Develop Reference

Search a string with spaces similar to full string in JAVA

I have an issue which I am facing I would really appreciate your help.
I am using java and connecting it to postgres DB. I tried writing a query with LIKE and it works, but what I am looking is regex that works similar to LIKE where white spaces are also counted.
For example lets say we have the following entries in our array from the results of the DB as
"ca ts", "cats", "ca ts"
etc. When I type
"c a ts"
in the search filter I should retrieve all the above from that array which has all the results from the database.

You may try with replacing spaces from input and search pattern:
String input = "a b c";
String searchPattern = "ab c";
Pattern pat = Pattern.compile(searchPattern.replace(" ", ""));
System.out.println(pat.matcher(input.replace(" ", "")).matches());

You don't need regex for it. EG here 'c a ts' is "like" 'ca ts':
b=# select replace('c a ts',' ','') like replace('ca ts',' ','') as example;
example
---------
t
(1 row)

How to extract the session id from an RTSP message's content?

I have a string like this:
RTSP/1.0 200 OK
CSeq: 3
Server: Ants Rtsp Server/1.0
Date: 21 Oct 2016 15:55:30 GMT
Session: 980603187; timeout=60
Transport: RTP/AVP/TCP;unicast;interleaved=0-1;ssrc=F006B800
I want to extract the session number(980603187)
Could someone please provide some help?

Simply use a regular expression with a group, then extract the value of the group as next:
String content ="RTSP/1.0 200 OK\n" +
"CSeq: 3\n" +
"Server: Ants Rtsp Server/1.0\n" +
"Date: 21 Oct 2016 15:55:30 GMT\n" +
"Session: 980603187; timeout=60\n" +
"Transport: RTP/AVP/TCP;unicast;interleaved=0-1;ssrc=F006B800\n";
Pattern pattern = Pattern.compile("Session: ([a-zA-Z0-9$\\-_.+]+)");
Matcher matcher = pattern.matcher(content);
if (matcher.find()) {
System.out.println(matcher.group(1));
}
Output:
980603187
Explanation:
Session: ([a-zA-Z0-9$\\-_.+]+)
Session: matches the characters Session: literally (case sensitive)
([a-zA-Z0-9$\\-_.+]+): Capturing group that matches with several consecutive ALPHA, DIGIT or SAFE characters (at least one) (cf RFC 2326 chapter 3.4 Session Identifiers)

Use Regex! Having String str = .., extract the number needed with the Regex capturing anything between Session: and ;:
Session: (.+);
Feel free to specify only letters \\w+ or digits \\d+. Mind the double escaping in Java. The first matched m.group(1) is your result:
Pattern p = Pattern.compile("Session: (.+);");
Matcher m = p.matcher(str);
if (m.find()) {
System.out.println(m.group(1));
}
Outputs 980603187. Check out the Regex101 for the explanation.
In come cases the ; timeout is optional and to need to amend the Regex used:
Session: (.+?)[\n;]

Once you have each header you can look up the specification in RFC 2336 which specifies the RTSP protocol.
First of all, you should split your string into lines. The lines end with CR/LF according to the specification. The first line indicates the response, the other should be header fields.
The definition is:
Session = "Session" ":" session-id [ ";" "timeout" "=" delta-seconds ]
where session-id is specified as:
session-id = 1*( ALPHA | DIGIT | safe )
which means you should not confuse it with a number. The definition of safe is
safe = "\$" | "-" | "_" | "." | "+"
and alpha means all upper- and lowercase numbers. This means it is possible to put in a base 64 url encoded binary session-id, by the way.
OK, now it becomes a question of looking for the session ID. You step through all lines (except the first one) and then look for the line that matches:
^Session[ \t]*:[ \t]*([a-zA-Z0-9\$\-_.+]+).*$
this will match only valid session headers / valid session identifiers. Note that the standard is vague about white-space, so I skipped over space and tab characters before and after the colon ':'. The session identifier is then in group 1 of the regular expression.
You can of course easily extend this by including the timeout in the regular expression, once you need it.
Note that you will have to double escape the backslash characters before using the regular expression in Java. It's also possible to use the Posix character classes defined in the Pattern class to make the regular expression more readable.

