I'm trying to find all entries that contain a backslash anywhere, like so:
SELECT * FROM animals WHERE bodyType LIKE '%\\%'
I also tried:
SELECT * FROM animals WHERE bodyType LIKE '%\\\\%'
and
SELECT * FROM animals WHERE bodyType LIKE '%\\\\\\\\%'
Neither worked. Anyone know how to do this?
I am running the commands in MySQL Quick Admin v1.5.4
This question has not answer.
You try to match searching of content with backslash on names of columns.
Your (a bit general) query
SELECT * FROM tableName WHERE columnName LIKE '%\\%'
will give you any result if content of any column will contain backslash. In this case your query is correct. But you cannot match name of any column.
I looked into some books I have and all they say the same: this will select all records written in column of chosen name that are containing backslash. But columns have to be chosen exactly (they cannot be selected by name with using of SQL query).
Related
I am a beginner so please help me.
There are 2 things you need to combine in this case.
Because you didn't provide enough information in your question we have to guess what you mean by name. I'm going to assume that you have a single name column, but that would be unusual.
With strings, to match a character column that is not an exact match, you need to use LIKE which allows for wildcards.
You also need to negate the match, or in other words show things that are NOT (something).
First to match names that START with 'A'.
SELECT * FROM table_name WHERE name LIKE 'A%';
This should get you all the PEOPLE who have names that "Start with A".
Some databases are case sensitive. I'm not going to deal with that issue. If you were using MySQL that is not an issue. Case sensitivity is not universal. In some RDBMS like Oracle you have to take some steps to deal with mixed case in a column.
Now to deal with what you actually want, which is NOT (starting with A).
SELECT * FROM table_name WHERE name NOT LIKE 'A%';
your question should have more detail however you can use the substr function
SELECT name FROM yourtable
WHERE SUBSTR(name,1,1) <> 'A'
complete list of mysql string functions here
mysql docs
NOT REGXP operator
MySQL NOT REGXP is used to perform a pattern match of a string expression expr against a pattern pat. The pattern can be an extended regular expression.
Syntax:
expr NOT REGEXP pat
Query:
SELECT * FROM emp_table WHERE emp_name NOT REGEXP '^[a]';
or
SELECT * FROM emp_table WHERE emp_name NOT REGEXP '^a';
I am trying to write a Query to find if a string contains part of the value in Column (Not to confuse with the query to find if a column contains part of a string).
Say for example I have a column in a table with values
ABC,XYZ
If I give search string
ABCDEFG
then I want the row with ABC to be displayed.
If my search string is XYZDSDS then the row with value XYZ should be displayed
The answer would be "use LIKE".
See the documentation: https://dev.mysql.com/doc/refman/5.0/en/string-comparison-functions.html
You can do WHERE 'string' LIKE CONCAT(column , '%')
Thus the query becomes:
select * from t1 where 'ABCDEFG' LIKE CONCAT(column1,'%');
If you need to match anywhere in the string:
select * from t1 where 'ABCDEFG' LIKE CONCAT('%',column1,'%');
Here you can see it working in a fiddle:
http://sqlfiddle.com/#!9/d1596/4
Select * from table where #param like '%' + col + '%'
First, you appear to be storing lists of things in a column. This is the wrong approach to storing values in the database. You should have a junction table, with one row per entity and value -- that is, a separate row for ABC and XYZ in your example. SQL has a great data structure for storing lists. It is called a "table", not a "string".
If you are stuck with such a format and using MySQL, there is a function that can help:
where find_in_set('ABC', col)
MySQL treats a comma delimited string as a "set" and offers this function. However, this function cannot use indexes, so it is not particularly efficient. Did I mention that you should use a junction table instead?
I have a field called 'areasCovered' in a MySQL database, which contains a string list of postcodes.
There are 2 rows that have similar data e.g:
Row 1: 'B*,PO*,WA*'
Row 2: 'BB*, SO*, DE*'
Note - The strings are not in any particular order and could change depending on the user
Now, if I was to use a query like:
SELECT * FROM technicians WHERE areasCovered LIKE '%B*%'
I'd like it to return JUST Row 1. However, it's returning Row 2 aswell, because of the BB* in the string.
How could I prevent it from doing this?
The key to using like in this case is to include delimiters, so you can look for delimited values:
SELECT *
FROM technicians
WHERE concat(', ', areasCovered, ', ') LIKE '%, B*, %'
In MySQL, you can also use find_in_set(), but the space can cause you problems so you need to get rid of it:
SELECT *
FROM technicians
WHERE find_in_set('B', replace(areasCovered, ', ', ',') > 0
Finally, though, you should not be storing these types of lists as strings. You should be storing them in a separate table, a junction table, with one row per technician and per area covered. That makes these types of queries easier to express and they have better performance.
You are searching wild cards at the start as well as end.
