Develop Reference

Build DB connection string from environment variables

I'm using the Play Framework (Java) and am not able to figure out how to use environment variables in my configuration file for building the database connection string. While I'm able to use environment variables like this (for user name and password):
default.username = ${?FU_MAIN_DB_USERNAME}
default.password = ${?FU_MAIN_DB_PASSWORD}
I'm not able to make it work in the url string. Perhaps this is a simple case of string processing in Scala that I'm missing, but since I'm working in Java, I could use some help.
So far, I have tried the url string in the following formats and failed:
Tried to add a $ to variable name to perform interpolation:
default.url = "jdbc:postgresql://$${?FU_MAIN_DB_HOST}:$${?FU_MAIN_DB_PORT}/$${?FU_MAIN_DB_NAME}";
But this doesn't substitute. Rather, it picks the string as such.
default.url = "jdbc:postgresql://${?FU_MAIN_DB_HOST}:${?FU_MAIN_DB_PORT}/${?FU_MAIN_DB_NAME}";
This too inserts the '$' and all verbatim. Then I thought maybe something like PHP-style will work
default.url = "jdbc:postgresql://${${?FU_MAIN_DB_HOST}}:${${?FU_MAIN_DB_PORT}}/${${?FU_MAIN_DB_NAME}}";
But no.
I also tried doing stuff like "jdbc:postgresql://".concat(${?FU_MAIN_DB_HOST}) ... but this also inserts '.concat' verbatim.
Finally, I tried concatenation using the '+' operator, but I'm told (by my IDE) that symbols like +: etc. are not allowed in the application.conf file.
How then, in God's name, am I supposed to do that?!

The double quotes turn off interpolation. But you need to do that for the : and the //.
Try
default.url = "jdbc:postgresql://"${?FU_MAIN_DB_HOST}":"${?FU_MAIN_DB_PORT}/${?FU_MAIN_DB_NAME}
Maybe you are better off to set the whole thing in one big environment variable instead.

java regex matcher results != to notepad++ regex find result

I am trying to extract data out of a website access log as part of a java program. Every entry in the log has a url. I have successfully extracted the url out of each record.
Within the url, there is a parameter that I want to capture so that I can use it to query a database. Unfortunately, it doesn't seem that the web developers used any one standard to write the parameter's name.
The parameter is usually called "course_id", but I have also seen "courseId", "course%3DId", "course%253Did", etc. The format for the parameter name and value is usually course_id=_22222_1, where the number I want is between the "_" and "_1". (The value is always the same, even if the parameter name varies.)
So, my idea was to use the regex /^.*course_id[^_]*_(\d*)_1.*$/i to find and extract the number.
In java, my code is
java.util.regex.Pattern courseIDPattern = java.util.regex.Pattern.compile(".*course[^i]*id[^_]*_(\\d*)_1.*", java.util.regex.Pattern.CASE_INSENSITIVE);
java.util.regex.Matcher courseIDMatcher = courseIDPattern.matcher(_url);
_courseID = "";
if(courseIDMatcher.matches())
{
_courseID = retrieveCourseID(courseIDMatcher.group(1));
return;
}
This works for a lot of the records. However, some records do not record the course_id, even though the parameter is in the url. One such example is the record:
/webapps/contentDetail?course_id=_223629_1&content_id=_3641164_1&rich_content_level=RICH&language=en_US&v=1&ver=4.1.2
However, I used notepad++ to do a regex replace on this (in fact, every) url using the regex above, and the url was successfully replaced by the course ID, implying that the regex is not incorrect.
Am I doing something wrong in the java code, or is the java matcher broken?

Website/URL Validation Regex in JAVA

I need a regex string to match URL starting with "http://", "https://", "www.", "google.com"
the code i tried using is:
//Pattern to check if this is a valid URL address
Pattern p = Pattern.compile("(http://|https://)(www.)?([a-zA-Z0-9]+).[a-zA-Z0-9]*.[a-z]{3}.?([a-z]+)?");
Matcher m;
m=p.matcher(urlAddress);
but this code only can match url such as "http://www.google.com"
I know this ma be a dupicate question but i have tried all of the regex provided and it does not suit my requirement. Willl someone please help me? Thank you.

