please can you help:
this is my string (the string length differs):
WWWWWWWWW/XXXXXXXXXX/YYYYYYYYRTDZZZZK/123/CQW/321/AAAAAAAAAA
What i want is to capture 123 (or whatever number is between the / /)
I tried this:
substr(MSG_TEXT,instr(MSG_TEXT, '/K')+2,INSTR(MSG_TEXT,'/CQW')-instr(msg_text,'/K')-2)
which does works, if there is no more K/ before it, somtimes there is.
so it didn't like this:
WWWWWK/WWWW/XXXXXXXXXX/YYYYYYYYRTDZZZZK/123/CQW/321/AAAAAAAAAA
I want it to start looking from RTD onwards, as it will be the first /K it finds
so I've tried this (trying to start about 50 characters in before RTD but after the first K/):
case when substr(msg_text,50,200) like '%RTD%' then substr(substr(MSG_TEXT,instr(MSG_TEXT, '/K')+2,INSTR(MSG_TEXT,'/CQW')-instr(msg_text,'/K')-2),50,200) end
but it returned null!
I know i'm on the right track, appreciate your assistance.
Related
It's a weird problem
to_be_stripped="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120"
And two strings below:
s1="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120\\[Content_Types].xml"
s2="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120\\_rels\.rels"
When I use the command below:
s1.strip(to_be_stripped)
s2.strip(to_be_stripped)
I get these outputs:
'[Content_Types].x'
'_rels\\.'
If I use lstrip(), they will be:
'[Content_Types].xml'
'_rels\\.rels'
Which is the right outputs.
However, if we replace all Project Known with zeus_pipeline:
to_be_stripped="D:\\Users\\UserKnown\\PycharmProjects\\zeus_pipeline\\PT\\collections\\120"
And:
s2="D:\\Users\\UserKnown\\PycharmProjects\\zeus_pipeline\\PT\\collections\\120\\_rels\.rels"
s2.lstrip(to_be_stripped)will be '.rels'
If I use / instead of \\, nothing goes wrong. I am wondering why this problem happens.
strip isn't meant to remove full strings exactly. Rather, you give it a string, and every character in that string is removed from the start and of the string to be stripped.
In your case, the variable to_be_stripped contains the characters m and l, so those are stripped from the end of s1. However, it doesn't contain the character x, so the stripping stops there and no characters beyond that are removed.
Check out this question. The accepted answer is probably more extensive than you need - I like another user's suggestion of using replace instead of strip. This would look like:
s1.replace(to_be_stripped, "")
I understand that the string_variable(start:length) can be used to get a substring of a string given a starting point and substring length, however, I am finding that I often need to get a substring between a 'start' and 'end' point.
While I know I could always do this:
SUBTRACT start FROM end GIVING len
string(start:len)
It seems cumbersome to have to do so every time when I am writing programs that use this functionality often. Is there perhaps a quicker/built-in way of achieving this?
How about?
move str (start-pos : end-pos - start-pos + 1) to ...
You can subtract the first from the last, but you need to add 1 to get the correct length.
STRING is a statement name, as is START, and END is reserved. LENGTH is a function name. I avoid those in anything that looks like code.
I want to remove all the characters from a string expect whatever character is between a certain set of characters. So for example I have the input of Grade:2/2014-2015 and I want the output of just the grade, 2.
I'm thinking that I need to use the FIND function to grab whatever is between the : and the / , this also needs to work with double characters such 10 however I believe that it would work so long as the defining values with the FIND function are correct.
Unfortunately I am totally lost on this when using the FIND function however if there is another function that would work better I could probably figure it out myself if I knew what function.
It's not particularly elegant but =MID(A1,FIND(":",A1)+1,FIND("/",A1) - FIND(":",A1) - 1) would work.
MID takes start and length,FIND returns the index of a given character.
Edit:
As pointed out, "Grade:" is fixed length so the following would work just as well:
=MID(A1,7,FIND("/",A1) - 7)
You could use LEFT() to remove "Grade:"
And then use and then use LEFTB() to remove the year.
Look at this link here. This is the way I would go about it.
=SUBSTITUTE(SUBSTITUTE(C4, "Grade:", ""), "/2014-2015", "")
where C4 is the name of your cell.
I just wrote a code with the criteria above, but it doesn't seem to work properly because I either miss a letter at the end or in the middle.
Could anyone please check out my code an tell me what I'm doing wrong. By the way I already checked other threads on this similar problem, but I'm not allowed to use regex or print function.
phrase=('my room is cold')
allSpaces=findstr(' ',phrase);
k=length(allSpaces)
acr=phrase(1:allSpaces(1):allSpaces(k)-1)
Output:
acr= mrms
Change last line to
acr = phrase([1 allSpaces+1])
That way you get the first letter, and then the first after each space.
I have a huge string. I need to extract a substring from that that huge string. The conditions are the string starts with either "TECHNICAL" or "JUSTIFY" or "ALIGN" and ends with a number( any number from 1 to 10) followed by period and then followed by space. so for example, I have
string x = "This is a test, again I am testing TECHNICAL: I need to extract this substring starting with testing. 8. This is test again and again and again and again.";
so I need this
TECHNICAL: I need to extract this substring starting with testing.
I was wondering if someone has elegant solution for that.
I was trying to use the regular expression, but I guess I could not figure out the right expresion.
any help will be appreciated.
Thanks in advance.
Try this: #"((?:TECHNICAL|JUSTIFY|ALIGN).*?)(?:[1-9]|10)\. "