You need only at end.
SELECT * FROM technicians WHERE areasCovered LIKE 'B*%'
Reference:
Normally I hate REGEXP. But ho hum:
SELECT * FROM technicians
WHERE concat(",",replace(areasCovered,", ",",")) regexp ',B{1}\\*';
To explain a bit:
Get rid of the pesky space:
select replace("B*,PO*,WA*",", ",",");
Bolt a comma on the front
select concat(",",replace("B*,PO*,WA*",", ",","));
Use a REGEX to match "comma B once followed by an asterix":
select concat(",",replace("B*,PO*,WA*",", ",",")) regexp ',B{1}\\*';
I could not check it on my machine, but it's should work:
SELECT * FROM technicians WHERE areasCovered <> replace(areaCovered,',B*','whatever')
In case the 'B*' does not exist, the areasCovered will be equal to replace(areaCovered,',B*','whatever'), and it will reject that row.
In case the 'B*' exists, the areCovered will NOT be eqaul to replace(areaCovered,',B*','whatever'), and it will accept that row.
You can Do it the way Programming Student suggested
SELECT * FROM technicians WHERE areasCovered LIKE 'B*%'
Or you can also use limit on query
SELECT * FROM technicians WHERE areasCovered LIKE '%B*%' LIMIT 1
%B*% contains % on each side which makes it to return all the rows where value contains B* at any position of the text however your requirement is to find all the rows which contains values starting with B* so following query should do the work.
SELECT * FROM technicians WHERE areasCovered LIKE 'B*%'
I'm trying to create a SQL query which will supply values for auto completion for a text field. Everything is working however I can't seem to create an SQL query which is exact enough for the purposes I want. I am using MySQL.
If there is a space (or multiple spaces) in the search term, I only want the query to do a LIKE comparison on the part of the string after the last space.
For example, say I have two possible values in the database:
Bolt
Bolts Large
Currently if the user types 'Bolt' then a space, both values above are returned using this query -
SELECT name FROM items WHERE name LIKE 'SEARCH_TERM%'
What I want is that if the user types 'Bolt' then a space, then only Bolt is returned from the database.
Effectively meaning that only the last part of the search term after the space is compared using LIKE, the results should match exactly up until the last space.
I've also tried:
SELECT name FROM items WHERE name LIKE 'SEARCH_TERM[a-z]%'
But that actually returns no results using the above scenario.
Is what I'm after possible? I've also tried to explore using Full Text Search but have had no look with that. I believe full text search is enabled on the name field, however I have limited experience with this. The query below didn't work.
SELECT name FROM items WHERE MATCH(name) AGAINST('SEARCH_TERM')
Any advice or points would be very appreciated.
The query
SELECT name FROM items WHERE name LIKE 'Bolt %'
doesn't return any record, because both 'Bolt' and 'Bolts Large' don't match 'Bolt %'.
SELECT name FROM items WHERE name LIKE 'Bolt%'
returns both records, because both 'Bolt' and 'Bolts Large' match 'Bolt%'.
To look for 'Bolt' and not 'Bolts', you must add a space to both your search string and the column string:
SELECT name FROM items WHERE concat(name, ' ') LIKE 'Bolt %'
returns 'Bolt' but not 'Bolts Large'.
SELECT name FROM items WHERE REPLACE(name, ' ', '') LIKE 'SEARCH_TERM%'
You could also use CONCAT and TRIM, or just trim
SELECT name FROM items WHERE name LIKE TRIM('SEARCH_TERM')
or your choice
SELECT name FROM items WHERE name LIKE CONCAT(TRIM('SEARCH_TERM'), '%')
SELECT name FROM items WHERE name LIKE CONCAT('%',TRIM('SEARCH_TERM'))
SELECT name FROM items WHERE name LIKE CONCAT('%',TRIM('SEARCH_TERM'), '%')
I use a string for store the days of the week, something like this:
MTWTFSS. And if I search for MF (Monday and Friday) then the query must return all the strings that contain MF (for example: MWF, MTWTFS, MF, and so on).
I don't know how to do this in SQL (MySQL).
use LIKE with %-wildcard between the single characters:
SELECT * FROM table WHERE column LIKE '%M%F%';
note that this will only work if the characters are in correct order - searching for FM instead of MF won't give you any result.
you'll also need to find a way to insert the %s to your search-term, but taht shouldn't be a big problem (sadly you havn't said wich programming-language you're using).
if the characters can be in random order, you'll have to built a query like:
SELECT * FROM table WHERE
column LIKE '%M%'
AND
column LIKE '%F%'
[more ANDs per character];
SELECT * FROM yourTable WHERE columnName LIKE '%MF%'
Learn more:
http://www.sqllike.com/
Can you not just say
SELECT * FROM blah WHERE weekday LIKE "%MF%"