.domainname.com/path1/2012/04/25/";
String textURL = "file.php?page=2";
URL url;
try {
url = new URL(contextURL);
URL reconUrl = new URL(url, textURL);
System.out.println(" url : " + reconUrl.toString());
} catch (MalformedURLException murle) {
murle.printStackTrace();
}

Do I need 3 try blocks for url encode?

I tried this:
String query = "http://maps.googleapis.com/maps/api/geocode/xml?address="+country+"+"+province+"+"+city+"&sensor=false";
try {
query = URLEncoder.encode(query, "UTF-8");
}
catch (Exception e) {
println("getLatLonFromAdress URLEncoder error: "+e);
return new float[] { -1f, -1f };
}
but it turns the url into:
http%3A%2F%2Fmaps.googleapis.com%2Fmaps%2Fapi%2Fgeocode%2Fxml%3Faddress%3DCanada%2BAlberta%2BGrande+Cache%26sensor%3Dfalse
So I only want to encode country, province and city. Is it bad to handle that in one try block? Like:
try {
country = URLEncoder.encode(country, "UTF-8");
province = URLEncoder.encode(province, "UTF-8");
city = URLEncoder.encode(city, "UTF-8");
}
catch (Exception e) {…}

The UnsupportedEncodingException will not be thrown if you are passing "UTF-8" as that will always be present. Therefor you can happily handle them all in 1 catch.

java concatenate two strings error

I have one function that returns me String :
public String getString(String password){
......
try {
.......
encodedPassword = Base64.encodeToString(msgDigest,1 );
} catch (UnsupportedEncodingException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (NoSuchAlgorithmException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return encodedPassword;
}
I want to add (concatenate) "=" String to returning string from function
I try using this:
encrptdPassword = getString("1234");
encrptdPassword = encrptdPassword+"=";
Or:
encrptdPassword = encrptdPassword .concat("=");
but I get result like two different objects (space or brake between)
I think problem is in Base64.encodeToString , but I must use 64 based string
Function getString returns me:
A6xnQhbz4Vx2HuGl4lXwZ5U2I8iziLRFnhP5eNfIRvQ
I want to add = to the returning string as:
A6xnQhbz4Vx2HuGl4lXwZ5U2I8iziLRFnhP5eNfIRvQ=
but I receive this on output
A6xnQhbz4Vx2HuGl4lXwZ5U2I8iziLRFnhP5eNfIRvQ =
Or:
A6xnQhbz4Vx2HuGl4lXwZ5U2I8iziLRFnhP5eNfIRvQ
=
...like 2 different strings.
Where I'm wrong?

I assume you're using Base64 from Apache Commons Codec.
The default constructor for this class uses "\r\n" as a line separator, which it adds to the end of every encoded line. If you don't want this, construct the object as:
new Base64(76, '');
If this isn't the class you're calling (it looks like from your code sample you're calling a static method), check the API and see if you can set a line separator for the conversion.

Isn't the 1 in Base64.encodeToString(msgDigest,1 ) padding?
If it's not, then you could just trim() the string to remove the whitespace.