I'm having an issue getting the correct text between characters. I'm currently trying to use SUBSTRING and CHARINDEX. The issue is that the data contains the same identifier. I would like the text that falls between the first '\' and the second '\', removing all the other text. None of the text fields have a fixed number of characters.
Thank you for your help its greatly appreciated.
Example
1. Location\Georgia\Atlanta
2. Country\USA\States\Minnesota
Final Result
1. Georgia
2. USA
My Current Attempt
SUBSTRING(Source,CHARINDEX('\',Source),CHARINDEX('\',Source) - CHARINDEX('\',Source))
I agree with Joel's comment. This does look like a violation of first normal form but to find the second instance of \ you can pass the location of the first one in as the third argument to CHARINDEX
WITH T(Source) AS
(
SELECT 'Location\Georgia\Atlanta' UNION ALL
SELECT 'Country\USA\States\Minnesota'
)
SELECT *,
SUBSTRING(Source, F,CHARINDEX('\',Source, F) - F)
FROM T
CROSS APPLY (SELECT 1 + CHARINDEX('\',Source)) CA(F)
WHERE Source LIKE '%\%\%'
Martin's answer is slick, this method works, but mostly just illustrates how slick his is:
WITH T(Source) AS
(
SELECT 'Location\Georgia\Atlanta' UNION
SELECT 'Country\USA\States\Minnesota'
)
SELECT *, SUBSTRING(Source,CHARINDEX('\',Source)+1,CHARINDEX('\',Source,CHARINDEX('\',Source)+1)-CHARINDEX('\',Source)-1)
FROM T
Related
Given this data:
Home: (708) 296-2112
I want everything to the right of the : character.
This is what I have so far, but I'm getting no results:
right(phone1, locate(':', phone1 + ':')-1) phone
If I use left instead of right, I get just "HOME" - just for testing purposes. I know I'm close, but I'm missing something.
You can use SUBSTRING (might be SUBSTR dependent on your version) instead:
SELECT SUBSTRING(phone1, LOCATE(':', phone1) + 1, LENGTH(phone1))
FROM yourtable
Here's a way to do it without hard-coding in Home:, so you can also use Office: or Mobile: or Fax:, or any other word followed by a colon.
This uses ADS's scripting ability to use a variable and the built-in System.iota single row table (similar to Oracle's dual). You can just use the last line, replacing test with the name of your column and system.iota with the name of your table.
declare test string;
set test = 'Home: (708) 296-2112';
select substring(test, position(':' in test) + 1, length(test)) from system.iota;
You were on the right track, but your algebra is off. You want to take the full length of the string offset by the position of the colon, minus one:
right(phone1, length(phone1) - locate(':', phone1) - 1)
You can use RIGHT function as follows:
RIGHT(phone1, LEN(phone1)-CHARINDEX(':', phone1))
I have a number of rows in a database column that look like this
AB01-52
AB01-52/1
AB01-53/2
AB01-54
I can get the 52 where there is no slash in the string but when I try to get it between the dash and the slash I either get an error or get the 52/1 for example
I have tried most of the researched solutions using substring and charindex but to no avail
Has anyone seen a solution for this.
Thanks in advance
Rick
You can subtract the characters positions (i.e. /) and use as length for substring()
substring(col, charindex('-', col)+1, len(col)-charindex('/', col)+1)
If the numbers have variable length then do the subtraction from both characters and use of length
substring(col, charindex('-', col)+1, charindex('/', col+'/')-charindex('-', col)-1)
Here's one approach....
with cte as(
select 'AB01-53/2' as C1
union
select 'AB01-54')
select left(right(c1,len(c1) - charindex('-',c1)),len(right(c1,len(c1) - charindex('-',c1))) - charindex('/',right(c1,len(c1) - charindex('-',c1)))+ 1)
from cte
If the field is ALWAYS AAAA-BB[/CCC], then simply:
SUNSTRING('AB01-52/1',6,2) ;
Of course, the 'AB01-52/1' may be substituted by a variable or column name.
