I'm currently trying to check if a string has a special character (value that is not 0 to 9 A to Z a to z), but the inhouse language that I'm currently using has a very limited function to do it (possible but it will take a lot of lines). but I am able to do a query on sql. Now I would like to ask if it is possible to query using the dual table on sql, My plan is to pass the string to variable and this variable will be use on my sql command. Thanks in advance.
Here is what you can use
SELECT REGEXP_INSTR('Test!ing','[^[:alnum:]]') FROM dual;
This will return a number other than 0 whenever your string has anything other than letters or numbers.
You can use TRANSLATE to remove all okay characters from the string. You get back a string containing only undesired characters - or an empty string when there are none.
select translate(
'AbcDefg1234%99.26éXYZ', -- your string
'.ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789',
'.') from dual;
returns: %.é
Related
I am trying to remove template text like &#x; or &#xx; or &#xxx; from long string
Note: x / xx / xxx - is number, The length of the number is unknown, The cell type is CLOB
for example:
SELECT 'H'ello wor±ld' FROM dual
A desirable result:
Hello world
I know that regexp_replace should be used, But how do you use this function to remove this text?
You can use
SELECT REGEXP_REPLACE(col,'&&#\d+;')
FROM t
where
& is put twice to provide escaping for the substitution character
\d represents digits and the following + provides the multiple occurrences of them
ending the pattern with ;
or just use a single ampersand ('&#\d+;') for the pattern as in the case of Demo , since an ampersand has a special meaning for Oracle, a usage is a bit problematic.
In case you wanted to remove the entities because you don't know how to replace them by their character values, here is a solution:
UTL_I18N.UNESCAPE_REFERENCE( xmlquery( 'the_double_quoted_original_string' RETURNING content).getStringVal() )
In other words, the original 'H'ello wor±ld' should be passed to XMLQUERY as '"H'ello wor±ld"'.
And the result will be 'H'ello wo±ld'
I have a database in an oracle, I filled in the "Number" fields with numbers starting from "A14602727" to "A14603000" but it turned out that I accidentally typed the symbol A (in Ukrainian) instead of A (in English). And now I when find the command:
select number from test where number 'А14602727'
...find me nothing. Is it possible somehow with the help of the command to replace all numbers from "A14602727" (Ukrainian) to "A14602727" (English) ?
I will be grateful for your help!)
You can use following trick to convert any character to its base character using accent-insesitive binary sorting.
select your_col,
utl_raw.cast_to_varchar2(nlssort(your_col, 'nls_sort=binary_ai')) converted_col
from your_table
Use it in update statement accordingly.
Cheers!!
You could use regexp_replace():
update test set number = regexp_replace(number, '^Ä', 'A');
Regexp '^Ä' represents character 'Ä' at the beginning of the string. I used 'Ä' to represent the A in Ukrainian: replace this with the correct character that you want to replace.
This can also be done, probably more efficiently, with substr and like:
update test set number = 'A' || substr(number, 2) where number like 'Ä%';
Use the simple Oracle REPLACE function:
UPDATE test
SET number = REPLACE(number, 'A', 'A');
Replace is taking 3 parameters:
String in which you will search for a string to replace (The number column)
The string you are searchin gto replace (Ukrainian A)
The string you will replace it with (English A)
Here you have a DEMO example.
Also please note that the query in your question:
select number from test where number 'А14602727';
Is not valid. It should be something like this:
select number from test where number like 'А14602727';
or like this:
select number from test where number = 'А14602727';
So first do check that!
Cheers!
I am trying to only return the second to last character from a string using MS SQL.
I've tried using MID and Substring but the length of the string isn't always the same for the column I am trying to return, So I can't do it that way.
So say I am returning the codes of something:
Code
'1234'
I want to just return '3' from that code.
How can I do this?
Cheers in advance :)
Use SUBSTRING and LEN. LEN gives you the length of the string, then subtract 1 to get the previous char:
SELECT SUBSTRING(Code, LEN(Code)-1,1)
How about
select left(right(code, 2), 1) from MyTable;
You might need to validate that the string actually has at least 2 chars, however.
SqlFiddle here
I am not sure how to do this, but I have a string of data. I need to isolate a number out of the string that can vary in length. The original string also varies in length. Let me give you an example. Here is a set of the original data string:
:000000000:370765:P:000001359:::3SA70000SUPPL:3SA70000SUPPL:
:000000000:715186816:P:000001996:::H1009671:H1009671:
For these two examples, I need 3SA70000SUPPL from the first and H1009671 from the second. How would I do this using SQL? I have heard that case statements might work, but I don't see how. Please help.
This works in Oracle 11g:
with tbl as (
select ':000000000:370765:P:000001359:::3SA70000SUPPL:3SA70000SUPPL:' str from dual
union
select ':000000000:715186816:P:000001996:::H1009671:H1009671:' str from dual
)
select REGEXP_SUBSTR(str, '([^:]*)(:|$)', 1, 8, NULL, 1) data
from tbl;
Which can be described as "look at the 8th occurrence of zero or more non-colon characters that are followed by a colon or the end of the line, and return the 1st subgroup (which is the data less the colon or end of the line).
From this post: REGEX to select nth value from a list, allowing for nulls
Sorry, just saw you are using DB2. I don't know if there is an equivalent regular expression function, but maybe it will still help.
For the fun of it: SQL Fiddle
first substring gets the string at ::: and second substring retrieves the string starting from ::: to :
declare #x varchar(1024)=':000000000:715186816:P:000001996:::H1009671:H1009671:'
declare #temp varchar(1024)= SUBSTRING(#x,patindex('%:::%', #x)+3, len(#x))
SELECT SUBSTRING( #temp, 0,CHARINDEX(':', #temp, 0))
I am trying to fetch an id from an oracle table. It's something like TN0001234567890345. What I want is to sort the values according to the right most 10 positions (e.g. 4567890345). I am using Oracle 11g. Is there any function to cut the rightmost 10 places in Oracle SQL?
You can use SUBSTR function as:
select substr('TN0001234567890345',-10) from dual;
Output:
4567890345
codaddict's solution works if your string is known to be at least as long as the length it is to be trimmed to. However, if you could have shorter strings (e.g. trimming to last 10 characters and one of the strings to trim is 'abc') this returns null which is likely not what you want.
Thus, here's the slightly modified version that will take rightmost 10 characters regardless of length as long as they are present:
select substr(colName, -least(length(colName), 10)) from tableName;
Another way of doing it though more tedious. Use the REVERSE and SUBSTR functions as indicated below:
SELECT REVERSE(SUBSTR(REVERSE('TN0001234567890345'), 1, 10)) FROM DUAL;
The first REVERSE function will return the string 5430987654321000NT.
The SUBSTR function will read our new string 5430987654321000NT from the first character to the tenth character which will return 5430987654.
The last REVERSE function will return our original string minus the first 8 characters i.e. 4567890345
SQL> SELECT SUBSTR('00000000123456789', -10) FROM DUAL;
Result: 0123456789
Yeah this is an old post, but it popped up in the list due to someone editing it for some reason and I was appalled that a regular expression solution was not included! So here's a solution using regex_substr in the order by clause just for an exercise in futility. The regex looks at the last 10 characters in the string:
with tbl(str) as (
select 'TN0001239567890345' from dual union
select 'TN0001234567890345' from dual
)
select str
from tbl
order by to_number(regexp_substr(str, '.{10}$'));
An assumption is made that the ID part of the string is at least 10 digits.