If you use apache-commons in your dependencies, then you can do it within one line:
StringUtils.substringBetween(string, "Session: ", ";");

Making a Regex More Dynamic

I posted this question a couple weeks ago pertaining to extracting a capture group using regex in Java, Extracting Capture Group Using Regex, and I received a working answer. I also posted this question a couple weeks ago pertaining to character replacement in Java using regex, Replace Character in Matching Regex, and received an even better answer that was more dynamic than the one I got from my first post. I'll quickly illustrate by example. I have a string like this that I want to extract the "ID" from:
String idInfo = "Any text up here\n" +
"Here is the id\n" +
"\n" +
"?a0 12 b5\n" +
"&Edit Properties...\n" +
"And any text down here";
And in this case I want the output to just be:
a0 12 b5
But it turns out the ID could be any number of octets (just has to be 1 or more octets), and I want my regex to be able to basically account for an ID of 1 octet then any number of subsequent octets (from 0 to however many). The person I received an answer from in my Replace Character in Matching Regex post did this for a similar but different use case of mine, but I'm having trouble porting this "more dynamic" regex over to the first use case.
Currently, I have ...
Pattern p = Pattern.compile("(?s)?:Here is the id\n\n\\?([a-z0-9]{2})|(?<!^)\\G:?([a-z0-9]{2})|.*?(?=Here is the id\n\n\\?)|.+");
Matcher m = p.matcher(certSerialNum);
String idNum = m.group(1);
System.out.println(idNum);
But it's throwing an exception. In addition, I would actually like it to use all known adjacent text in the pattern including "Here is the id\n\n\?" and "\n&Edit Properties...". What corrections do I need to get this working?

Seems like you want something like this,
String idInfo = "Any text up here\n" +
"Here is the id\n" +
"\n" +
"?a0 12 b5\n" +
"&Edit Properties...\n" +
"And any text down here";
Pattern regex = Pattern.compile("Here is the id\\n+\\?([a-z0-9]{2}(?:\\s[a-z0-9]{2})*)(?=\\n&Edit Properties)");
Matcher matcher = regex.matcher(idInfo);
while(matcher.find()){
System.out.println(matcher.group(1));
}
Output:
a0 12 b5
DEMO

Google Directory Service query with space and wildcard: behavior not matching expectations

I ran into some unexpected behavior when trying to combine whitespaces in a Directory Service API search query with a wildcard character:
Code
Directory ds = new GoogleDirectoryServiceManager().getDirectoryService("admin#randomdomain.com");
Directory.Users usersClient = ds.users();
String lastNameBuffer = term;
StringBuilder sb = new StringBuilder();
if(term.contains(" ")){
sb.append("\'");
sb.append(lastNameBuffer);
sb.append("*\'");
}else{
sb.append(lastNameBuffer);
sb.append("*");
}
queryString = "familyName:" + sb.toString();
users = usersClient.list()
.setDomain("randomdomain.com")
.setQuery(queryString)
.setMaxResults(MAX_DIRECTORY_RESULTS)
.setFields("users(name,primaryEmail,thumbnailPhotoUrl)")
.execute();
I added the single quotes to search through the directory with the spaces. Following is an extract from the Google Directory API:
Surround with single quotes ' if the query contains whitespace. Escape single quotes in queries with \', for example 'Valentine\'s Day'.
Input
"johns" --> Returns all users with Johnson name (familyName: johns*) GOOD
"van bur" --> Does not return any users (familyName: 'van bur*') NOT GOOD
"van buren" --> Returns all users with Van Buren name (familyName: 'van buren*') GOOD
tl;dr
Is it not possible to combine a search query with spaces with a wildcard? Or does the wildcard symbol (*) have to be escaped?
What query do I need to use to successfully return a list of users in case #2?

During my attempts with the API explorer embedded in the documentation, I have noted that the following query, the same as you without the single quotes, would work as expected on my test domain. Can you test it ?
familyName:van bur

Java String Replace Regex

I am doing some string replace in SQL on the fly.
MySQLString = " a.account=b.account ";
MySQLString = " a.accountnum=b.accountnum ";
Now if I do this
MySQLString.replaceAll("account", "account_enc");
the result will be
a.account_enc=b.account_enc
(This is good)
But look at 2nd result
a.account_enc_num=a.account_enc_num
(This is not good it should be a.accountnum_enc=b.accountnum_enc)
Please advise how can I achieve what I want with Java String Replace.
Many Thanks.

From your comment:
Is there anyway to tell in Regex only replace a.account=b.account or a.accountnum=b.accountnum. I do not want accountname to be replace with _enc
If I understand correctly you want to add _enc part only to account or accountnum. To do this you can use
MySQLString = MySQLString.replaceAll("\\baccount(num)?\\b", "$0_enc");
(num)? mean that num is optional so regex will accept account or accountnum
\\b at start mean that there can be no letters, numbers or "_" before account so it wont accept (affect) something like myaccount, or my_account.
\\b at the end will prevent other letters, numbers or "_" after account or accountnum.

It's hard to extrapolate from so few examples, but maybe what you want is:
MySQLString = MySQLString.replaceAll("account\\w*", "$0_enc");
which will append _enc to any sequence of letters, digits, and underscores that starts with account.

try
String s = " a.accountnum=b.accountnum ".replaceAll("(account[^ =]*)", "$1_enc");
it means replace any sequence characters which are not ' ' or '=' which starts the word "account" with the sequence found + "_enc".
$1 is a reference to group 1 in regex; group 1 is the expression in parenthesis (account[^ =]+), i.e. our sequence
See http://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html for details