You need to make (http://|https://) part in your regex as optional one.
^(http:\/\/|https:\/\/)?(www.)?([a-zA-Z0-9]+).[a-zA-Z0-9]*.[a-z]{3}.?([a-z]+)?$
DEMO

You can use the Apache commons library(org.apache.commons.validator.UrlValidator) for validating a url:
String[] schemes = {"http","https"}.
UrlValidator urlValidator = new UrlValidator(schemes);
And use :-
urlValidator.isValid(your url)
Then there is no need of regex.
Link:-
https://commons.apache.org/proper/commons-validator/apidocs/org/apache/commons/validator/routines/UrlValidator.html

If you use Java, I recommend use this RegEx (I wrote it by myself):
^(https?:\/\/)?(www\.)?([\w]+\.)+[‌​\w]{2,63}\/?$
"^(https?:\\/\\/)?(www\.)?([\\w]+\\.)+[‌​\\w]{2,63}\\/?$" // as Java-String
to explain:
^ = line start
(https?://)? = "http://" or "https://" may occur.
(www.)? = "www." may orrur.
([\w]+.)+ = a word ([a-zA-Z0-9]) has to occur one or more times. (extend here if you need special characters like ü, ä, ö or others in your URL - remember to use IDN.toASCII(url) if you use special characters. If you need to know which characters are legal in general: https://kb.ucla.edu/articles/what-characters-can-go-into-a-valid-http-url
[‌​\w]{2,63} = a word ([a-zA-Z0-9]) with 2 to 63 characters has to occur exactly one time. (a TLD (top level domain (for example .com) can not be shorter than 2 or longer than 63 characters)
/? = a "/"-character may occur. (some people or servers put a / at the end... whatever)
$ = line end
-
If you extend it by special characters it could look like this:
^(https?:\/\/)?(www\.)?([\w\Q$-_+!*'(),%\E]+\.)+[‌​\w]{2,63}\/?$
"^(https?:\\/\\/)?(www\.)?([\\w\\Q$-_+!*'(),%\\E]+\\.)+[‌​\\w]{2,63}\\/?$" // as Java-String
The answer of Avinash Raj is not fully correct.
^(http:\/\/|https:\/\/)?(www.)?([a-zA-Z0-9]+).[a-zA-Z0-9]*.[a-z]{3}.?([a-z]+)?$
The dots are not escaped what means it matches with any character. Also my version is simpler and I never heard of a domain like "test..com" (which actually matches...)
Demo: https://regex101.com/r/vM7wT6/279
Edit:
As I saw some people needing a regex which also matches servers directories I wrote this:
^(https?:\/\/)?([\w\Q$-_+!*'(),%\E]+\.)+(\w{2,63})(:\d{1,4})?([\w\Q/$-_+!*'(),%\E]+\.?[\w])*\/?$
while this may not be the best one, since I didn't spend too much time with it, maybe it helps someone. You can see how it works here: https://regex101.com/r/vM7wT6/700
It also matches urls like "hello.to/test/whatever.cgi"

Java compatible version of #Avinash's answer would be
//Pattern to check if this is a valid URL address
Pattern p = Pattern.compile("^(http://|https://)?(www.)?([a-zA-Z0-9]+).[a-zA-Z0-9]*.[a-z]{3}.?([a-z]+)?$");
Matcher m;
m=p.matcher(urlAddress);
boolean matches = m.matches();

pattern="w{3}\.[a-z]+\.?[a-z]{2,3}(|\.[a-z]{2,3})"
this will only accept addresses like e.g www.google.com & www.google.co.in

//I use that
static boolean esURL(String cadena){
boolean bandera = false;
bandera = cadena.matches("\\b(https://?|ftp://|file://|www.)[-a-zA-Z0-9+&##/%?=~_|!:,.;]*[-a-zA-Z0-9+&##/%=~_|]");
return bandera;
}

URI.getPath() does not return anything after '#'

I have a URI with path like this :
ftp://test:test#someftp/ready123/users/abc/#0#.
But the method getPath() on the URI returns this:
/ready123/users/abc/.
I need the whole path to be returned like this :
/ready123/users/abc/#0#
... So that I can change the working directory(CWD) to the folder #0#. The code is in a generic method and is jarred up which gets used by many other applications. So I have to be very careful when I make changes. I believe anything after # is considered as a fragment, but in this case it is actually the name of a folder.
How do I get the path /ready123/users/abc/#0# from the URI object ?

