I would like to write a regular expression that starts with the string "wp" and ends with the string "php" to locate a file in a directory. How do I do it?
Example file: wp-comments-post.php
This should do it for you ^wp.*php$
Matches
wp-comments-post.php
wp.something.php
wp.php
Doesn't match
something-wp.php
wp.php.txt
^wp.*\.php$ Should do the trick.
The .* means "any character, repeated 0 or more times". The next . is escaped because it's a special character, and you want a literal period (".php"). Don't forget that if you're typing this in as a literal string in something like C#, Java, etc., you need to escape the backslash because it's a special character in many literal strings.
Example:
ajshdjashdjashdlasdlhdlSTARTasdasdsdaasdENDaknsdklansdlknaldknaaklsdn
1) START\w*END
return: STARTasdasdsdaasdEND - will give you words between START and END
2) START\d*END
return: START12121212END - will give you numbers between START and END
3) START\d*_\d*END
return: START1212_1212END - will give you numbers between START and END having _
Related
I am a complete Reg-exp noob, so please bear with me. Tried to google this, but haven't found it yet.
What would be an appropriate way of writing a Regular expression matching files starting with a dot, such as .buildpath or .htaccess?
Thanks a lot!
In most regex languages, ^\. or ^[.] will match a leading dot.
The ^ matches the beginning of a string in most languages. This will match a leading .. You need to add your filename expression to it.
^\.
Likewise, $ will match the end of a string.
You may need to substitute the \ for the respective language escape character. However, under Powershell the Regex I use is: ^(\.)+\/
Test case:
"../NameOfFile.txt" -match '^(\\.)+\\\/'
works, while
"_./NameOfFile.txt" -match '^(\\.)+\\\/'
does not.
Naturally, you may ask, well what is happening here?
The (\\.) searches for the literal . followed by a +, which matches the previous character at least once or more times.
Finally, the \\\/ ensures that it conforms to a Window file path.
It depends a bit on the regular expression library you use, but you can do something like this:
^\.\w+
The ^ anchors the match to the beginning of the string, the \. matches a literal period (since an unescaped . in a regular expression typically matches any character), and \w+ matches 1 or more "word" characters (alphanumeric plus _).
See the perlre documentation for more info on Perl-style regular expressions and their syntax.
It depends on what characters are legal in a filename, which depends on the OS and filesystem.
For example, in Windows that would be:
^\.[^<>:"/\\\|\?\*\x00-\x1f]+$
The above expression means:
Match a string starting with the literal character .
Followed by at least one character which is not one of (whole class of invalid chars follows)
I used this as reference regarding which chars are disallowed in filenames.
To match the string starting with dot in java you will have to write a simple expression
^\\..*
^ means regular expression is to be matched from start of string
\. means it will start with string literal "."
.* means dot will be followed by 0 or more characters
I have two types of URL's which I would need to clean, they look like this:
["//xxx.com/se/something?SE_{ifmobile:MB}{ifnotmobile:DT}_A_B_C_D_E_F_G_H"]
["//www.xxx.com/se/car?p_color_car=White?SE_{ifmobile:MB}{ifnotmobile:DT}_A_B_C_D_E_F_G_H"]
The outcome I want is;
SE_{ifmobile:MB}{ifnotmobile:DT}_A_B_C_D_E_F_G_H"
I want to remove the brackets and everything up to SE, the URLS differ so I want to remove:
First URL
["//xxx.com/se/something?
Second URL:
["//www.xxx.com/se/car?p_color_car=White?
I can't get my head around it,I've tried this .*\/ . But it will still keep strings I don't want such as:
(1 url) =
something?
(2 url) car?p_color_car=White?
You can use
regexp_replace(FinalUrls, r'.*\?|"\]$', '')
See the regex demo
Details
.*\? - any zero or more chars other than line breakchars, as many as possible and then ? char
| - or
"\]$ - a "] substring at the end of the string.
Mind the regexp_replace syntax, you can't omit the replacement argument, see reference:
REGEXP_REPLACE(value, regexp, replacement)
Returns a STRING where all substrings of value that match regular
expression regexp are replaced with replacement.
You can use backslashed-escaped digits (\1 to \9) within the
replacement argument to insert text matching the corresponding
parenthesized group in the regexp pattern. Use \0 to refer to the
entire matching text.
