I've lots of string values containing single quotes which I need to insert to a column in REDSHIFT table.
I used both /' and '' to escape the single quote in INSERT statement.
e.g.
INSERT INTO table_Temp
VALUES ('1234', 'O\'Niel'), ('3456', 'O\'Brien')
I also used '' instead of \' but it keeps giving me error that "VALUES list must of same length" i.e. no: of arguments for each record >2.
Can you let know how to have this issue resolved?
The standard in SQL is double single quotes:
INSERT INTO table_Temp (col1, col2) -- include the column names
VALUES ('1234', 'O''Niel'), ('3456', 'O''Brien');
You should also include the column names corresponding to the values being inserted. That is probably the cause of your second error.
You could use CHR(39) and concat the strings. Your name would look like below:
('O' || CHR(39)||'Brian')
I think it may depend on your environment. I'm using Periscope Data's redshift SQL editor, and \ worked as an escape character. '' and \\ did not work.
I was facing similar problem , I was needing send a kind of JSON structure to then decode it into my query but there was a program receiving my string and this program was escaping my escapes, so the query fails, finally I found this :
Put $$ in dollar-quoted string in PostgreSQL
mentioning quote_literal(42.5)
https://www.postgresql.org/docs/current/functions-string.html#FUNCTIONS-STRING-OTHER
This resolves my issue . an example
String is
'LocalTime={US/Central}; NumDays={1}; NumRows={3}; F_ID={[Apple, Orange, Bannana]}'
Select
Param, value , replace(quote_literal(replace(replace(Value,'[',''),']','')),',',quote_literal(',')) ValueList
FROM (
select
SPLIT_PART(split,'=',1) as Param,
replace( replace(SPLIT_PART(split,'=',2),'{',''),'}','') as Value
FROM
(
select
trim(split_part(freeform.txt, ';', number.n)) as split
from
( select
'LocalTime={US/Central}; NumDays={1}; NumRows={3}; F_ID={[Apple, Orange, Bannana]}' as txt
) freeform,
( select 1 as n union all
select 2 union all
select 3 union all
select 4 union all
select 5 union all
select 6 union all
select 7 union all
select 8 union all
select 9 union all
select 10
) number
where split <> ''
) as MY_TMP
) as valuePart
use \\' to escape '
s = s.replace("'", "\\'")
Related
I need to find (exclude in fact) any results that contain '%' sign, wherever in a string field. That would mean ... WHERE string LIKE '%%%'. Googling about escaping gave me the following ideas. The first throws syntax error, the second returns rows but there are records actually contain '%'.
1st:
SELECT * FROM table
WHERE string NOT LIKE '%!%%' ESCAPE '!'
///tried with different escape characters
2nd:
SELECT * FROM table
WHERE string NOT LIKE '%[%]%'
Trying on GCP BigQuery.
Try:
SELECT *
FROM table
WHERE string NOT LIKE '%!%%' {ESCAPE '!'}
With curly braces as shown in microsoft sql server docs
Or also:
WITH indata(s) AS (
SELECT 'not excluded'
UNION ALL SELECT '%excluded'
UNION ALL SELECT 'Ex%cluded'
UNION ALL SELECT 'Excluded%'
)
SELECT * FROM indata WHERE INSTR(s,'%') = 0;
-- out s
-- out --------------
-- out not excluded
find (exclude in fact) any results that contain '%'
Consider below simple approach
select *
from your_table
where not regexp_contains(string , '%')
Some of the column values in my table are enclosed within double quotes and I need to remove those double quotes. The problem is, I am not able to view the double quotes symbol in SQL Developer (version 4.2.0.16.260). But when I copy the data to Notepad++, there I am getting double quotes. I tried instr to get the position of double quotes, its returning the result 0. Also I tried to replace double quotes with blank, but they are not being removed.
Below is the data I see when copied to Notepad++. 2 records are displayed. I'm trying to take the distinct value from these 2 values. But I am not able to.
"Testdata
"
Testdata
The column value in the database table does not have double quotes.
When you copy the data from the results grid, SQL Developer is adding them as part of the copy operation, to help you out. (There's probably way to ask it not to, but I can't immediately see one.) It's doing that because the first value you're getting has a trailing new line character. I can duplicate what you're seeing if I do:
select 'Testdata' || chr(10) from dual
union all
select 'Testdata' from dual;
If I run as a script, the script output window shows:
'TESTDATA
---------
Testdata
Testdata
Here the newline is lost, and copy and pasting from that doesn't preserve it. If I run as a statement the data in the query result window looks the same:
but if I copy that data from the grid and paste it (anywhere, not just in Notepad++) I also see:
"Testdata
"
Testdata
... so the newline is preserved, and is enclosed in double-quotes so whatever it is pasted into (I'm guessing this is targeted at Excel) understands that it is a single value, including that newline character.