From the JavaDoc of URI class:
URI syntax and components At the highest level a URI reference
(hereinafter simply "URI") in string form has the syntax
[scheme:]scheme-specific-part[#fragment]
where square brackets [...] delineate optional components and the
characters : and # stand for themselves.
As such if you want to retrieve everything after the first # you need to use URI.getFragment()

You need to replace the hash character with %23
URI uir = new URI("ftp://test:test#someftp/ready123/users/abc/%230%23");
System.out.println(uir.getPath()); // returns '/ready123/users/abc/#0#'

How do I open the default mail program with a Subject and Body in a cross-platform way?

How do I open the default mail program with a Subject and Body in a cross-platform way?
Unfortunately, this is for a a client app written in Java, not a website.
I would like this to work in a cross-platform way (which means Windows and Mac, sorry Linux). I am happy to execute a VBScript in Windows, or AppleScript in OS X. But I have no idea what those scripts should contain. I would love to execute the user's default program vs. just searching for Outlook or whatever.
In OS X, I have tried executing the command:
open mailto:?subject=MySubject&body=TheBody
URL escaping is needed to replace spaces with %20.
Updated On Windows, you have to play all sorts of games to get start to run correctly. Here is the proper Java incantation:
class Win32 extends OS {
public void email(String subject, String body) throws Exception {
String cmd = "cmd.exe /c start \"\" \"" + formatMailto(subject, body) + "\"";
Runtime.getRuntime().exec(cmd);
}
}

In Java 1.6 you have a stardard way to open the default mailer of the platform:
the Desktop.mail(URI) method.The URI can be used to set all the fields of the mail (sender, recipients, body, subject).
You can check a full example of desktop integration in Java 1.6 on Using the Desktop API in Java SE 6

start works fine in Windows (see below). I would use Java's built in UrlEscape then just run a second replacement for '+' characters.
start mailto:"?subject=My%20Subject&body=The%20Body"

Never use Runtime.exec(String) on Mac OS X or any other operating system. If you do that, you'll have to figure out how to properly quote all argument strings and so on; it's a pain and very error-prone.
Instead, use Runtime.exec(String[]) which takes an array of already-separated arguments. This is much more appropriate for virtually all uses.

1. Add a Subject Line
You can prefill the subject line in the email by adding the subject preceded by '?subject=' after the email address.
So the link now becomes:
Email Us
2. Send to Multiple Recipients
Mail can be sent to additional recipients either as carbon copies (cc) or blind carbon copies (bcc).
This is done in a similar way, by placing '?cc=someoneelse#theirsite.com' after the initial address.
So the link looks like this:
Email Us
cc can simply be replaced by bcc if you wish to send blind carbon copies.
This can be very useful if you have links on pages with different subjects. You might have the email on each page go to the appropriate person in a company but with a copy of all mails sent to a central address also.
You can of course specify more than one additional recipient, just separate your list of recipients with a comma.
Email Us
Sourced from Getting More From 'mailto' which now 404s. I retrieved the content from waybackmachine.
3. Combining Code
You can combine the various bits of code above by the addition of an '&' between each.
Thus adding
me#mysite.com?subject=Hello&cc=you#yoursite.com&bcc=her#hersite.com
would send an email with the subject 'Hello' to me, you and her.
4. Write the Email
You can also prefill the body of the email with the start of a message, or write the whole message if you like! To add some thing to the body of the email it is again as simple as above - '?body=' after the email address. However formatting that email can be a little tricky. To create spaces between words you will have to use hex code - for example '%20' between each word, and to create new lines will mean adding '%0D'. Similarly symbols such as $ signs will need to be written in hex code.
If you also wish to add a subject line and send copies to multiple recipients, this can make for a very long and difficult to write bit of code.
It will send a message to three people, with the subject and the message filled in, all you need to do is add your name.
Just look at the code!
<a href="mailto:abbeyvet#outfront.net?CC=spooky#outfront.net
&BCC=thomasbrunt#outfront.net&Subject=Please%2C%20I%20insist
%21&Body=Hi%0DI%20would%20like%20to%20send%20you%20
%241000000%20to%20divide%20as%20you%20see%20fit%20among
%20yourselves%20and%20all%20the%20moderators.%0DPlease%
20let%20me%20know%20to%20whom%20I%20should%20send
%20the%20check.">this link</a>
Note: Original source URL where I found this is now 404ing so I grabbed to content from waybackmachine and posted it here so it doesn't get lost. Also, the OP stated it was not for a website, which is what these examples are, but some of these techniques may still be useful.