I may have found an answer but it involves a 2 stage process
First add an additional slash to the column that contains the data where there is no slash at the moment ie AB01-52 becomes AB01-52/ the others like AB01-1023/01 remain the same
stage 2 use this code:
substring(right(col1,len(col1)-charindex('-',col1)),0,charindex('/',right(col1,len(col1)-charindex('-',col1))))
this allows me to get the middle string. The use of left and right by various people who contributed helped to get to this - however I still thing there should be an answer without stage 1
Thanks you all for your ideas which got me to this point
I have code that looks like this:
left outer join
gme_batch_header bh
on
substr(ln.lot_number,instr(ln.lot_number,'(') + 1,
instr(ln.lot_number,')') - instr(ln.lot_number,'(') - 1)
=
bh.batch_no
It works fine, but I have come across a few lot numbers that have two sections of strings that are between parenthesis. How would I compare what is between the second set of parenthesis? Here is an example of the data in the lot number field:
E142059-307-SCRAP-(74055)
This one works with the code,
58LF-3-B-2-2-2 (SCRAP)-(61448)
This one tries comparing SCRAP with the batch no, which isn't correct. It needs to be the 61448.
The result is always the last item in parenthesis.
After more research, I actually got it to work with this code:
substr(ln.lot_number,instr(ln.lot_number,'(',-1) + 1, instr(ln.lot_number,')',-1) - instr(ln.lot_number,'(',-1) - 1)
Assuming SQL2005+, and it is always the last occurrence you want, then I would suggest finding the last instance of a ( in your query and substring to there. To get the last instance you could use something like:
REVERSE(SUBSTRING(REVERSE(lot_number),0,CHARINDEX('(',REVERSE(lot_number))))
If your version of Oracle supports regular expressions try this:
substr(regexp_substr(ln.lot_number,'[0-9]+\)$'),1,length(regexp_substr(ln.lot_number,'[0-9]+\)$'))-1)
Explanation:
regexp_substr(scrap_row,'[0-9]+\)$' ==> find me just numbers in the string that ends in ). This returns the numbers but it includes the closing parenthesis.
To remove the closing parenthsis, just send it through substring and extract first number through the length of the number stopping at 1 character from the end of the string.
Query for analysis:
with scrap
as (select '58LF-3-B-2-2-2 (SCRAP)-(61448)' as scrap_row from dual)
select scrap_row,
regexp_substr(scrap_row,'[0-9]+\)$') as regex_substring,
length(regexp_substr(scrap_row,'[0-9]+\)$')) as length_regex_substring,
substr(regexp_substr(scrap_row,'[0-9]+\)$'),1,length(regexp_substr(scrap_row,'[0-9]+\)$'))-1) as regex_sans_parenthesis
from scrap
If you have 11g, this will do it pretty simply by using the subgroup argument of regexp_substr() and constructing the regex appropriately:
SQL> with tbl(data) as
(
select 'E142059-307-SCRAP-(74055)' from dual
union
select '58LF-3-B-2-2-2 (SCRAP)-(61448)' from dual
)
select data from tbl
where regexp_substr(data, '\((\d+)\)$', 1, 1, NULL, 1)
= '61448';
DATA
------------------------------
58LF-3-B-2-2-2 (SCRAP)-(61448)
The regular expression can be read as:
\( - Search for a literal left paren
( - Start a remembered subgroup
\d+ - followed by 1 more more digits
) - End remembered subgroup
\) - followed by a literal right paren
$ - at the end of the line.
The regexp_substr function arguments are:
Source - the source string
Pattern - The regex pattern to look for
position - Position in the string to start looking for the pattern
occurrence - If the pattern occurs multiple times, which occurrence you want
match_params - See the docs, not used here
subexpression - which subexpression to use (the remembered group)
So in English, look for a series of 1 or more digits surrounded by parens, where it occurs at the end of the line and save the digit part only to use to compare. IMHO a lot easier to follow/maintain than nested instr(), substr().
For re-useability, make a function called get_last_number_in_parens() that contains this code and uses an argument of the string to search. This way that logic is encapsulated and can be re-used by folks that may not be so comfortable with regular expressions, but can benefit from the power! One place to maintain code too. Then call like this:
select data from tbl
where get_last_number_in_parens(data) = '61448';
How easy is that?!
Hello you can check with this code. It works whaever the condition may be
SELECT SUBSTR('58LF-3-B-2-2-2-(61448)',instr('58LF-3-B-2-2-2-(61448)','(',-1)+1,LENGTH('58LF-3-B-2-2-2-(61448)')-instr('58LF-3-B-2-2-2-(61448)','(',-1)-1)
FROM dual;
SELECT SUBSTR('58LF-3-B-2-2-2 (SCRAP)-(61448)',instr('58LF-3-B-2-2-2 (SCRAP)-(61448)','(',-1)+1,LENGTH('58LF-3-B-2-2-2 (SCRAP)-(61448)')-instr('58LF-3-B-2-2-2 (SCRAP)-(61448)','(',-1)-1)
FROM dual;
Output
==================================
61448
==================================
How can I find the first word and second word in a string separated by unknown number of spaces in SQL Developer? I need to run a query to get the expected result.
