I have a table named
logs
and column named
url
whose value is like this
https://api.abc.com:443/xyz/live/dashboard_critical/v1?cust_id=-1111%7C-1111%7C-1111%7C-1111%7C-1111%7C-1111%7C-1111%7C-1111%7C-1111%7C-1111%7C-1111&model_id=&startdate=2021-06-29&enddate=2021-07-28&wcombo=&site_id=&instance=&unit=&combo=&source=&priority=&verification=&critical=Yes%7Ctrue&percentile=95&startevent=receiveddatetime&endevent=uploadtohostdatetime
I trying to extract all strings after '/' OR '?' OR '%7C' OR '&'
I am able to extract it individually but not all 4 together
SELECT regexp_extract_all(url, '\d+[/]*')
FROM logs.url
WHERE request_verb='GET'
AND REGEXP_LIKE(url, 'https://api.abc.com:443/xyz/live/');
I am looking for output like below
[https:,api.abc.com:443,xyz,live,dashboard_critical,v1,cust_id=-1111,-1111,-1111,-1111,-1111,-1111,-1111,-1111,-1111,-1111,-1111,model_id=,startdate=2021-06-29,enddate=2021-07-28,wcombo=,site_id=,instance=,unit=,combo=,source=,priority=,verification=,critical=Yes,true,percentile=95,startevent=receiveddatetime,endevent=uploadtohostdatetime]
You can use regexp_split(str, regexp) function, as a regexp pattern concatenate all values by wich string should be splitted using | (OR in regexp), it will produce array required. Note: some characters have special meaning in Presto CLI or regexp and need shielding.
select regexp_split(url,'/+|\?|%%7C|&')
Regexp meaning:
/+ - one or more slash
| - OR
\? - literally ?, needs shielding by backslash because ? has special meaning in regexp
| - OR
%%7C - literally %7C. % has special meaning in Presto CLI and presto is using double-character notation for shielding.
| - OR
& - literally &
After you got an ARRAY, you can concatenate it or explode, etc
Related
We have a problem with a regular expression on hive.
We need to exclude the numbers with +37 or 0037 at the beginning of the record (it could be a false result on the regex like) and without letters or space.
We're trying with this one:
regexp_like(tel_number,'^\+37|^0037+[a-zA-ZÀÈÌÒÙ ]')
but it doesn't work.
Edit: we want it to come out from the select as true (correct number) or false.
To exclude numbers which start with +01 0r +001 or +0001 and having only digits without spaces or letters:
... WHERE tel_number NOT rlike '^\\+0{1,3}1\\d+$'
Special characters like + and character classes like \d in Hive should be escaped using double-slash: \\+ and \\d.
The general question is, if you want to describe a malformed telephone number in your regex and exclude everything that matches the pattern or if you want to describe a well-formed telephone number and include everything that matches the pattern.
Which way to go, depends on your scenario. From what I understand of your requirements, adding "not starting with 0037 or +37" as a condition to a well-formed telephone number could be a good approach.
The pattern would be like this:
Your number can start with either + or 00: ^(\+|00)
It cannot be followed by a 37 which in regex can be expressed by the following set of alternatives:
a. It is followed first by a 3 then by anything but 7: 3[0-689]
b. It is followed first by anything but 3 then by any number: [0-24-9]\d
After that there is a sequence of numbers of undefined length (at least one) until the end of the string: \d+$
Putting everything together:
^(\+|00)(3[0-689]|[0-24-9]\d)\d+$
You can play with this regex here and see if this fits your needs: https://regex101.com/r/KK5rjE/3
Note: as leftjoin has pointed out: To use this regex in hive you might need to additionally escape the backslashes \ in the pattern.
You can use
regexp_like(tel_number,'^(?!\\+37|0037)\\+?\\d+$')
See the regex demo. Details:
^ - start of string
(?!\+37|0037) - a negative lookahead that fails the match if there is +37 or 0037 immediately to the right of the current location
\+? - an optional + sign
\d+ - one or more digits
$ - end of string.
Hi I tried using Regex_replace and it is still not working.
select CASE WHEN sbbb <> ' ' THEN regexp_replace(sbbb,'[a-zA-Z _-#]','']
ELSE sbbb
AS ABCDF
from Table where sccc=1;
This is the query which I am using to remove alphabets and specials characters from string and have only numbers. but it doesnot work. Query returns me the complete string with numbers,characters and special characters .What is wrong in the above query
I am working on a sql query. There is a column in database which contains characters,special characters and numbers. I want to only keep the numbers and remove all the special characters and alphabets. How can I do it in query of DB2. If a use PATINDEX it is not working. please help here.
The allowed regular expression patterns are listed on this page
Regular expression control characters
Outside of a set, the following must be preceded with a backslash to be treated as a literal
* ? + [ ( ) { } ^ $ | \ . /
Inside a set, the follow must be preceded with a backslash to be treated as a literal
Characters that must be quoted to be treated as literals are [ ] \
Characters that might need to be quoted, depending on the context are - &
So for you, this should work
regexp_replace(sbbb,'[a-zA-Z _\-#]','')
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.
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 have a simple search query:
<cfquery name="_qSearch" dbtype="Query">
SELECT
*
FROM MyQoQ
WHERE
DESCRIPTION LIKE '%#URL.searchString#%'
</cfquery>
This query works excellently for most values. However, if someone searches for a value like "xxx[en", it bombs with the error message The pattern of the LIKE conditional is malformed..
Is there any way around this, since the bracket has a special use in CFQUERY?
QoQ shares a feature of TSQL (MS SQL Server) whereby it's not just % and _ that are wildcards in LIKE - it also supports regex-style character classes, as in[a-z] for any lowercase letter.
To escape these values and match the literal equivalents, you can use a character class itself, i.e. [[] will match a literal [, and of course you probably also want to escape any % and _ in the user input - you can do all three like so:
'%#Url.SearchString.replaceAll('[\[%_]','[$0]')#%'
That is just a simple regex replace (using String.replaceAll) to match all instances of [ or % or _ and wrap each one in [..] - the $0 on the replacement side represents the matched text.