I have a field with this value:
TEST:ATEST:TESTA
And I want to replace "TEST" with "NEW", I have tried this query:
UPDATE `table` SET `field` = REPLACE(`field`, 'TEST', 'NEW') WHERE `field` REGEXP 'TEST';
The result was:
NEW:ANEW:NEWA
Q: How could I do the replacement query so the result would be like this:
NEW:ATEST:TESTA
It is a bit of a pain, but you can do it this way:
UPDATE `table`
SET field = substr(REPLACE(concat(':', field, ':'), ':TEST:', ':NEW:'),
2, length(REPLACE(concat(':', field, ':'), ':TEST:', ':NEW:')) - 2)
WHERE concat(':', field, ':') LIKE '%:TEST:%';
I prefer LIKE to REGEXP because there is the hope of being able to use an index. That is not a possibility in this case, but there is the hope.
This is delimiting the values with colons at the beginning and the end, and only replacing fully delimited values. The trick is to then remove the additional colons.
You can try http://sqlfiddle.com/#!9/4e66b/3
so the update query is (if table name = table1, field name = field1, and there is unique column id):
UPDATE `table1`
INNER JOIN
(SELECT id,
#set_idx:=FIND_IN_SET('TEST',REPLACE(field1,':',',')),
#set_size:=LENGTH(field1)-LENGTH(REPLACE(field1,':',''))+1,
CASE
WHEN #set_idx=1 THEN CONCAT('NEW',SUBSTRING(field1, 4))
WHEN #set_idx>1 THEN CONCAT(SUBSTRING_INDEX(field1, ':',#set_idx-1),':NEW', IF(#set_size>#set_idx,CONCAT(':',SUBSTRING_INDEX(field1, ':',-(#set_size-#set_idx))),''))
END as new
FROM table1
WHERE `field1` REGEXP '(^TEST$)|(^TEST:)|(:TEST$)|(:TEST:)'
) t
ON t.id = table1.id
SET table1.field1 = t.new;
I expected to get full name without middle initial if there is no middle name, but if column "M_NAME" is Null, the select statement returns empty. How can i solve this?
SELECT CONCAT(`Employee`.`F_NAME`, ' ', LEFT(`Employee`.`M_NAME`, 1), '. ', `Employee`.`L_NAME`) FROM `ccms`.`Employee` WHERE HR_ID = '223';
CONCAT returns NULL if any argument is null. You can solve this by making sure that no null argument is null by wrapping any nullable column in either IFNULL or COALESCE (the latter can take more than two arguments).
SELECT
CONCAT(
IFNULL(F_NAME, ''),
' ',
IFNULL(CONCAT(LEFT(M_NAME, 1), '. '), ''),
IFNULL(L_NAME, '')
)
FROM
ccms.Employee
WHERE
HR_ID = '223';
What this does is replace NULL column values with an empty string, which is probably your intent. Note that I updated the selection of M_NAME so that the period is only added if the value is not null by using this very behavior.
EDIT: You can use backticks and qualify column names if you want, but it's not necessary for this exact query.
Use COALESCE()
Returns the first non-NULL value in the list, or NULL if there are no non-NULL values.
SELECT CONCAT(`Employee`.`F_NAME`, ' ', COALESCE(LEFT(`Employee`.`M_NAME`, 1), ''), '. ', `Employee`.`L_NAME`) FROM `ccms`.`Employee` WHERE HR_ID = '223';
Use IFNULL().
Any place you have a possibly NULL field, wrap it with IFNULL(whatever, '') and then you'll get an empty string instead of a result-killing NULL.
another simple solution is use CONCAT_WS. It will work defenitely.
SELECT
CONCAT_WS(''
F_NAME,
'',
LEFT(M_NAME, 1),
L_NAME) FROM
ccms.Employee WHERE HR_ID = '223';
I have html content in the post_content column.
I want to search and replace A with B but only the first time A appears in the record as it may appear more than once.
The below query would obviously replace all instances of A with B
UPDATE wp_posts SET post_content = REPLACE (post_content, 'A', 'B');
This should actually be what you want in MySQL:
UPDATE wp_post
SET post_content = CONCAT(REPLACE(LEFT(post_content, INSTR(post_content, 'A')), 'A', 'B'), SUBSTRING(post_content, INSTR(post_content, 'A') + 1));
It's slightly more complicated than my earlier answer - You need to find the first instance of the 'A' (using the INSTR function), then use LEFT in combination with REPLACE to replace just that instance, than use SUBSTRING and INSTR to find that same 'A' you're replacing and CONCAT it with the previous string.
