Got ids stored in DB with Json format like this
'["1454","474","545"]'
I can build list IDs :
SELECT replace
(replace(
replace(
replace('["1454","474","545"]','[','\'')
,']','\'')
,'"','')
,',','\',\'')
mySql returns '1454','474','545'
But when I try to list DB records from this build list of IDs :
SELECT col FROM table WHERE col in (REPLACE
(REPLACE(
REPLACE(
REPLACE('["1454","474","545"]','[','\'')
,']','\'')
,'"','')
,',','\',\''));
mySql says "0 records" even if I add a "SELECT" before the first "REPLACE"
Any help ?
Try below query, you need to add select in your in clause also:
SELECT col FROM table WHERE col in (select REPLACE
(REPLACE(
REPLACE(
REPLACE('["1454","474","545"]','[','\'')
,']','\'')
,'"','')
,',','\',\''));
Alas, you cannot use in with a comma delimited string. It takes a list of elements, but not within a string. So, this works as you expect:
where x in (1,2,3)
This does not work as you expect (although it does work as I expect0;
where x in ('1,2,3')
This looks for one value of x that is the string '1,2,3'.
The solution is to use the MySQL function find_in_set():
SELECT col
FROM table
WHERE find_in_set(col, REPLACE(REPLACE(REPLACE('["1454","474","545"]', '[','\''
), ']', '\''
), '"', ''
), ',', '\',\''
);
To be honest, though, you might be better off with something like:
where '["1454","474","545"]' like concat('%;', col, '&%')
Related
i have found some wrong text in 120 record in a table so i have in a varchar field:
'Name rubbish rubbish2 more rubbish'
i want to keep 'Name' en remove all after Name
i have tried this :
SELECT REPLACE(field_name, 'rubbish','') as test
from table
where field_name like '%rubbish%'
but this will ofcourse remove only 'rubbish' not the rest.
I think ther must be a way to remove everything after 5 digits!?
Txs
To remove everything after the first space character:
update mytable set
field_name = substr(field_name, 1, instr(field_name, ' '))
where field_name like '%rubbish%';
See SQLFiddle.
In MySQL, if you want to keep everything before the first space, then you can use substring_index():
update t
set col = substring_index(col, ' ', 1)
where col like '% %';
If you have some set pattern, such as the string 'rubbish', then you can use that. So, this keeps everything before "rubbish":
update t
set col = substring_index(col, 'rubbish', 1)
where col like '%rubbish%';
You can also use this logic in a SELECT statement:
select substring_index(col, 'rubbish', 1)
. . .
If the string does not contain "rubbish", then everything is returned.
I have a table which i am using to query and getting its one column which matches regular expression which is (\/.+\/\?).
Content of the resulted column is like:
/Anything here/?
Example output:
\abc\cdf\?....
\ab\?....
\abc\cdf\?....
\sb\?....
where '....' can be anything
Desired result i want is unique values before \? such that rows with duplicate regexp matched content are shown once only like here (\abc\cdf\?.... showing twice instead of onece)
\abc\cdf\?....
\ab\?....
\sb\?....
OR
\abc\cdf\?
\ab\?
\sb\?
I have looked very much but couldn't find anything there is regexp_substr in oracle but that is not working in SQL.
Please if someone could help me with the sql query that would be awesome.
If you want everything before the last \, then you can use substring_index() and some string manipulation:
select substring_index(col, '\\',
length(col) - length(replace(col, '\\', ''))
) as firstpart,
count(*)
from table t
group by substring_index(col, '\\',
length(col) - length(replace(col, '\\', ''))
);
I am writing the mysql query where I need to compare data entered by user with the sku column in database tab
Query is like
SELECT *
FROM `is_product_info`
WHERE REPLACE( '-', '', sku ) LIKE '%ETI2006%'
so that if sku in database contains "-" in sku, I am replacing all hyphens with "" before comparing.
so whether sju no contains ETI2006 or ETI-2006, it will come in output
I think you may just like this
SELECT * FROM is_product_info WHERE REPLACE( sku , '-', '' ) LIKE '%ETI2006%'
I didn't try on a running MySQL but this should work:
SELECT *
FROM `is_product_info`
WHERE sku REGEXP REPLACE('ETI-2006','-','-?');
I made the mistake in replace syntax, it works with the below one:
SELECT * FROM is_product_info WHERE REPLACE( sku , '-', '' ) LIKE '%ETI2006%'
Using replace() on every row is what's slowing it down.
All of these approaches should be faster - see which one works best for you:
Option 1 - use LIKE twice:
WHERE sku LIKE '%ETI-2006%'
OR sku LIKE '%ETI2006%'
Option 2 - Use RLIKE once:
WHERE sku RLIKE 'ETI-?2006'
my raw query look something like this-
UPDATE main,category,sub_category
SET main.biz_keyword = (category.category','sub_category.sub_cat_name','main.biz_keyword)
so the result something like main.biz_keyword='Doctor,General Physician,Physician'
I know this is wrong query but you got the Idea what I am looking for,
So my question is that I can do this by single query?
You might want to have a look at using CONCAT_WS(separator,str1,str2,...)
CONCAT_WS() stands for Concatenate With Separator and is a special
form of CONCAT(). The first argument is the separator for the rest of
the arguments. The separator is added between the strings to be
concatenated.
is this something you want to achieve?
Update TableName
set biz_keyword = category.category + ',' + sub_category.sub_cat_name + ',' + main.biz_keyword
Maybe you're looking for something like this?
UPDATE
main
SET
biz_keyword = CONCAT_WS(', ',
(SELECT category FROM category WHERE ... ),
(SELECT sub_cat_name FROM sub_category WHERE ... ),
biz_keyword)
The following query
SELECT ASSOCIATED_RISK
FROM PROJECT_ISSUES
WHERE FIND_IN_SET('98',ASSOCIATED_RISK);
returns output as
96,98
90,98
but if I use
SELECT ASSOCIATED_RISK
FROM PROJECT_ISSUES
WHERE FIND_IN_SET('96,98',ASSOCIATED_RISK);
it doesn't returns anything.In this case I would like to retrieve the first row.
96,98
Use the AND clause, like this:
SELECT ASSOCIATED_RISK
FROM PROJECT_ISSUES
WHERE FIND_IN_SET('96',ASSOCIATED_RISK)
AND FIND_IN_SET('98',ASSOCIATED_RISK)
Your query is failing because FIND_IN_SET() does not work properly if the first argument contains a comma (",") character. Reference: http://dev.mysql.com/doc/refman/5.0/en/string-functions.html#function_find-in-set. In your case, the first argument is '96,98', so it fails.
Your comment:
is there any other way I can get it in single query instead of framing multiple find_in_set and concat them
As an alternative solution, you can use locate on your ASSOCIATED_RISK value.
Example:
locate( replace( '96,98', ',', '' ), replace( ASSOCIATED_RISK, ',', '' ) )
Edit:
As per Aziz Shaikh comment, we can see that there is a possibility of true result though the search string not existing in the target string.
As an alternative solution, you can replace the search string from target string with an empty string and compare the lengths. If original string's length is grater than new replaced string, then it is a found true result.
Example:
-- this should be greater than 0 for a found true
length( ASSOCIATED_RISK ) > length( replace( ASSOCIATED_RISK, '96,98', '' ) )
This will Give result. see the difference.
SELECT ASSOCIATED_RISK FROM PROJECT_ISSUES WHERE FIND_IN_SET(ASSOCIATED_RISK,'96,98');