Im trying to take the distinct value from these 2 values
The problem is that they are not, in fact, distinct; one has a newline, the other does not.
If you want to ignore that and treat them as the same you can trim off the trailing newline:
select distinct rtrim(col, chr(10))
from your_table;
Demo with the same sample data:
-- CTE for sample data
with your_table (col) as (
select 'Testdata' || chr(10) from dual
union all
select 'Testdata' from dual
)
select col
from your_table;
COL
---------
Testdata
Testdata
-- CTE for sample data
with your_table (col) as (
select 'Testdata' || chr(10) from dual
union all
select 'Testdata' from dual
)
select distinct rtrim(col, chr(10)) as col
from your_table;
COL
---------
Testdata
Please try this out
SELECT REPLACE(TRIM(column_name), CHR(13)||CHR(10))
FROM table_name;
Getting Examples from similar Stack Overflow threads,
Remove all characters after a specific character in PL/SQL
and
How to Select a substring in Oracle SQL up to a specific character?
I would want to retrieve only the first characters before the occurrence of a string.
Example:
STRING_EXAMPLE
TREE_OF_APPLES
The Resulting Data set should only show only STRING_EXAM and TREE_OF_AP because PLE is my delimiter
Whenever i use the below REGEXP_SUBSTR, It gets only STRING_ because REGEXP_SUBSTR treats PLE as separate expressions (P, L and E), not as a single expression (PLE).
SELECT REGEXP_SUBSTR('STRING_EXAMPLE','[^PLE]+',1,1) from dual;
How can i do this without using numerous INSTRs and SUBSTRs?
Thank you.
The problem with your query is that if you use [^PLE] it would match any characters other than P or L or E. You are looking for an occurence of PLE consecutively. So, use
select REGEXP_SUBSTR(colname,'(.+)PLE',1,1,null,1)
from tablename
This returns the substring up to the last occurrence of PLE in the string.
If the string contains multiple instances of PLE and only the substring up to the first occurrence needs to be extracted, use
select REGEXP_SUBSTR(colname,'(.+?)PLE',1,1,null,1)
from tablename
Why use regular expressions for this?
select substr(colname, 1, instr(colname, 'PLE')-1) from...
would be more efficient.
with
inputs( colname ) as (
select 'FIRST_EXAMPLE' from dual union all
select 'IMPLEMENTATION' from dual union all
select 'PARIS' from dual union all
select 'PLEONASM' from dual
)
select colname, substr(colname, 1, instr(colname, 'PLE')-1) as result
from inputs
;
COLNAME RESULT
-------------- ----------
FIRST_EXAMPLE FIRST_EXAM
IMPLEMENTATION IM
PARIS
PLEONASM
I have to select only the IDs which have only even digits (an ID looks like: p19 ,p20 etc). That is, p20 is good (both 2 and 0 are even digits); p18 is not.
I thought to use substr to get each number from the IDs and then see if it's even .
select from profs
where to_number(substr(id_prof,2,2))%2=0 and to_number(substr(id_prof,3,2))%2=0;
IF you need all rows consist of 'p' in beginning and even digits on tail It should look like:
select *
from profs
where regexp_like (id_prof, '^p[24680]+$');
with
profs ( prof_id ) as (
select 'p18' from dual union all
select 'p24' from dual union all
select 'p53' from dual
)
-- End of test data; what is above this line is NOT part of the solution.
-- The solution (SQL query) begins here.
select *
from profs
where length(prof_id) = length(translate(prof_id, '013579', '0'));
PROF_ID
-------
p24
This solution should work faster than anything using regular expressions. All it does is to replace 0 with itself and DELETE all odd digits from the input string. (The '0' is included due to a strange but documented behavior of translate() - the third argument can't be empty). If the length of the input string doesn't change after the translation, that means the input string didn't have any odd digits.
where mod(to_number(regexp_replace(id_prof, '[^[:digit:]]', '')),2) = 0
I have (and don't own, so I can't change) a table with a layout similar to this.