See my test below:
SET #string = 'this is A string with A replace and An Answer';
SELECT #string as actual_string
, CONCAT(REPLACE(LEFT(#string, INSTR(#string, 'A')), 'A', 'B'), SUBSTRING(#string, INSTR(#string, 'A') + 1)) as new_string;
Produces:
actual_string new_string
--------------------------------------------- ---------------------------------------------
this is A string with A replace and An Answer this is B string with A replace and An Answer
Alternatively, you could use the functions LOCATE(), INSERT() and CHAR_LENGTH() like this:
INSERT(originalvalue, LOCATE('A', originalvalue), CHAR_LENGTH('A'), 'B')
Full query:
UPDATE wp_posts
SET post_content = INSERT(originalvalue, LOCATE('A', originalvalue), CHAR_LENGTH('A'), 'B');
With reference to https://dba.stackexchange.com/a/43919/200937 here is another solution:
UPDATE wp_posts
SET post_content = CONCAT( LEFT(post_content , INSTR(post_content , 'A') -1),
'B',
SUBSTRING(post_content, INSTR(post_content , 'A') +1))
WHERE INSTR(post_content , 'A') > 0;
If you have another string, e.g. testing then you need to change the +1 above to the according string length. We can use LENGTH() for this purpose. By the way, leave the -1 untouched.
Example: Replace "testing" with "whatever":
UPDATE wp_posts
SET post_content = CONCAT( LEFT(post_content , INSTR(post_content , 'testing') -1),
'whatever',
SUBSTRING(post_content, INSTR(post_content , 'testing') + LENGTH("testing"))
WHERE INSTR(post_content , 'testing') > 0;
By the way, helpful to see how many rows will be effected:
SELECT COUNT(*)
FROM post_content
WHERE INSTR(post_content, 'A') > 0;
If you are using an Oracle DB, you should be able to write something like :
UPDATE wp_posts SET post_content = regexp_replace(post_content,'A','B',1,1)
See here for more informations : http://docs.oracle.com/cd/B19306_01/server.102/b14200/functions130.htm
Note : you really should take care of post_content regarding security issue since it seems to be an user input.
Greg Reda's solution did not work for me on strings longer than 1 character because of how the REPLACE() was written (only replacing the first character of the string to be replaced). Here is a solution that I believe is more complete and covers every use case of the problem when defined as How do I replace the first occurrence of "String A" with "String B" in "String C"?
CONCAT(LEFT(buycraft, INSTR(buycraft, 'blah') - 1), '', SUBSTRING(buycraft FROM INSTR(buycraft, 'blah') + CHAR_LENGTH('blah')))
This assumes that you are sure that the entry ALREADY CONTAINS THE STRING TO BE REPLACED! If you try replacing 'dog' with 'cat' in the string 'pupper', it will give you 'per', which is not what you want. Here is a query that handles that by first checking to see if the string to be replaced exists in the full string:
IF(INSTR(buycraft, 'blah') <> 0, CONCAT(LEFT(buycraft, INSTR(buycraft, 'blah') - 1), '', SUBSTRING(buycraft FROM INSTR(buycraft, 'blah') + CHAR_LENGTH('blah'))), buycraft)
The specific use case here is replacing the first instance of 'blah' inside column 'buycraft' with an empty string ''. I think a pretty intuitive and natural solution:
Find the index of the first occurrence of the string that is to be replaced.
Get everything to the left of that, not including the index itself (thus '-1').
Concatenate that with whatever you are replacing the original string with.
Calculate the ending index of the part of the string that is being replaced. This is easily done by finding the index of the first occurrence again, and adding the length of the replaced string. This will give you the index of the first char after the original string
Concatenate the substring starting at the ending index of the string
An example walkthrough of replacing "pupper" in "lil_puppers_yay" with 'dog':
Index of 'pupper' is 5.
Get left of 5-1 = 4. So indexes 1-4, which is 'lil_'
Concatenate 'dog' for 'lil_dog'
Calculate the ending index. Start index is 5, and 5 + length of 'pupper' = 11. Note that index 11 refers to 's'.
Concatenate the substring starting at the ending index, which is 's_yay', to get 'lil_dogs_yay'.
All done!
Note: SQL has 1-indexed strings (as an SQL beginner, I didn't know this before I figured this problem out). Also, SQL LEFT and SUBSTRING seem to work with invalid indexes the ideal way (adjusting it to either the beginning or end of the string), which is super convenient for a beginner SQLer like me :P
Another Note: I'm a total beginner at SQL and this is pretty much the hardest query I've ever written, so there may be some inefficiencies. It gets the job done accurately though.
I made the following little function and got it:
CREATE DEFINER=`virtueyes_adm1`#`%` FUNCTION `replace_first`(
`p_text` TEXT,
`p_old_text` TEXT,
`p_new_text` TEXT
)
RETURNS text CHARSET latin1
LANGUAGE SQL
NOT DETERMINISTIC
CONTAINS SQL
SQL SECURITY DEFINER
COMMENT 'troca a primeira ocorrencia apenas no texto'
BEGIN
SET #str = p_text;
SET #STR2 = p_old_text;
SET #STR3 = p_new_text;
SET #retorno = '';
SELECT CONCAT(SUBSTRING(#STR, 1 , (INSTR(#STR, #STR2)-1 ))
,#str3
,SUBSTRING(#STR, (INSTR(#str, #str2)-1 )+LENGTH(#str2)+1 , LENGTH(#STR)))
INTO #retorno;
RETURN #retorno;
END
Years have passed since this question was asked, and MySQL 8 has introduced REGEX_REPLACE:
REGEXP_REPLACE(expr, pat, repl[, pos[, occurrence[, match_type]]])
Replaces occurrences in the string expr that match the regular
expression specified by the pattern pat with the replacement string
repl, and returns the resulting string. If expr, pat, or repl is NULL,
the return value is NULL.