I want to judge if a positive number string is end with ".0", so I wrote the following sql:
select '12310' REGEXP '^[0-9]*\.0$'. The result is true however. I wonder why I got the result, since I use "\" before "." to escape.
So I write another one as select '1231.0' REGEXP '^[0-9]\d*\.0$', but this time the result is false.
Could anyone tell me the right pattern?
Dot (.) in regexp has special meaning (any character) and requires escaping if you want literally dot:
select '12310' REGEXP '^[0-9]*\\.0$';
Result:
false
Use double-slash to escape special characters in Hive. slash has special meaning and used for characters like \073 (semicolon), \n (newline), \t (tab), etc. This is why for escaping you need to use double-slash. Also for character class digit use \\d:
hive> select '12310.0' REGEXP '^\\d*?\\.0$';
OK
true
Also characters inside square brackets do not need double-slash escaping: [.] can be used instead of \\.
If you know it is a number string, why not just use:
select ( val like '%.0' )
You need regular expression if you want to validate that the string has digits everywhere else. But if you only need to check the last two characters, like is sufficient.
As for your question . is a wildcard in regular expressions. It matches any character.
I am trying to call back all rows who's email starts with a character which is not AlphaNumeric.
The line I am trying to use in the statement is
REGEXP_LIKE (SUBSTR(hcp_email.email_address,1,2), '![%a-zA-Z%]')
This does not bring back the relevant lines.
I am able to bring back results with the below but this is not as practical as using a catch all range of text and numbers to ignore.
REGEXP_LIKE (SUBSTR(hcp_email.email_address, 1,2), '[":., ]')
Ideally I would like to use a NOT LIKE statement with a range a-z 0-9.
No need to embed a SUBSTR call. Just anchor the regex to the start of the string, and look at the first character. This example uses the POSIX shorthand for readability.
where regexp_like(hcp_email.email_address, '^[^[:alnum:]]');
EDIT - Added regex explanation
^ - Anchor to the start of the string
[ - Start a character class
^ - Inside of a character class this means NOT
[:alnum:] - POSIX shorthand for an alphanumeric (A-Za-z0-9)
] - End character class (the character class describes 1 character)
I'm trying to figure out the base regex to capture the middle of a google url out of a sql database.
For example, a few links:
https://www.google.com/cars/?year=2016&model=dodge+durango&id=1234
https://www.google.com/cars/?year=2014&model=jeep+cherokee+crossover&id=6789
What would be the regex to capture the text to get dodge+durango , or jeep+cherokee+crossover ? (It's alright that the + still be in there.)
My Attempts:
1)
\b[=.]\W\b\w{5}\b[+.]?\w{7}
, but this clearly does not work as this is a hard coded scenario that would only work like something for the dodge durango example. (would extract "dodge+durango)
2) Using positive lookback ,
[^+]( ?=&id )
but I am not fully sure how to use this, as this only grabs one character behind the & symbol.
How can I extract a string of (potentially) any length with any amount of + delimeters between the "model=" and "&id" boundaries?
seems like you could use regexp_replace and access match groups:
regexp_replace(input, 'model=(.*?)([&\\s]|$)', E'\\1')
from here:
The regexp_replace function provides substitution of new text for
substrings that match POSIX regular expression patterns. It has the
syntax regexp_replace(source, pattern, replacement [, flags ]). The
source string is returned unchanged if there is no match to the
pattern. If there is a match, the source string is returned with the
replacement string substituted for the matching substring. The
replacement string can contain \n, where n is 1 through 9, to indicate
that the source substring matching the n'th parenthesized
subexpression of the pattern should be inserted, and it can contain \&
to indicate that the substring matching the entire pattern should be
inserted. Write \ if you need to put a literal backslash in the
replacement text. The flags parameter is an optional text string
containing zero or more single-letter flags that change the function's
behavior. Flag i specifies case-insensitive matching, while flag g
specifies replacement of each matching substring rather than only the
first one
I may be misunderstanding, but if you want to get the model, just select everything between model= and the ampersand (&).
regexp_matches(input, 'model=([^&]*)')
model=: Match literally
([^&]*): Capture
[^&]*: Anything that isn't an ampersand
*: Unlimited times