According to the Atlassian documentation, you can search for words contained in a text field like so:
description ~ term
It also indicates you can group terms like so:
description ~ (term OR "different thing")
But when I try that, it shows that the syntax is wrong.
I want to do this, but with grouping (because I will have more than 2 terms, and I'd like to keep the query easy to read and short-ish):
description ~ term OR description ~ "different thing"
description doesn’t support lists...
description doesn’t support = operator....
I think you already have the best answer to your question with JQL, by using....
description ~ “term” OR description ~ “different thing”
Although, a list method would work for the priority field, the list option isn’t available for every field...
This appears to do what you want:
description ~ ("term OR different thing")
Actually the multi-text match search phrase that worked for me was this example:
text ~ "(uc55 OR uc60 OR UC65)"
Related
I've looked at lots of examples for TRIM and REPLACE on the internet and for some reason I keep getting errors when I try.
I need to strip suffixes from my Netsuite item record names in a saved item search. There are three possible suffixes: -T, -D, -S. So I need to turn 24335-D into 24335, and 24335-S into 24335, and 24335-T into 24335.
Here's what I've tried and the errors I get:
Can you help me please? Note: I can't assume a specific character length of the starting string.
Use case: We already have a field on item records called Nickname with the suffixes stripped. But I've ran into cases where Nickname is incorrect compared to Name. Ex: Name is 24335-D but Nickname is 24331-D. I'm trying to build a saved search alert that tells me any time the Nickname does not equal suffix-stripped Name.
PS: is there anywhere I can pay for quick a la carte Netsuite saved search questions like this? I feel bad relying on free technical internet advice but I greatly appreciate any help you can give me!
You are including too much SQL - a formulae is like a single result field expression not a full statement so no FROM or AS. There is another place to set the result column/field name. One option here is Regex_replace().
REGEXP_REPLACE({name},'\-[TDS]$', '')
Regex meaning:
\- : a literal -
[TDS] : one of T D or S
$ : end of line/string
To compare fields a Formulae (Numeric) using a CASE statement can be useful as it makes it easy to compare the result to a number in a filter. A simple equal to 1 for example.
CASE WHEN {custitem_nickname} <> REGEXP_REPLACE({name},'\-[TDS]$', '') then 1 else 0 end
You are getting an error because TRIM can trim only one character : see oracle doc
https://docs.oracle.com/javadb/10.8.3.0/ref/rreftrimfunc.html (last example).
So try using something like this
TRIM(TRAILING '-' FROM TRIM(TRAILING 'D' FROM {entityid}))
And always keep in mind that saved searches are running as Oracle SQL queries so Oracle SQL documentation can help you understand how to use the available functions.
For my database, I have a list of company numbers where some of them start with two letters. I have created a regex which should eliminate these from a query and according to my tests, it should. But when executed, the result still contains the numbers with letters.
Here is my regex, which I've tested on https://www.regexpal.com
([^A-Z+|a-z+].*)
I've tested it against numerous variations such as SC08093, ZC000191 and NI232312 which shouldn't match and don't in the tests, which is fine.
My sql query looks like;
SELECT companyNumber FROM company_data
WHERE companyNumber ~ '([^A-Z+|a-z+].*)' order by companyNumber desc
To summerise, strings like SC08093 should not match as they start with letters.
I've read through the documentation for postgres but I couldn't seem to find anything regarding this. I'm not sure what I'm missing here. Thanks.
The ~ '([^A-Z+|a-z+].*)' does not work because this is a [^A-Z+|a-z+].* regex matching operation that returns true even upon a partial match (regex matching operation does not require full string match, and thus the pattern can match anywhere in the string). [^A-Z+|a-z+].* matches a letter from A to Z, +,|or a letter fromatoz`, and then any amount of any zero or more chars, anywhere inside a string.
You may use
WHERE companyNumber NOT SIMILAR TO '[A-Za-z]{2}%'
See the online demo
Here, NOT SIMILAR TO returns the inverse result of the SIMILAR TO operation. This SIMILAR TO operator accepts patterns that are almost regex patterns, but are also like regular wildcard patterns. NOT SIMILAR TO '[A-Za-z]{2}%' means all records that start with two ASCII letters ([A-Za-z]{2}) and having anything after (%) are NOT returned and all others will be returned. Note that SIMILAR TO requires a full string match, same as LIKE.
Your pattern: [^A-Z+|a-z+].* means "a string where at least some characters are not A-Z" - to extend that to the whole string you would need to use an anchored regex as shown by S-Man (the group defined with (..) isn't really necessary btw)
I would probably use a regex that specifies want the valid pattern is and then use !~ instead.
where company !~ '^[0-9].*$'
^[0-9].*$ means "only consists of numbers" and the !~ means "does not match"
or
where not (company ~ '^[0-9].*$')
Not start with a letter could be done with
WHERE company ~ '^[^A-Za-z].*'
demo: db<>fiddle
The first ^ marks the beginning. The [^A-Za-z] says "no letter" (including small and capital letters).
Edit: Changed [A-z] into the more precise [A-Za-z] (Why is this regex allowing a caret?)
Is it possible to conditionally replace parts of strings in MySQL?
Introduction to a problem: Users in my database stored articles (table called "table", column "value", each row = one article) with wrong links to images. I'd like to repair all of them at once. To do that, I have to replace all of the addresses in "href" links that are followed by images, i.e.,
<img src="link2">
should by replaced by
<img src="link2">
My idea is to search for each "href" tag and if the tag is followed by and "img", than I'd like to obtain "link2" from the image and use it replace "link1".
I know how to do it in bash or python but I do not have enough experience with MySQL.