REGEXP_REPLACE() takes these optional arguments:
pos: The position in expr at which to start the search. If omitted, the default is 1.
occurrence: Which occurrence of a match to replace. If omitted, the default is 0 (which means “replace all occurrences”).
match_type: A string that specifies how to perform matching. The meaning is as described for REGEXP_LIKE().
So, assuming you can use regular expressions in your case:
UPDATE wp_posts SET post_content = REGEXP_REPLACE (post_content, 'A', 'B', 1, 1);
Unfortunately for those of us on MariaDB, its REGEXP_REPLACE flavor is missing the occurrence parameter. Here's a regex-aware version of Andriy M's solution, conveniently stored as a reusable function as suggested by Luciano Seibel:
DELIMITER //
DROP FUNCTION IF EXISTS replace_first //
CREATE FUNCTION `replace_first`(
`i` TEXT,
`s` TEXT,
`r` TEXT
)
RETURNS text CHARSET utf8mb4
BEGIN
SELECT REGEXP_INSTR(i, s) INTO #pos;
IF #pos = 0 THEN RETURN i; END IF;
RETURN INSERT(i, #pos, CHAR_LENGTH(REGEXP_SUBSTR(i, s)), r);
END;
//
DELIMITER ;
It's simpler
UPDATE table_name SET column_name = CONCAT('A',SUBSTRING(column_name, INSTR(column_name, 'B') + LENGTH('A')));
For MYSQL version pre-5.6 and 8.0, I've used this pattern to fix my issue, it's a bit gross, but I hope it helps some of you guys:
SET #string = 'I love shop it is a terrific shop, I love eveything about it';
SET #shop_code = 'shop';
SET #shop_date = CONCAT(#shop_code, '__', DATE_FORMAT(NOW(), '%Y_%m_%d__%Hh%im%ss'));
SET #part1 = SUBSTRING_INDEX(#string, #shop_code, 1);
SET #shop_nb = ROUND( (LENGTH(#string) - LENGTH(REPLACE(#string, #shop_code,''))) / LENGTH(#shop_code) );
SET #part2 = SUBSTRING_INDEX(#string, #shop_code, -#shop_nb);
SET #string = CONCAT(#part1, #shop_date, #part2);
SELECT #string;
To keep the sample of gjreda a bit more simple use this:
UPDATE wp_post
SET post_content =
CONCAT(
REPLACE(LEFT(post_content, 1), 'A', 'B'),
SUBSTRING(post_content, 2)
)
WHERE post_content LIKE 'A%';
I'm don't have a lot of knowledge of MySql (or SQL in general) so sorry for the noobness.
I'm trying to update a bunch of String entries this way:
Lets say we have this:
commands.firm.pm.Stuff
Well I want to convert that into:
commands.firm.pm.print.Stuff
Meaning, Add the .print after pm, before "Stuff" (where Stuff can be any Alphanumerical String).
How would I do this with a MySql Query? I'm sure REGEXP has to be used, but I'm not sure how to go about it.
Thanks
Try something like this. It finds the last period and inserts your string there:
select insert(s, length(s) - instr(reverse(s), '.') + 1, 0, '.print')
from (
select 'commands.firm.pm.Stuff' as s
) a
To update:
update MyTable
set MyColumn = insert(MyColumn, length(MyColumn) - instr(reverse(MyColumn), '.') + 1, 0, '.print')
where MyColumn like 'commands.firm.pm.%'
Perhaps use a str_replace to replace commands.firm.pm to commands.firm.pm.print
$original_str = "commands.firm.pm.15hhkl15k0fak1";
str_replace("commands.firm.pm", "commands.firm.pm.print", $original_str);
should output: commands.firm.pm.print.15hhkl15k0fak1
then update your table with the new value...How to do it all in one query (get column value and do the update), I do not know. All I can think of is you getting the column value in one query, doing the replacement above, and then updating the column with the new value in a second query.
To update rows that end in '.Stuff' only:
UPDATE TableX
SET Column = CONCAT( LEFT( CHAR_LENGTH(Column) - CHAR_LENGTH('.Stuff') )
, '.print'
, '.Stuff'
)
WHERE Column LIKE '%.Stuff'
To update all rows - by appending .print just before the last dot .:
UPDATE TableX
SET Column = CONCAT( LEFT( CHAR_LENGTH(Column)
- CHAR_LENGTH(SUBSTRING_INDEX(Column, '.', -1))
)
, 'print.'
, SUBSTRING_INDEX(Column, '.', -1)
)
WHERE Column LIKE '%.%'