how to use like function which contains space as one of the character in msql.
In my database, i am having column name as "contractor_id".
This is my code for selecting column name from table.
SELECT COLUMN_NAME as a FROM INFORMATION_SCHEMA.COLUMNS WHERE table_name = '$table' AND column_name LIKE '%".actor id."%'.
Just leave out the quotation marks like
SELECT COLUMN_NAME as a
FROM INFORMATION_SCHEMA.COLUMNS
WHERE table_name = '$table'
AND column_name LIKE '%actor id%'
Or, in case you are generally looking for a column name with a blank in it use
... column_name like '% %'
Related
I have to retrieve distinct entities from a column, all of which start with a vowel. The query looks like this :
Select DISTINCT column_name
FROM table_name
WHERE column_name LIKE '[aeiou]%';
It's not giving compilation errors, but it's not returning anything. Any ideas?
You can use regular expressions:
Select DISTINCT column_name
FROM table_name
WHERE column_name REGEXP '^[aeiou]';
LIKE wildcard patterns do not support character classes, except in a couple of databases that extend the definitions of the LIKE pattern.
Also, you might want:
WHERE column_name REGEXP '^[aeiouAEIOU]'
If you have a case-sensitive collation.
I have a table with 300+ column. Looking for a specific column is like nightmare. Is there any query where If I would like to search for columns stats with 'grand' can be listed...
You can use show columns from table with where. Try the following query,
SHOW COLUMNS FROM tablename WHERE field like 'grand%';
Just put in your tablename after FROM and it would work.
select * from myTable where mycolumn like 'grand%'
Try this.
SELECT
table_name,
column_name,
data_type,
ordinal_position
FROM INFORMATION_SCHEMA.COLUMNS
WHERE table_schema = 'DatabaseName' --- the database you want to search
AND table_name = 'yourTableName'
AND column_name LIKE '%Grand' ;
This one worked for me. Its shows all columns and table in entire database
SELECT TABLE_NAME, COLUMN_NAME
FROM INFORMATION_SCHEMA.COLUMNS
WHERE column_name OR table_name LIKE '%sale%';
Hello i need query for select all records from database containig &var separted by semicolons.
where &var is any number bellow which start with 1, like 123,166,1444
my varchar column look like this "123;166;234;1444"
the column may look and like this 233;166;12, which also containg numbers which start with 1.
select * from table_name where column_name like '1%' or column_name like '%;1%'
replace table_name with your tablename . Also replace column_name with your column name
i try to find %column_name% from another table_name i see the result
but he seems to ignore the %%
Why it does not work for me?
SELECT column_name FROM table_name WHERE NOT LIKE '%(SELECT column_name FROM table_name)%'
thanks,
Apologies if I do not act according to the rules, this is my first question.
Try this:
SELECT column_name FROM table_name WHERE NOT LIKE (SELECT CONCAT('%', column_name, '%') FROM table_name)
You can't just include a query in a string, so put the wildcards next to the column name in a query and return that to the NOT LIKE.
SELECT column_name
FROM table_name T
WHERE NOT EXISTS ( SELECT column_name from other_table O
WHERE O.column_name LIKE CONCAT('%',T.column_name,'%')
)
I have 2-3 different column names that I want to look up in the entire database and list out all tables which have those columns. Is there any easy script?
To get all tables with columns columnA or ColumnB in the database YourDatabase:
SELECT DISTINCT TABLE_NAME
FROM INFORMATION_SCHEMA.COLUMNS
WHERE COLUMN_NAME IN ('columnA','ColumnB')
AND TABLE_SCHEMA='YourDatabase';
SELECT TABLE_NAME, COLUMN_NAME
FROM INFORMATION_SCHEMA.COLUMNS
WHERE COLUMN_NAME LIKE '%wild%';
More simply done in one line of SQL:
SELECT * FROM information_schema.columns WHERE column_name = 'column_name';
SELECT DISTINCT TABLE_NAME, COLUMN_NAME
FROM INFORMATION_SCHEMA.COLUMNS
WHERE column_name LIKE 'employee%'
AND TABLE_SCHEMA='YourDatabase'
In older MySQL versions or some MySQL NDB Cluster versions that do not have information_schema, you can dump the table structure and search the column manually.
mysqldump -h$host -u$user -p$pass --compact --no-data --all-databases > some_file.sql
Now search the column name in some_file.sql using your preferred text editor, or use some nifty AWK scripts.
And a simple sed script to find the column. Just replace COLUMN_NAME with yours:
sed -n '/^USE/{h};/^CREATE/{H;x;s/\nCREATE.*\n/\n/;x};/COLUMN_NAME/{x;p};' <some_file.sql
USE `DATABASE_NAME`;
CREATE TABLE `TABLE_NAME` (
`COLUMN_NAME` varchar(10) NOT NULL,
You can pipe the dump directly in sed, but that's trivial.
For those searching for the inverse of this, i.e. looking for tables that do not contain a certain column name, here is the query...
SELECT DISTINCT TABLE_NAME FROM information_schema.columns WHERE
TABLE_SCHEMA = 'your_db_name' AND TABLE_NAME NOT IN (SELECT DISTINCT
TABLE_NAME FROM information_schema.columns WHERE column_name =
'column_name' AND TABLE_SCHEMA = 'your_db_name');
This came in really handy when we began to slowly implement use of InnoDB's special ai_col column and needed to figure out which of our 200 tables had yet to be upgraded.
Use this one line query. Replace desired_column_name by your column name.
SELECT TABLE_NAME FROM information_schema.columns WHERE column_name = 'desired_column_name';
If you want to "get all tables only", then use this query:
SELECT TABLE_NAME
FROM INFORMATION_SCHEMA.TABLES
WHERE TABLE_NAME like '%'
and TABLE_SCHEMA = 'tresbu_lk'
If you want "to get all tables with columns", then use this query:
SELECT DISTINCT TABLE_NAME, COLUMN_NAME
FROM INFORMATION_SCHEMA.COLUMNS
WHERE column_name LIKE '%'
AND TABLE_SCHEMA='tresbu_lk'
select distinct table_name
from information_schema.columns
where column_name in ('ColumnA')
and table_schema='YourDatabase';
and table_name in
(
select distinct table_name
from information_schema.columns
where column_name in ('ColumnB')
and table_schema='YourDatabase';
);
That ^^ will get the tables with ColumnA and ColumnB instead of ColumnA or ColumnB like the accepted answer
SELECT DISTINCT TABLE_NAME FROM INFORMATION_SCHEMA.COLUMNS WHERE COLUMN_NAME LIKE '%city_id%' AND TABLE_SCHEMA='database'
The problem with information_schema is that it can be terribly slow. It is faster to use the SHOW commands.
After you select the database you first send the query SHOW TABLES. And then you do SHOW COLUMNS for each of the tables.
In PHP that would look something like
$res = mysqli_query("SHOW TABLES");
while($row = mysqli_fetch_array($res))
{ $rs2 = mysqli_query("SHOW COLUMNS FROM ".$row[0]);
while($rw2 = mysqli_fetch_array($rs2))
{ if($rw2[0] == $target)
....
}
}