SELECT update_log, update_idccode, update_filenumber, update_filetype, update_timedate
FROM updates_log
WHERE substring(update_idccode,0,7) ='idc2997%' AND update_filetype = 'E'
ORDER BY update_log DESC
I am trying to get this to get the first 7 characters of my update_idccode table column. I cannot get it to work. Any thoughts?
You're pulling out 7 chars from your column, and comparing them against an 8-char string. In other words, it is imposssible for a 7-char word to ever be identical to an 8-char word.
Try
WHERE substring(update_idccode,0,7) = 'idc2997' ...
instead. As an alternative,
WHERE update_idccode LIKE 'idc2997%' ...
would also work. The % suggests you may have been trying this, but % is only relevant as a wildcard in LIKE comparisons, not = equality testing.
Related
I got this question when I use mysql search something. here is the detailed information.
say I got a table named test with a column named content. in a specific record, the content column holds:
["
/^\w{2,}/","
/^[a-z][a-z0-9]+$/","
/^[a-z0-9]+$/","
/^[a-z]\d+$/"]
there is a linefeed character in the end of the lines(last line excluded)
so when I used the like syntax to search this record, I wrote a SQL like this
select * from test where `content` like
'[\"\n/^\\\\w{2,}/\",\"\n/^[a-z][a-z0-9]+$/\",\"\n/^[a-z0-9]+$/\",\"\n/^[a-z]\\\\d+$/\"]'
and it returned the right result. but when I changed the like to = and this SQL statement didn't work, after I tried several times, I got this SQL statement that worked:
select * from test where `content` =
'[\"\n/^\\w{2,}/\",\"\n/^[a-z][a-z0-9]+$/\",\"\n/^[a-z0-9]+$/\",\"\n/^[a-z]\\d+$/\"]'
it worked. so here is the question:
why on earth the like and = have different escape strategy? in the like statement I have to use \\\\w,\\\\d while in the = statement \\w,\\d just doing fine?
MySQL LIKE operator to select data based on patterns.
The LIKE operator is commonly used to select data based on patterns. Using the LIKE operator in the right way is essential to increase the query performance.
The LIKE operator allows you to select data from a table based on a specified pattern. Therefore, the LIKE operator is often used in the WHERE clause of the SELECT statement.
MySQL provides two wildcard characters for using with the LIKE operator, the percentage % and underscore _.
The percentage (%) wildcard allows you to match any string of zero or more characters.
The underscore (_) wildcard allows you to match any single character.
Comparison operations result in a value of 1 (TRUE), 0 (FALSE), or NULL. These operations work for both numbers and strings. Strings are automatically converted to numbers and numbers to strings as necessary.
The following relational comparison operators can be used to compare not only scalar operands, but row operands:
= > < >= <= <> !=
Note: = is Equal operator and LIKE for Simple pattern matching
1. select count(*) from tableX where code = "XYZ";
2. select count(*) from tableX where code like "%XYZ";
Result for query 1 is 18734. <== Not Correct
Result for query 2 is 93003. <== Correct
We know that query 2's count is correct based on independent verification.
We expect these two queries to have the exact same count for each because we know that no rows in tableX have a code that ends with "XYZ", so the wildcard at the beginning shouldn't affect the query.
Why would these queries produce different counts?
We have already researched the differences between "=" comparison and "like" string comparison, but based on all our verification checks, we still don't understand why this would give us different counts
We have confirmed the following:
There are no leading or trailing characters in the "code" field
There are no hidden characters (tried all found here: How can I find non-ASCII characters in MySQL?)
The collation is "utf8_unicode_ci"
We are using MySQL version 5.5.40-0ubuntu0.12.04.1.
Try this in order to get your answer:
SELECT code
FROM tableX
WHERE code LIKE "%XYZ"
AND code <> "XYZ"
LIMIT 10
My guess is that some of your codes end with a lowercase xyz, and since LIKE is case-insensitive, it matched these where = did not.
where code = "XYZ"; gives exact match whereas where code LIKE "%XYZ"; includes partial match as well. In your case, there could be an extra space present which is giving wrong count. Consider trimming before comparing like
where UPPER(TRIM(code)) = 'XYZ';
We restarted the server that the database resides on, we re-ran the queries, and now they all are producing the expected, correct results...
We'll have to look into possibilities for why this "fixed" the issue.
I have this query:
UPDATE tbl SET fldname = replace(fldname,'St','Stats');
However if anything begins with the letters St, such as Stump, it would then look like Statsump. How can I just have the St replaced with the exact match? Thanks!
AFAIK MySQL does not directly support regexp replacing (http://bugs.mysql.com/bug.php?id=27389), unless you use user defined functions.
You could try adding a WHERE char_length(fldname) == 5, though - this would match only the strings of length 5, and the only string of length 5 containing 'Stats' is... 'Stats'.
Or of course ditch the whole thing and just do SET fldname='St' WHERE fldname == 'Stats', which would probably be the sane thing to do unless your query is more complex than your example.
I have a database of words for dictionary lookup purposes. What I need to be able to do with mysql is allow a user to input to variables (alpha) and my script will return every word that starts with both of those variables and everything in between.
Let's say the two variables are:
$letters1 = abor
$letters2 = accr
I want to get every word that starts with abor through accr. I need to return every word that would fit between those two starting points. So an example SQL statement that I know does not work but might help you understand what I am asking:
SELECT word from table1 WHERE word LIKE '%abor%' THROUGH '%accr%' ORDER BY word ASC
I know that THROUGH is not an operator but that's the general idea of what I need to accomplish.
If you merely want words that start with letters between the two variables, you can use MySQL's BETWEEN ... AND ... operator:
SELECT word FROM table1 WHERE word BETWEEN 'abor' AND 'accr' ORDER BY word
Consider the string "55,33,255,66,55"
I am finding ways to count number of occurence of a specific characters ("55" in this case) in this string using mysql select query.
Currently i am using the below logic to count
select CAST((LENGTH("55,33,255,66,55") - LENGTH(REPLACE("55,33,255,66,55", "55", ""))) / LENGTH("55") AS UNSIGNED)
But the issue with this one is, it counts all occurence of 55 and the result is = 3,
but the desired output is = 2.
Is there any way i can make this work correct? please suggest.
NOTE : "55" is the input we are giving and consider the value "55,33,255,66,55" is from a database field.
Regards,
Balan
You want to match on ',55,', but there's the first and last position to worry about. You can use the trick of adding commas to the frot and back of the input to get around that:
select LENGTH('55,33,255,66,55') + 2 -
LENGTH(REPLACE(CONCAT(',', '55,33,255,66,55', ','), ',55,', 'xxx'))
Returns 2
I've used CONCAT to pre- and post-pend the commas (rather than adding a literal into the text) because I assume you'll be using this on a column not a literal.
Note also these improvements:
Removal of the cast - it is already numeric
By replacing with a string one less in length (ie ',55,' length 4 to 'xxx' length 3), the result doesn't need to be divided - it's already the correct result
2 is added to the length because of the two commas added front and back (no need to use CONCAT to calculate the pre-replace length)
Try this:
select CAST((LENGTH("55,33,255,66,55") + 2 - LENGTH(REPLACE(concat(",","55,33,255,66,55",","), ",55,", ",,"))) / LENGTH("55") AS UNSIGNED)
I would do an sub select in this sub select I would replace every 255 with some other unique signs and them count the new signs and the standing 55's.
If(row = '255') then '1337'
for example.