#"^(?=.*[0-9]+.*)(?=.*[a-zA-Z]+.*)[0-9a-zA-Z]{6,}$"
I am using this regular expression for password validation which gives one upper case, one lowercase and a number. But what I want is a special character in it but it should optional but above mentioned must be mandatory.
This will allow these special characters: - (hyphen), * (asterisk) and _ (underscore).
^(?=[-_*]*)(?=.*[0-9]+.*)(?=.[a-zA-Z]+.)[-*_0-9a-zA-Z]{6,}$
If you want to add your own special characters, add them into this part of the regex [-*_0-9a-zA-Z] (inside the square bracket)
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'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
In the following sql, what the use of escape is ?
select * from dual where dummy like 'funny&_' escape '&';
SQL*Plus ask for the value of _ whether escape is specified or not.
The purpose of the escape clause is to stop the wildcard characters (eg. % or _) from being considered as wildcards, as per the documentation
The reason why you're being prompted for the value of _ is because you're using &, which is also usually the character used to prompt for a substitution variable.
To stop the latter from happening, you could:
change to a different escape character
prior to running your statement, run set define off if you're using SQL*Plus (or as a script in a GUI, eg. Toad) or turn off the substitution variable prompting if you're using a GUI.
change the define character to something different by running set define <character>
The escape character is used to indicate that the underscore should be matched as an actual character, rather than as a single-character wildcard. This is explained in the documentation.
You can include the actual characters % or _ in the pattern by using the ESCAPE clause, which identifies the escape character. If the escape character precedes the character % or _ in the pattern, then Oracle interprets this character literally in the pattern rather than as a special pattern-matching character.
If you didn't have the escape clause then the underscore would match any single character, so where dummy like 'funny_' would match 'funnyA', 'funnyB', etc. and not just an actual underscore.
The escape character you've chosen is & which is the default SQL*Plus client substitution variable marker. It has nothing to do with the escape clause, and using that is causing the &_ part of the pattern to be interpreted as a substitution variable called _, hence your being prompted. As it isn't related, the escape clause has no effect on that.
The simplest thing is probably to choose a different escape character. If you want to use that specific escape character and not be prompted, disable or change the substitution character:
set define off
select * from dual where dummy like 'funny&_' escape '&';
set define on
That will then match rows where dummy contains exactly the string 'funny_'. (It's therefore equivalent to where dummy = 'funny_', as there are no unescaped wildcards, making the like pattern matching redundant). It will not match any that start with that pattern (it's sort of like using regexp_like with start and end anchors, and you might be expecting it to work as if you hadn't supplied anchors, but it doesn't). You would need to add a % wildcard for that:
set define off
select * from dual where dummy like 'funny&_%' escape '&';
set define on
And if you want to match any that don't start with funny_ but have it somewhere in the middle of the value, you would need to add another wildcard before it too:
set define off
select * from dual where dummy like '%funny&_%' escape '&';
set define on
You haven't shown any sample data or expected results to it isn't clear which pattern you need.
SQL Fiddle doesn't have substitution variables but here's an example showing how those three patterns match various values.
The syntax for the SQL LIKE Condition is:
expression LIKE pattern [ ESCAPE 'escape_character' ]
Parameters or Arguments
expression : A character expression such as a column or field.
pattern : A character expression that contains pattern matching. The patterns that you can choose from are:
Wildcard | Explanation
---------+-------------
% | Allows you to match any string of any length (including zero length)
_ | Allows you to match on a single character
escape_character: Optional. It allows you to test for literal instances of a wildcard character such as % or _.
Source : http://www.techonthenet.com/sql/like.php
I would like to have a regular expression to make an Oracle SQL REGEXP_LIKE query that checks
if a string starts with one alphanumeric character
if the string ends with one alphanumeric character
if the "body" of the string contains only alphanumeric character OR these authorized characters (written) : hyphen (dash), dot, apostrophe, space
if the authorised characters are NOT adjacent (to avoid something like "he--'''l..'-lo")
I started with this :
^[a-zA-Z0-9]+(a-zA-Z0-9\-\.'|([^\-\.'])\1)*[a-zA-Z0-9]$
I used backslash to escape assuming that dot and hyphen are metacharacters
I think this is what you want:
^[a-zA-Z0-9]+([-.' ][a-zA-Z0-9]|[a-zA-Z0-9])*\w?$
It looks for
at least 1 alphanumeric (alnum),
followed by
either an authorized character followed by an alphanumeric or just an alphanumeric, repeated any number of times (including 0).
optionally followed by
an alnum
This meets your specification. I'm not sure if starts with one alnum and ends with one alnum means that there must be at least 2 alnums, or if they can be the same. If there must be at least 2 of them, remove the last ? (which make the last alnum optional).
Regards
assuming you meant "authorised characters are NOT adjacent to each other"
try something along these lines
^[a-zA-Z0-9]+([a-zA-Z0-9]+[\-\.' ]?)*[a-zA-Z0-9]$
so that the repeating middle part always has one alphanumeric character followed by zero to one special characters.
When I search one ldap server using the following filter
(cn=%*)
It return all results under the base dn? LDAP treat '%' specially? But I haven't found any description about it.
What is your directory server ?
Are you sure tha '%' is not replace by your command line interpreter or your compiler ?
According to RFC2254 % is not a special character
If a value should contain any of the following characters
Character ASCII value
---------------------------
* 0x2a
( 0x28
) 0x29
\ 0x5c
NUL 0x00
the character must be encoded as the backslash '\' character (ASCII
0x5c) followed by the two hexadecimal digits representing the ASCII
value of the encoded character. The case of the two hexadecimal
digits is not significant.
This simple escaping mechanism eliminates filter-parsing ambiguities
and allows any filter that can be represented in LDAP to be
represented as a NUL-terminated string. Other characters besides the
ones listed above may be escaped using this mechanism, for example,
non-printing characters.
For example, the filter checking whether the "cn" attribute contained
a value with the character "" anywhere in it would be represented as
"(cn=\2a*)".
Note that although both the substring and present productions in the
grammar above can produce the "attr=*" construct, this construct is
used only to denote a presence filter.