I had to re-implement URLencode
because Java's would use + for space
and Mail took those literally.
I don't know if Java has some built-in method for urlencoding the string, but this link http://www.permadi.com/tutorial/urlEncoding/ shows some of the most common chars to encode:
; %3B
? %3F
/ %2F
: %3A
# %23
& %24
= %3D
+ %2B
$ %26
, %2C
space %20 or +
% %25
< %3C
> %3E
~ %7E
% %25

I don't know if Java has some built-in method for urlencoding the string, but this link http://www.permadi.com/tutorial/urlEncoding/ shows some of the most common chars to encode:
For percent-encoding mailto URI hnames and hvalues, I use the rules at http://shadow2531.com/opera/testcases/mailto/modern_mailto_uri_scheme.html#encoding. Under http://shadow2531.com/opera/testcases/mailto/modern_mailto_uri_scheme.html#implementations, there's a Java example that may help.
Basically, I use:
private String encodex(final String s) {
try {
return java.net.URLEncoder.encode(s, "utf-8").replaceAll("\\+", "%20").replaceAll("\\%0A", "%0D%0A");
} catch (Throwable x) {
return s;
}
}
The string that's passed in should be a string with \r\n, and stray \r already normalized to \n.
Also note that just returning the original string on an exception like above is only safe if the mailto URI argument you're passing on the command-line is properly escaped and quoted.
On windows that means:
Quote the argument.
Escape any " inside the quotes with \.
Escape any \ that precede a " or the end of the string with \.
Also, on windows, if you're dealing with UTF-16 strings like in Java, you might want to use ShellExecuteW to "open" the mailto URI. If you don't and return s on an exception (where some hvalue isn't completely percent-encoded, you could end up narrowing some wide characters and losing information. But, not all mail clients accept unicode arguments, so ideally, you want to pass a properly percent-encoded-utf8 ascii argument with ShellExecute.
Like 'start', ShellExecute with "open" should open the mailto URI in the default client.
Not sure about other OS's.

Mailto isn't a bad route to go. But as you mentioned, you'll need to make sure it is encoded correctly.
The main problem with using mailto is with breaking lines. Use %0A for carriage returns, %20 for spaces.
Also, keep in mind that the mailto is considered the same as a URL of sorts and therefore will have the same limitations for length. See
http://support.microsoft.com/kb/208427, note the maximum URL length of 2083 characters. This is confirmed for mailto as well
in this article: http://support.microsoft.com/kb/279460/en-us. Also, some mail clients can also have a limit (I believe older versions of Outlook Express had a limit of something much smaller like 483 characters or something. If you expect to have a longer string than that then you'll need to look at alternatives.
BTW, you shouldn't have to resort to kicking out a script to do that as long as you can shell out a command from Java (I don't know if you can since I don't do Java).

You may use this...
main(string[] args){
String forUri = String.format("mailto:?subject=%s&body=%s", urlEncode(sub), urlEncode(mailBody));
Desktop.getDesktop().mail(new URI(forUri));
}
private static final String urlEncode(String str) {
try {
return URLEncoder.encode(str, "UTF-8").replace("+", "%20");
} catch (UnsupportedEncodingException e) {
throw new RuntimeException(e);
}
}
Also for formatting read A simple way of sending emails in Java: mail-to links

I have implemented this, and it works well on OS X. (Ryan's mention of the max URL length has not been codified.)
public void email(String subject, String body) throws Exception {
String cmd = "open mailto:";
cmd += "?subject=" + urlEncode(subject);
cmd += "&body=" + urlEncode(body);
Runtime.getRuntime().exec(cmd);
}
private static String urlEncode(String s) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < s.length(); i++) {
char ch = s.charAt(i);
if (Character.isLetterOrDigit(ch)) {
sb.append(ch);
}
else {
sb.append(String.format("%%%02X", (int)ch));
}
}
return sb.toString();
}
I had to re-implement URLencode because Java's would use + for space and Mail took those literally. Haven't tested on Windows yet.