ID | CATEGORIES
---------------
1 | c1
2 | c2,c3
3 | c3,c2
4 | c3
5 | c4,c8,c5,c100
I need to return the rows that contain a specific category id. I starting by writing the queries with LIKE statements, because the values can be anywhere in the string
SELECT id FROM table WHERE categories LIKE '%c2%';
Would return rows 2 and 3
SELECT id FROM table WHERE categories LIKE '%c3%' and categories LIKE '%c2%'; Would again get me rows 2 and 3, but not row 4
SELECT id FROM table WHERE categories LIKE '%c3%' or categories LIKE '%c2%'; Would again get me rows 2, 3, and 4
I don't like all the LIKE statements. I've found FIND_IN_SET() in the Oracle documentation but it doesn't seem to work in 10g. I get the following error:
ORA-00904: "FIND_IN_SET": invalid identifier
00904. 00000 - "%s: invalid identifier"
when running this query: SELECT id FROM table WHERE FIND_IN_SET('c2', categories); (example from the docs) or this query: SELECT id FROM table WHERE FIND_IN_SET('c2', categories) <> 0; (example from Google)
I would expect it to return rows 2 and 3.
Is there a better way to write these queries instead of using a ton of LIKE statements?
You can, using LIKE. You don't want to match for partial values, so you'll have to include the commas in your search. That also means that you'll have to provide an extra comma to search for values at the beginning or end of your text:
select
*
from
YourTable
where
',' || CommaSeparatedValueColumn || ',' LIKE '%,SearchValue,%'
But this query will be slow, as will all queries using LIKE, especially with a leading wildcard.
And there's always a risk. If there are spaces around the values, or values can contain commas themselves in which case they are surrounded by quotes (like in csv files), this query won't work and you'll have to add even more logic, slowing down your query even more.
A better solution would be to add a child table for these categories. Or rather even a separate table for the catagories, and a table that cross links them to YourTable.
You can write a PIPELINED table function which return a 1 column table. Each row is a value from the comma separated string. Use something like this to pop a string from the list and put it as a row into the table:
PIPE ROW(ltrim(rtrim(substr(l_list, 1, l_idx - 1),' '),' '));
Usage:
SELECT * FROM MyTable
WHERE 'c2' IN TABLE(Util_Pkg.split_string(categories));
See more here: Oracle docs
Yes and No...
"Yes":
Normalize the data (strongly recommended) - i.e. split the categorie column so that you have each categorie in a separate... then you can just query it in a normal faschion...
"No":
As long as you keep this "pseudo-structure" there will be several issues (performance and others) and you will have to do something similar to:
SELECT * FROM MyTable WHERE categories LIKE 'c2,%' OR categories = 'c2' OR categories LIKE '%,c2,%' OR categories LIKE '%,c2'
IF you absolutely must you could define a function which is named FIND_IN_SET like the following:
CREATE OR REPLACE Function FIND_IN_SET
( vSET IN varchar2, vToFind IN VARCHAR2 )
RETURN number
IS
rRESULT number;
BEGIN
rRESULT := -1;
SELECT COUNT(*) INTO rRESULT FROM DUAL WHERE vSET LIKE ( vToFine || ',%' ) OR vSET = vToFind OR vSET LIKE ('%,' || vToFind || ',%') OR vSET LIKE ('%,' || vToFind);
RETURN rRESULT;
END;
You can then use that function like:
SELECT * FROM MyTable WHERE FIND_IN_SET (categories, 'c2' ) > 0;
For the sake of future searchers, don't forget the regular expression way:
with tbl as (
select 1 ID, 'c1' CATEGORIES from dual
union
select 2 ID, 'c2,c3' CATEGORIES from dual
union
select 3 ID, 'c3,c2' CATEGORIES from dual
union
select 4 ID, 'c3' CATEGORIES from dual
union
select 5 ID, 'c4,c8,c5,c100' CATEGORIES from dual
)
select *
from tbl
where regexp_like(CATEGORIES, '(^|\W)c3(\W|$)');
ID CATEGORIES
---------- -------------
2 c2,c3
3 c3,c2
4 c3
This matches on a word boundary, so even if the comma was followed by a space it would still work. If you want to be more strict and match only where a comma separates values, replace the '\W' with a comma. At any rate, read the regular expression as:
match a group of either the beginning of the line or a word boundary, followed by the target search value, followed by a group of either a word boundary or the end of the line.
As long as the comma-delimited list is 512 characters or less, you can also use a regular expression in this instance (Oracle's regular expression functions, e.g., REGEXP_LIKE(), are limited to 512 characters):
SELECT id, categories
FROM mytable
WHERE REGEXP_LIKE('c2', '^(' || REPLACE(categories, ',', '|') || ')$', 'i');
In the above I'm replacing the commas with the regular expression alternation operator |. If your list of delimited values is already |-delimited, so much the better.