I want to use REGEXP_INSTR function in a query to search for any match for user input but I don't know how to write the regular expression that for example will match any value that includes the word car followed by unspecified numbers of letters/numbers/spaces and then the word Paterson. can any one please help me with writing this regEx?
Ok, so let's break this down.
"any value that includes the word car"
I surmise from this that the word car doesn't need to be at the start of the string, therefore i would start the format string with...
'^.*'
Here the '^' character means the start of the string, the '.' means any character and '*' means 0 or more of the preceding character. So zero or more of any character after the start of the string.
Then the word 'car', so...
'^.*car'
Next up...
"followed by unspecified numbers of letters/numbers/spaces"
I'm guessing that unspecified means zero or more. This is very similar to what we did to identify any characters that might come before 'car'. Where the '.' means any character
'^.*car.*'
However, if unspecified means one or more, then you can use '+' in place of '*'
"then the word Paterson"
I'm going to assume that as this is the end of the description, there are no more characters after 'Paterson'.
'^.*car.*Paterson$'
The '$' symbol means that the 'n' of 'Paterson' must be at the end of the string.
Code example:
select
REGEXP_INSTR('123456car1234ABCDPaterson', '^.*car.*Paterson$') as rgx
from dual
Output
RGX
----------
1
Related
I need to find rows where the phone number field contains unexpected characters.
Most of the values in this field look like:
123456-7890
This is expected. However, we are also seeing character values in this field such as * and #.
I want to find all rows where these unexpected character values exist.
Expected:
Numbers are expected
Hyphen with numbers is expected (hyphen alone is not)
NULL is expected
Empty is expected
Tried this:
WHERE phone_num is not like ' %[0-9,-,' ' ]%
Still getting rows where phone has numbers.
from https://regexr.com/3c53v address you can edit regex to match your needs.
I am going to use example regex for this purpose
select * from Table1
Where NOT REGEXP_LIKE(PhoneNumberColumn, '^[+]*[(]{0,1}[0-9]{1,4}[)]{0,1}[-\s\./0-9]*$')
You can use translate()
...
WHERE translate(Phone_Number,'a1234567890-', 'a') is NOT NULL
This will strip out all valid characters leaving behind the invalid ones. If all the characters are valid, the result would be NULL. This does not validate the format, for that you'd need to use REGEXP_LIKE or something similar.
You can use regexp_like().
...
WHERE regexp_like(phone_num, '[^ 0123456789-]|^-|-$')
[^ 0123456789-] matches any character that is not a space nor a digit nor a hyphen. ^- matches a hyphen at the beginning and -$ on the end of the string. The pipes are "ors" i.e. a|b matches if pattern a matches of if pattern b matches.
Oracle has REGEXP_LIKE for regex compares:
WHERE REGEXP_LIKE(phone_num,'[^0-9''\-]')
If you're unfamiliar with regular expressions, there are plenty of good sites to help you build them. I like this one
I have a requirement where the persons can have more than one first name and i need to convert the name into first characters from the 2 or more names with capital letters.
Examples:
Srinivas Kalyan ,Sai Kishore
if the above is one having 4 first names then i need to display as below
S K S K
The comma is also be replaced and take all the first characters
2. Srinivas-Kalyan Sai Kishore
For the above name the value should be as
S S K
since Srinivas-Kalyan is considered as one name
Also the name can be in small letters
srinivas kalyan sai kishore
For this
S K S K
This has to be in oracle
Tried the below regex_replace which is working fine in sql developer but in the application it is changing into space
replace(trim(regexp_replace(to_char(regexp_replace(initcap(regexp_replace(regexp_replace
(FIRST_NAME, '[0-9]', ''), '( *[[:punct:]])', '')), '([[:lower:]]| )')), '(.)', '\1 ')),',',null)
The missing letter in your output is caused by this regex for character removal:
( *[[:punct:]])
This will turn Kalyan ,Sai into KalyanSai, which will be treated as one word by the rest of the process, and so you will not have the S of Sai in your output.
