I am using UPDATE to insert simple text into a table where the field is MEDIUMTEXT (nullable field).
It is strange that it does not work when the field is null initially. If I manually enter at least a one character/space, then it's working.
I want to append the new text into existing text in the field.
UPDATE pen SET
PEN_STATUS = #PenStat,
PEN_STATUS_CHANGE_REASON = CONCAT(PEN_STATUS_CHANGE_REASON,'\n',ChangeDate,':',EmployeeID,':',ChangeReason)
WHERE PEN_ID = PenID;
Why is this?
CONCAT does not handle NULL values. As explained in the MySQL manual:
CONCAT() returns NULL if any argument is NULL.
You want to use COALESCE to handle that use case, like :
UPDATE pen SET
PEN_STATUS = #PenStat,
PEN_STATUS_CHANGE_REASON = CONCAT(
COALESCE(PEN_STATUS_CHANGE_REASON, ''),
'\n',
ChangeDate,
':',
EmployeeID,
':',
ChangeReason
)
WHERE PEN_ID = PenID;
Presumably, because something is NULL. Try using CONCAT_WS() instead:
UPDATE pen
SET PEN_STATUS = #PenStat,
PEN_STATUS_CHANGE_REASON = CONCAT_WS('\n',
PEN_STATUS_CHANGE_REASON,
CONCAT_WS(':', ChangeDate, EmployeeID, ChangeReason
)
)
WHERE PEN_ID = PenID;
CONCAT_WS() ignores NULL arguments. Plus, the separator only needs to be listed once.
I feel like I've done this before but missing something. I'm going through a database I inherited and want to see a bunch of DISTINCT values. I was thinking I could do something like this instead of writing cf_840 (or whatever number) a bunch of times and just change the actual field name in one spot...
SET #var = 'cf_840';
SELECT DISTINCT #var, COUNT(#var) AS counter
FROM vtiger_leadscf
GROUP BY #var
ORDER BY #var;
But this isn't working right and I feel like I'm missing something simple but can't find the right thing to search on SO.
You cannot use variables to directly specify fields outside of dynamically constructing a query; at best you can choose a value from a field conditionally, like CASE #var WHEN 840 THEN cf_840 WHEN 1 THEN cf_1 .... etc END AS fieldVal
Otherwise, you need to dynamically construct a query string with the field name "baked" into the query that gets executed.
C# style: var query = String.Format("SELECT {0}, COUNT(DISTINCT {0}) AS counter FROM .... blah blah blah", fieldName);
SQL Proc Style: SET query := 'SELECT ' + fieldName + ', COUNT(DISTINCT ' + fieldName.....and so on, then PREPARE and EXECUTE.
Edit: I'm not sure why you're selecting #var if you're getting the count of its values. I am guessing you want the field name included in the result, so perhaps the examples would be better as SELECT '{0}' as theField, COUNT(DISTINCT {0}) AS counter ... and 'SELECT ''' + fieldName + ''' AS fieldName, COUNT(DISTINCT ' + fieldName ....`
Also, you should not need GROUP BY or ORDER BY clauses, these queries will have only one result row.
To use a variable in the current query, declare it in the body of the SELECT
SELECT DISTINCT #var := 'cf_840' AS fieldName, COUNT(#var) AS counter
FROM vtiger_leadscf
GROUP BY fieldName
ORDER BY counter;
I have a field urn_sem.studentid that I'd like to replace a few characters in; for example:
ABC/2011/BCOMH_NC/I/12 → ABC/2011/BCOMH/I/12
ABC/2011/BCOMH_NC/I/24 → ABC/2011/BCOMH/I/24
I've tried this query:
SELECT REPLACE(studentid, 'KNC/2011/BCOMH_NC/', ' KNC/2011/BCOMH/')
FROM urn_sem
but it doesn't show the new value.
Do you want this:
update urn_sem
set studentid = REPLACE(studentid, 'KNC/2011/BCOMH_NC/', ' KNC/2011/BCOMH/')
where studentid like '%KNC/2011/BCOMH_NC/%'
The WHERE clause is optional. It ensures that the replace is only on rows that change.
And this sample query does not work?
SELECT REPLACE (studentid, '_', '') FROM urn_sem
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 '%.%'