I have the following regular expression:
WHERE A.srvc_call_id = '40750564' AND REGEXP_LIKE (A.SRVC_CALL_DN, '[^TEST]')
The row that contains 40750564 has "TEST CALL" in the column SRVC_CALL_DN and REGEXP_LIKE doesn't seem to be filtering it out. Whenever I run the query it returns the row when it shouldn't.
Is my regex pattern wrong? Or does SQL not accept [^whatever]?
The carat anchors the expression to the start of a string. By enclosing the letters T, E, S & T in square brackets you're searching, as barsju suggests for any of these characters, not for the string TEST.
You say that SRVC_CALL_DN contains the string 'TEST CALL', but you don't say where in the string. You also say that you're looking for where this string doesn't match. This implies that you want to use not regexp_like(...
Putting all this together I think you need:
AND NOT REGEXP_LIKE (A.SRVC_CALL_DN, '^TEST[[:space:]]CALL')
This excludes every match from your query where the string starts with 'TEST CALL'. However, if this string may be in any position in the column you need to remove the carat - ^.
This also assumes that the string is always in upper case. If it's in mixed case or lower, then you need to change it again. Something like the following:
AND NOT REGEXP_LIKE (upper(A.SRVC_CALL_DN), '^TEST[[:space:]]CALL')
By upper-casing SRV_CALL_DN you ensure that you're always going to match but ensure that your query may not use an index on this column. I wouldn't worry about this particular point as regular expressions queries can be fairly poor at using indexes anyway and it appears as though SRVC_CALL_ID is indexed.
Also if it may not include 'CALL' you will have to remove this. It is best when using regular expressions to make your match pattern as explicit as possible; so include 'CALL' if you can.
Try with '^TEST' or '^TEST.*'
Your regexp means any string not starting with any of the characters: T,E,S,T.
But your case is so simple, starts with TEST. Why not use a simple like:
LIKE 'TEST%'
Related
Am trying to determine how one attempts to identify, in Snowflake SQL, if a product code begins with three letters.
Suggestions?
I did just try: LEFT(P0.PRODUCTCODE,3) NOT LIKE '[a-zA-Z]%' but it didn't work.
Thanks folks
You can use REGEXP_LIKE to return a boolean value indicating whether or not your string matched the pattern you're interested in.
In your case, something like REGEXP_LIKE(string_field_here, '[a-zA-Z]{3}.*')
Breaking down the regular expression pattern:
[a-zA-Z]: Only match letter characters, both upper and lowercase
{3}: Require three of those letters
.*: Allow any number of any characters after those three letters
Note: in many cases, you would need to specifically indicate the beginning/ending of the string in the pattern, but Snowflake's implementation handles that for you. From the docs:
The function implicitly anchors a pattern at both ends (i.e. ''
automatically becomes '^$', and 'ABC' automatically becomes '^ABC$').
To match any string starting with ABC, the pattern would be 'ABC.*'.
You can try running these examples:
SELECT REGEXP_LIKE('abc', '[a-zA-Z]{3}.*') AS _abc,
REGEXP_LIKE('123', '[a-zA-Z]{3}.*') AS _123,
REGEXP_LIKE('abc123', '[a-zA-Z]{3}.*') AS _abc123,
REGEXP_LIKE('123abc', '[a-zA-Z]{3}.*') AS _123abc
Using Regular Extractor in JMeter, I need to get the value of "fullBkupUNIXTime" from the below response,
{"fullBackupTimeString":["Mon 10 Apr 2017 14:14:36"],"fullBkupUNIXTime":["1491833676"],"fullBackupDirName":["10_04_2017_0636"]}
I tried with Ref Name as time and
Regular Expression: "fullBkupUNIXTime": "([0-9])" and "(.+?)"
and pass them as input for 2nd request ${time}
The above 2 two doesn't work out for me.
Please Help me out of this.
First of all: why not just use this thing?
Then, if you firm with your RegExp adventure to get happen.
First expression is not going to work because you've defined it to match exactly one [0-9] charcter.
Add the appropriate repetition character, like "fullBkupUNIXTime": "([0-9]+)".
And basically it make sense to tell the engine to stop at first narrowest match too: "fullBkupUNIXTime": "([0-9]+?)"
Next, make sure you're handling space chars between key and value and colon mark properly. Better mark them explicitly, if any, with \s
And last but not least: make sure you're properly handle multiple lines (if appropriate, of course). Add the (?m) modifier to your expression.
And/or (?im) to be not case-sensitive, in addition.
[ is a reserve character in regex, you need to escape it, in your case use:
Regular Expression fullBkupUNIXTime":\["(\d+)
Template: $1$
Match No.: 1
I have a text field which acts a filter field. It checks for equals, contains and starts with. My problem is, with out changing any of my code, can I check for the 'does not contain', 'does not start with' and so on by just using the string i'm passing with something like '!' operator or "<>"?
for example:
I want to get all the records that do not have 'a' in them, so can I pass the string as "!a" or "<>a" or something so that I can get the required records? (I know these two don't work cause I tried.)