I would suggest this shorter expression:
trim(upper(regexp_replace(
regexp_replace(first_name, '([[:alpha:]])(-|[[:alpha:]])+', '\1'),
'.*?([[:alpha:]]|$)', ' \1'
)))
Explanation
([[:alpha:]])(-|[[:alpha:]])+ -> \1
This replaces any sequence of letters (or hyphen) with the first of those. So it reduces words to their initial letter.
.*?([[:alpha:]]|$) -> (space)\1
This replaces anything that precedes the next letter, with a space. As no letter will be skipped (because .*? is non-greedy) this effectively replaces all non-letter sequences with a space. To also replace the non-letters at the very end (which do not precede a letter), the special case $ (end-of-string) is added as alternative.
After these two steps there just remains to:
Upper case everything
Remove blanks at the start and end of the result with trim
I find the advantage of this method is that it does not use any other class than [[:alpha]]. Digits, punctuation, lowercase -- or whatever else -- does not need to be identified explicitly, as it is just the negation of [[:alpha]]. The only exception that has to be made, is for the hyphen.
As some names might include some other non-letters, like a quote, you might want to add such characters as well in the first regular expression.
I have a questions regarding below data.
You clearly can see each EMP_IDENTIFIER has connected with EMP_ID.
So I need to pull only identifier which is 10 characters that will insert another column.
How would I do that?
I did some traditional way, using INSTR, SUBSTR.
I just want to know is there any other way to do it but not using INSTR, SUBSTR.
EMP_ID(VARCHAR2)EMP_IDENTIFIER(VARCHAR2)
62049 62049-2162400111
6394 6394-1368000222
64473 64473-1814702333
61598 61598-0876000444
57452 57452-0336503555
5842 5842-0000070666
75778 75778-0955501777
76021 76021-0546004888
76274 76274-0000454999
73910 73910-0574500122
I am using Oracle 11g.
If you want the second part of the identifier and it is always 10 characters:
select t.*, substr(emp_identifier, -10) as secondpart
from t;
Here is one way:
REGEXP_SUBSTR (EMP_IDENTIFIER, '-(.{10})',1,1,null,1)
That will give the 1st 10 character string that follows a dash ("-") in your string. Thanks to mathguy for the improvement.
Beyond that, you'll have to provide more details on the exact logic for picking out the identifier you want.
Since apparently this is for learning purposes... let's say the assignment was more complicated. Let's say you had a longer input string, and it had several groups separated by -, and the groups could include letters and digits. You know there are at least two groups that are "digits only" and you need to grab the second such "purely numeric" group. Then something like this will work (and there will not be an instr/substr solution):
select regexp_substr(input_str, '(-|^)(\d+)(-|$)', 1, 2, null, 2) from ....
This searches the input string for one or more digits ( \d means any digit, + means one or more occurrences) between a - or the beginning of the string (^ means beginning of the string; (a|b) means match a OR b) and a - or the end of the string ($ means end of the string). It starts searching at the first character (the second argument of the function is 1); it looks for the second occurrence (the argument 2); it doesn't do any special matching such as ignore case (the argument "null" to the function), and when the match is found, return the fragment of the match pattern included in the second set of parentheses (the last argument, 2, to the regexp function). The second fragment is the \d+ - the sequence of digits, without the leading and/or trailing dash -.
This solution will work in your example too, it's just overkill. It will find the right "digits-only" group in something like AS23302-ATX-20032-33900293-CWV20-3499-RA; it will return the second numeric group, 33900293.
I have a query (sql) to pull out a street name from a string. It's looking for the last occurrence of a digit, and then pulling the proceeding text as the street name. I keep getting the oracle
"argument '0' is out of range"
error but I'm struggling to figure out how to fix it.
the part of the query in question is
substr(address,regexp_instr(address,'[[:digit:]]',1,regexp_count(address,'[[:digit:]]'))+2)
any help would be amazing. (using sql developer)
The fourth parameter of regexp_instr is the occurrence:
occurrence is a positive integer indicating which occurrence of
pattern in source_string Oracle should search for. The default is 1,
meaning that Oracle searches for the first occurrence of pattern.
