I have a mysql statement like this
$sql_insertwotd = "SELECT * FROM table WHERE word != '' AND word NOT LIKE '%\_%' ORDER BY RAND() LIMIT 1";
for some reason it it will still pull up a word with an underscore. I read this is the proper format to clarify an underscore with a .
I dont want it to select a word that contains an underscore
You have to use TWO BACKSLASHES (\\) and it should work
SELECT * FROM table WHERE word != '' AND word NOT LIKE '%\\_%'
ORDER BY RAND() LIMIT 1
You may need an explicit escape:
word NOT LIKE '%/_%' ESCAPE '/'
https://dev.mysql.com/doc/refman/5.0/en/string-comparison-functions.html
By default \ is the escape character and your query should work as is.
Related
Field X in table may contain special characters e.g hello!World and I would like to know if there is a way to match that with HelloWorld (Ignore case and special characters).
SELECT * FROM table WHERE X='Helloworld'
http://sqlfiddle.com/#!9/2afa1/1
if you need exaclty match of string:
SELECT *
FROM table1
WHERE x REGEXP '^hello[[:punct:],[:space:]]world$';
And if hello world could be a part of larger string:
SELECT *
FROM table1
WHERE x REGEXP 'hello[[:punct:],[:space:]]world';
What you can do is to replace all special characters like this:
SELECT * FROM table WHERE LOWER(REPLACE(X, '!', '')) = LOWER('HelloWorld');
Chain those replacements if you have to replace more:
SELECT * FROM table WHERE LOWER(REPLACE(REPLACE(X, '!', ''), '?', '')) = LOWER('HelloWorld');
If I understood your question right, you need to filter out non-ASCII characters? Please confirm whether this is true. In order to do that, have a look at REGEXP matching as in the comment link and this question.
Try something like
SELECT * FROM `table ` WHERE `X` REGEXP 'Helloworld';
REGEXP 'hello[^[:alpha:]]*world'
Notes:
This finds the string in the middle of other stuff; add ^ and $ to anchor to ends.
This assumes the non-alpha character(s) are between hello and world, not some other spot in the string.
This relies on the relevant collation to do (or not do) case folding.
How do I remove all superfluous full-stop . and semi-colon ; characters from end of last name field values in SQL?
One way to check of the last character is a "full stop" or "semicolon" is to use a substring function to get the last character, and compare that to the characters you are looking for. (There are several ways to do this, for example, using LIKE or REGEXP operator.
If that last character matches, then lop off that last character. One way to do that is to use a substring function. (Use the CHAR_LENGTH function to return the number of characters in the string.)
For example, something like this:
UPDATE mytable t
SET t.last_name = SUBSTR(t.last_name,1,CHAR_LENGTH(t.last_name)-1)
WHERE SUBSTRING(t.last_name,CHAR_LENGTH(t.last_name),1) IN ('.',';')
But, I'd strongly recommend that you test those expressions using a SELECT statement, before running an UPDATE statement.
SELECT t.last_name AS old_val
, SUBSTR(t.last_name,1,CHAR_LENGTH(t.last_name)-1) AS new_val
FROM mytable t
WHERE SUBSTRING(t.last_name,CHAR_LENGTH(t.last_name),1) IN ('.',';')
Substring rows that have a semi-colon or dot :
update emp
set ename = substring(ename, 1, char_length(ename) - 1)
where ename REGEXP '[.;]$';
I need a SELECT query in MYSQL that will retrieve all rows in one table witch field values contain "?" char with one condition: the char is not the last character
Example:
ID Field
1 123??see
2 12?
3 45??78??
Returning rows would then be those from ID 1 and 3 that match the condition given
The only statement I have is:
SELECT *
FROM table
WHERE Field LIKE '%?%'
But, the MySQL query does not solve my problem..
The LIKE expressions also support a wildcard "_" which matches exactly one character.
So you can write an expression like the example below, and know that your "?" will not be the last character in the string. There must be at least one more character.
WHERE intrebare LIKE '%?_%'
Re comment from #JohnRuddell,
Yes, that's true, this will match the string "??" because a "?" exists in a position that is not the last character.
It depends whether the OP means for that to be a match or not. The OP says the string "45??78??" is a match, but it's not clear if they would intend that "4578??" to be a match.
An alternative is to use a regular expression, but this is a little more tricky because you have to escape a literal "?", so it won't be interpreted as a regexp metacharacter. Then also escape the escape character.
WHERE intrebare REGEXP '\\?[^?]'
you can just add an additional where where the last character is not a ?
SELECT *
FROM intrebari
WHERE intrebare LIKE '%?%' AND intrebare NOT LIKE '%?'
you could also do it like this
SELECT *
FROM intrebari
WHERE intrebare LIKE '%?%' AND RIGHT(intrebare,1) <> '?'
DEMO
I have a column that contains strings as:
aaa_1
aaa_11
I need to query strings that ends with _1. I tried the following:
select col from table where col like %_1;
But the query gives me the strings that ends with _1 and _11. How can I correct this ?
Try this:
select col from table where col like '%\_1'
character _ is a jolly, like %, but it matches only a single character, so you have to escape it with \
See here: http://dev.mysql.com/doc/refman/5.0/en/string-comparison-functions.html#operator_like
You'll want to use something more substring related.
Try a where clause like:
SELECT col from table WHERE RIGHT(col, 2) = '_1'
You should escape % and _ by adding backslash \ as they are wildcards in mysql:
http://dev.mysql.com/doc/refman/5.0/en/string-comparison-functions.html
String | Description
\% | Matches one “%” character
\_ | Matches one “_” character
Many of these will pick up things like %_12 or %_111121, etc. Make sure you test out various cases to be sure these SELECT statements are giving you the proper subset.
SELECT col FROM table WHERE col REGEXP '_1[[:>:]]'
You can make a quick use of this query to filter out strings ending with specific character(s).
The below query output will give all names ending with 'er'.
select column_name
from table
where column_name regexp '.*er$';
I want to fetch all records which has one column contained % sign in mysql
we can do this using mysql using like
for ex..
select * from table where column like '%%';
it returns all records..
Please suggest
Use a backslash to escape the percent:
select * from table where column like '%\%%';
will match any row containing a percent character
% is a special character, try using escape characters to find it. Right now you're just telling mysql to look for a string using 2 wildcard characters (%) as opposed to the actual '%' character. Try using
select * from table where column = 'a%' ESCAPE 'a'
Basically telling mySQL to "Look for the string 'a%', but remove the char a in front of it.
EDIT: Another option is just using
select * from table where column = '\%'
Doing the same thing on later mySQL versions. The backslash is the "standard" escape character.
EDIT 2: Or to actually answer your question:
select * from table where column = '%\%%'
You need to escape the literal % sign with a \ e.g.
select * from table where column like '\%%';