You have to use the keyword NOT, e.g.
SELECT * FROM Table WHERE Foo NOT LIKE '%bar%'
Please refer http://www.sqlservercentral.com/Forums/Topic1213442-338-1.aspx
Can solve with case in where clause
I need to match two ipaddress/hostname with a regular expression:
Like 20.20.20.20
should match with 20.20.20.20
should match with [http://20.20.20.20/abcd]
should not match with 20.20.20.200
should not match with [http://20.20.20.200/abcd]
should not match with [http://120.20.20.20/abcd]
should match with AB_20.20.20.20
should match with 20.20.20.20_AB
At present i am using something like this regular expression: "(.*[^(\w)]|^)20.20.20.20([^(\w)].*|$)"
But it is not working for the last two cases. As the "\w" is equal to [a-zA-Z0-9_]. Here I also want to eliminate the "_" underscore. I tried different combination but not able to succeed. Please help me with this regular expression.
(.*[_]|[^(\w)]|^)10.10.10.10([_]|[^(\w)].*|$)
I spent some more time on this.This regular expression seems to work.
I don't know which language you're using, but with Perl-like regular expressions you could use the following, shorter expression:
(?:\b|\D)20\.20\.20\.20(?:\b|\D)
This effectively says:
Match word boundary (\b, here: the start of the word) or a non-digit (\D).
Match IP address.
Match word boundary (\b, here: the end of the word) or a non-digit (\D).
Note 1: ?: causes the grouping (\b|\D) not to create a backreference, i.e. to store what it has found. You probably don't need the word boundaries/non-digits to be stored. If you actually need them stored, just remove the two ?:s.
Note 2: This might be nit-picking, but you need to escape the dots in the IP address part of the regular expression, otherwise you'd also match any other character at those positions. Using 20.20.20.20 instead of 20\.20\.20\.20, you might for example match a line carrying a timestamp when you're searching through a log file...
2012-07-18 20:20:20,20 INFO Application startup successful, IP=20.20.20.200
...even though you're looking for IP addresses and that particular one (20.20.20.200) explicitly shouldn't match, according to your question. Admittedly though, this example is quite an edge case.
I'd like to create a regular expression such that when I compare the a string against an array of strings, matches are returned with the regex ignoring certain characters.
Here's one example. Consider the following array of names:
{
"Andy O'Brien",
"Bob O'Brian",
"Jim OBrien",
"Larry Oberlin"
}
If a user enters "ob", I'd like the app to apply a regex predicate to the array and all of the names in the above array would match (e.g. the ' is ignored).
I know I can run the match twice, first against each name and second against each name with the ignored chars stripped from the string. I'd rather this by done by a single regex so I don't need two passes.
Is this possible? This is for an iOS app and I'm using NSPredicate.
EDIT: clarification on use
From the initial answers I realized I wasn't clear. The example above is a specific one. I need a general solution where the array of names is a large array with diverse names and the string I am matching against is entered by the user. So I can't hard code the regex like [o]'?[b].
Also, I know how to do case-insensitive searches so don't need the answer to focus on that. Just need a solution to ignore the chars I don't want to match against.
Since you have discarded all the answers showing the ways it can be done, you are left with the answer:
NO, this cannot be done. Regex does not have an option to 'ignore' characters. Your only options are to modify the regex to match them, or to do a pass on your source text to get rid of the characters you want to ignore and then match against that. (Of course, then you may have the problem of correlating your 'cleaned' text with the actual source text.)
If I understand correctly, you want a way to match the characters "ob" 1) regardless of capitalization, and 2) regardless of whether there is an apostrophe in between them. That should be easy enough.
1) Use a case-insensitivity modifier, or use a regexp that specifies that the capital and lowercase version of the letter are both acceptable: [Oo][Bb]
2) Use the ? modifier to indicate that a character may be present either one or zero times. o'?b will match both "o'b" and "ob". If you want to include other characters that may or may not be present, you can group them with the apostrophe. For example, o['-~]?b will match "ob", "o'b", "o-b", and "o~b".
So the complete answer would be [Oo]'?[Bb].
Update: The OP asked for a solution that would cause the given character to be ignored in an arbitrary search string. You can do this by inserting '? after every character of the search string. For example, if you were given the search string oleary, you'd transform it into o'?l'?e'?a'?r'?y'?. Foolproof, though probably not optimal for performance. Note that this would match "o'leary" but also "o'lea'r'y'" if that's a concern.
In this particular case, just throw the set of characters into the middle of the regex as optional. This works specifically because you have only two characters in your match string, otherwise the regex might get a bit verbose. For example, match case-insensitive against:
o[']*b
You can add more characters to that character class in the middle to ignore them. Note that the * matches any number of characters (so O'''Brien will match) - for a single instance, change to ?:
o[']?b
You can make particular characters optional with a question mark, which means that it will match whether they're there or not, e.g:
/o\'?b/
Would match all of the above, add .+ to either side to match all other characters, and a space to denote the start of the surname:
/.+? o\'?b.+/
And use the case-insensitivity modifier to make it match regardless of capitalisation.