In this case, if an address has no digits within, the regexp_count will return 0, that's not a valid occurrence.
A simpler solution, which does not require separate treatment for addresses without a house number, is this:
with t (address) as (
select '422 Hickory Str.' from dual union all
select 'One US Bank Plaza' from dual
)
select regexp_substr(address, '\s*([^0-9]*)$', 1, 1, null, 1) as street from t;
The output looks like this:
STREET
-------------------------
Hickory Str.
One US Bank Plaza
The third argument to regexp_substr is the first of the three 1's. It means start the search at the first character of address. The second 1 means find the first occurrence of the search pattern. The null means no special match modifiers (such as case insensitive - nothing like that needed here). The last 1 means "return the first SUBEXPRESSION from the match pattern". Subexpressions are parts of the match expression enclosed in parentheses.
The match pattern has a $ at the end - meaning "anchor at the end of the input string" ($ means the end of the string). Then [...] means match any of the characters in square brackets, but the ^ in [^...] changes it to match any character OTHER THAN what is in the square brackets. 0-9 means all characters between 0 and 9; so [^0-9] means match any character(s) OTHER THAN digits, and the * after that means "any number of such characters" (between 0 and everything in the input string). \s is "blank space" - if there are any blank spaces following a possible number in the address, you don't want them included right at the beginning of the street name. The subexpression is just [^0-9]* meaning the non-digits, not including any spaces before them (because the \s* is outside the left parenthesis).
My example illustrates a potential problem though - sometimes an address does, in fact, have a "number" in it, but spelled out as a word instead of using digits. What I show is in fact a real-life address in my town.
Good luck!
looking for the last occurrence of a digit, and then pulling the proceeding text as the street name
You could simply do:
SELECT REGEXP_REPLACE( address, '^(.*)\d+\D*$', '\1' )
AS street_name
FROM address_table;
I try to extract the '930' from 'EM 930' with following Regexp
REGEXP_REPLACE(info,'^[:space:]*[ABCDEFGHIJKLMNOPQRSTUVWXYZ]*[:space:]*([0-9]+)[:space:]*$','\1')
But it returns me the original string.
An idea why ?
Subsidiary question:
Why does the "\1" returned the original string when the pattern is not matched ? I expected it to return NULL, as in my other regexp experiences (eg Perl).
Who I can re-write this in a performant way so that I get of wel the matched string of well NULL ?
Your space character class was not exactly correct. If we change [:space:] to [[:space:]], your regexp_replace works as you expect:
REGEXP_REPLACE(info, '^[[:space:]]*[ABCDEFGHIJKLMNOPQRSTUVWXYZ]*[[:space:]]*([0-9]+)[[:space:]]*$','\1')
For the sake of succinctness, we could use the upper character class, [[:upper:]], for [ABCDEFGHIJKLMNOPQRSTUVWXYZ]. This changes the function invocation to:
regexp_replace(info, '^[[:space:]]*[[:upper:]]*[[:space:]]*([0-9]+)[[:space:]]*$','\1')
Or escape characters could be used in lieu of character classes:
\s space
\w word character
\d digit character
regexp_replace(info, '^\s*\w*\s*(\d+)\s*$','\1')
Explanation:
Since your malformed character class, [:space:], does not match the space that exists between 'EM' and '930', your search by parameter does not match any characters in the source parameter.
Your search by parameter, '^[[:space:]]*[ABCDEFGHIJKLMNOPQRSTUVWXYZ]*[[:space:]]*([0-9]+)[[:space:]]*$', is anchored to the beginning and end of the column, info, thus it can only match the column, info, one time at most.
In your case, there is no match and the character group, '\1' which is associated with '([0-9]*)', has no value.
Consequently, no characters are replaced and you are left with original value of the column, info, 'EM 930'.
interesting variations to better understand this function:
-If your corrected function invocation had no pattern_to_replace_by parameter, '\1', then a NULL would be returned:
regexp_replace(info, '^\s*\w*\s*(\d+)\s*$' ) FROM dual;
-Since you have a pattern_to_replace_by parameter, '\1', and now it has the matching character group, the repeating digit group is returned:
930