String:
Hello Monkey this is me
Different sentences have different number of spaces between the first and second word and I need a generic query to get the result.
Expected Result:
Hello
Monkey
I have managed to find the first word using substr and instr. However, I do not know how to find the second word due to the unknown number of spaces between the first and second word.
select substr((select ltrim(sentence) from table1),1,
(select (instr((select ltrim(sentence) from table1),' ',1,1)-1)
from table1))
from table1
Since you seem to want them as separate result rows, you could use a simple common table expression to duplicate the rows, once with the full row, then with the first word removed. Then all you have to do is get the first word from each;
WITH cte AS (
SELECT value FROM table1
UNION ALL
SELECT SUBSTR(TRIM(value), INSTR(TRIM(value), ' ')) FROM table1
)
SELECT SUBSTR(TRIM(value), 1, INSTR(TRIM(value), ' ') -1) word
FROM cte
Note that this very simple example assumes that there is a second word, if there isn't, NULL will be returned for both words.
An SQLfiddle to test with.
While Joachim Isaksson's answer is a robust and fast approach, you can also consider splitting the string and selecting from the resulting pieces set. This is just meant as hint for another approach, if your requirements alter (e.g. more than two string pieces).
You could split finally by the regex /[ ]+/, and so getting the words between the blanks.
Find more about splitting here: How do I split a string so I can access item x?
This will strongly depend on the SQL dialect you are using.
Try this with REGEXP_SUBSTR:
SELECT
REGEXP_SUBSTR(sentence,'\w+\s+'),
REGEXP_SUBSTR(sentence,'\s+(\w+)\s'),
REGEXP_SUBSTR(sentence,'\s+(\w+)\s+(\w+)'),
REGEXP_SUBSTR（REGEXP_SUBSTR(sentence,'\s+(\w+)\s+(\w+)'),'\w+$'),
REGEXP_SUBSTR(sentence,'\s+(\w+)\s+$')
FROM table1;
result:
1 2 3 4 5
Hello Monkey Monkey this this is_me
Learn more about REGEXP_SUBSTR reference to Using Regular Expressions With Oracle Database
Test use SqlFiddle： http://sqlfiddle.com/#!4/8e9ef/9
If you only want to get the first and the second word, use REGEXP_INSTR to get second word start position :
SELECT
REGEXP_SUBSTR(sentence,'\w+\s+') AS FIRST,
REGEXP_SUBSTR(sentence,'\w+\s',REGEXP_INSTR(sentence,'\w+\s+')+length(REGEXP_SUBSTR(sentence,'\w+\s+'))) AS SECOND
FROM table1;
I know I'm close to figuring this out but need a little help. What I'm trying to do is all grab a column from a particular table, but chop off the first 4 characters. For example if in a column the value is "KPIT08L", the result I was is 08L. Here is what I have so far but not getting the desired results.
SELECT LEFT(FIELD_NAME, 4)
FROM TABLE_NAME
First up, left will give you the leftmost characters. If you want the characters starting at a specific location, you need to look into mid:
select mid (field_name,5) ...
Secondly, if you value performance,portability and scalability at all, this sort of "sub-column" manipulation should generally be avoided. It's usually far easier (and faster) to patch columns together than to split them apart.
In other words, keep the first four characters in their own column and the rest in a separate column, and do your selects on the relevant one. If you're using anything less than a full column, then it's technically not one attribute of the row.
Try with
SELECT MID(FIELD_NAME, 5) FROM TABLE_NAME
Mid is very powerfull, it let you select the starting point and all the remainder, or,
if specified, the length desidered as in
SELECT MID(FIELD_NAME, 5, 2) FROM TABLE_NAME ' gives 08 in your example text
SELECT RIGHT(FIELD_NAME,LEN(FIELD_NAME)-4)
FROM TABLE_NAME;
If it is for a generic string then the above one will work...
Don't have Access at my current location, but please try this.
SELECT RIGHT(FIELD_NAME, LEN(FIELD_NAME)-4)
FROM TABLE_NAME
The LEFT(FIELD_NAME, 4) will return the first 4 caracters of FIELD_NAME.
What you need to do is :
SELECT MID(FIELD_NAME, 5)
FROM TABLE_NAME
If you have a FIELD_NAME of 10 caracters, the function will return the 6 last caracters (chopping the first 4)!