To be specific, my table contains references to images like
<a href="www.a.cz/b/c"><img class="image image-thumbnail " src="www.d.cz/e/f.jpg" ...
I'd like to replace the first adress (href) by the image link. To get
<a href="www.d.cz/e/f.jpg"><img class="image image-thumbnail " src="www.d.cz/e/f.jpg" ...
Is it possible to make a query (queries?) like
UPDATE `table`
SET value = REPLACE(value, 'www.a.cz/b/c', 'XXX')
WHERE `value` LIKE '%www.a.cz/b/c%'
where XXX differs every time and its value is obtained from the database? Moreover, "www.a.cz/b/c" varies.
To make things complicated, not all of the images have the "href" link and not all of the links refer to images. There are three possibilities:
"href" followed by "img" -> replace
"href" not followed by "img" -> keep original link (probably a link to another page)
"img" without "href" -> do nothing (there is no wrong link to replace)
Of course, some of the images may have a correct link. In this case it may be also replaced (original and new will be the same).
Database info from phpMyAdmin
Software: MariaDB
Software version: 10.1.32-MariaDB - Source distribution
Protocol version: 10
Server charset: UTF-8 Unicode (utf8)
Apache
Database client version: libmysql - 5.6.15
PHP extension: mysqli
Thank you in advance
SELECT
regexp_replace(
value,
'^<a href="([^"]+)"><img class="([^"]+)" src="([^"]+)"(.*)$',
'<a href="\\3"><img class="\\2" src="\\3"\\4'
)
FROM
yourTable
The replacement only happens if the pattern is matched.
^ at the start means start of the string
([^"]+) means one of more characters, excluding "
(.*) means zero or more of any character
$ at the end means end of the string
The replacement takes the 3rd "pattern enclosed in braces" (back-reference) and puts it where the 1st "pattern enclosed in braces" (back-reference) was.
The 2nd, 3rd and 4th back-references are replaced with themselves (no change).
https://dbfiddle.uk/?rdbms=mariadb_10.2&fiddle=96aef2214f844a1466772f41415617e5
If you have strings that don't exactly match the pattern, it will do nothing. Extra spaces will trip it up, for example.
In which case you need to work out a new regular expression that always matches all of the strings you want to work on. Then you can use the \\n back-references to make replacements.
For example, the following deals with extra spaces in the href tag...
SELECT
regexp_replace(
value,
'^<a[ ]+href[ ]*=[ ]*"([^"]+)"><img class="([^"]+)" src="([^"]+)"(.*)$',
'<a href="\\3"><img class="\\2" src="\\3"\\4'
)
FROM
yourTable
EDIT:
Following comments clarifying that these are actually snippets from the MIDDLE of the string...
https://dbfiddle.uk/?rdbms=mariadb_10.2&fiddle=48ce1cc3df5bf4d3d140025b662072a7
UPDATE
yourTable
SET
value = REGEXP_REPLACE(
value,
'<a href="([^"]+)"><img class="([^"]+)" src="([^"]+)"',
'<a href="\\3"><img class="\\2" src="\\3"'
)
WHERE
value REGEXP '<a href="([^"]+)"><img class="([^"]+)" src="([^"]+)"'
(Though I prefer the syntax RLIKE, it's functionally identical.)
This will also find an replace that pattern multiple times. You're not clear if that's desired or possible.
Solved, thanks to #MatBailie , but I had to modified his answer. The final query, including the update, is
UPDATE `table`
SET value = REGEXP_REPLACE(value, '(.*)<a href="([^"]+)"><img class="([^"]+)" src="([^"]+)"(.*)', '\\1<a href="\\4"><img class="\\3" src="\\4"\\5'
)
A wildcard (.*) had to be put at the beginning of the search because the link is included in an article (long text) and, consequently, the arguments of the replace pattern are increased.
I'm using a manual verification point - it works well - but when it appears in the log its name -or title- appears empty, what should i do to make the manual VP's title appears in the log, thanks in advance
Abed.
The syntax is
vpManual("Your VP name here", expectedData, actualData).performTest();
The VP name should not contain spaces, dots or other "strange" characters, keep on letters, digits and underscore _
If you use a string variable for the name, check it is not null or empty.
Also note the name must be unique in the script. If you provide a little example of your problem I can try answering with more details.
Take a look at the RFT docs
I am having an issue using special characters. I am parsing a text file separated by tabs. I want to have the program add a "*" to the first word in the line if a certain parameter is true.
if ($Var < $3) $1 = \*$1
Now every time I run it I get the error that it is not the end of the line.
2 things, but without more context to test with we really can't help you much.
$Var will only have meaning if you have set it above like Var=3. Then I don't think gawk will evaluate your $3 to the value of $3. The other side of that expression < $3 WILL expand to the value of the 3rd field. If you're getting $Var from the shell environment, you need to let the gawk script 'see' that value, i.e.
awk '{ ..... if ('"$Var"' < $3) $1= "*" $1 .....}
If you want the string literal '*' pre-pended, you're better off doing $1 = "*" $1
Without sample inputs, sample expected output, actual output and error messages, we'll be playing 20 questions here. If these comments don't solve your problem, please edit your question above to include these items.
P.S. Welcome to StackOverflow and let me remind you of three things we usually do here: 1) As you receive help, try to give it too, answering questions in your area of expertise 2) Read the FAQs, http://tinyurl.com/2vycnvr , 3) When you see good Q&A, vote them up by using the gray triangles, http://i.imgur.com/kygEP.png , as the credibility of the system is based on the reputation that users gain by sharing their knowledge. Also remember to accept the answer that better solves your problem, if any, by pressing the checkmark sign , http://i.imgur.com/uqJeW.png