I am trying to find a number in a column (a varchar column) using LIKE, I am using a statement like this SELECT * FROM MYTABLE WHERE mynumberslist LIKE '%76%'. The data is stored like this: nums=;76;78;80;81;176; So how do I go about searching for 76 because if I just do LIKE %76% I will get 76 and 176 returned is there anyway around this? I did not design the database, and have no control over the data that is stored, I am only responsible for writing a program that gets the data out.
It is as simple as just
LIKE '%;76;%'
ps: send hello to the database schema author and donate him some money to buy any DB-design-related book
Use regex matching:
SELECT *
FROM MYTABLE
WHERE mynumberslist RLIKE '\b76\b'
This will match even if you don't have a trailing ; at the end - it matched the word 76 (\b means "word boundary" in regex)
SELECT *
FROM MYTABLE
WHERE ';' + mynumberslist + ';' LIKE '%;76;%';
Related
I have value in my db
like
1,12,13,25,44,414,2114
I have to find exact 14 from the db.
but it also return the value 414 and 2114
but i want exact 14.
How can i achieve this through sql query Please help
i have tried this but didn't worked.
Select * from tb_name where columnNAme like '%value%'
If you want search exactly 14, then you need search %,14,%, but 14 may appears at start or at end of string, so you need add commas to column also.
Well, you can use:
select * from tb_name where concat(',',columnNAme ,',') like '%,14,%'
Side note, comma separated values in one column is bad database design
Select * from tb_name where columnNAme like 'value'
% - The percent sign represents zero, one, or multiple characters
SELET * FROM database WHERE column LIKE 14 should work fine, just double checked it on my database.
If this doesn't work can you show your query please?
Above answer is correct, I didn't see your query before I posted this.
I've been to the regexp page on the MySQL website and am having trouble getting the query right. I have a list of links and I want to find invalid links that do not contain a period. Here's my code that doesn't work:
select * from `links` where (url REGEXP '[^\\.]')
It's returning all rows in the entire database. I just want it to show me the rows where 'url' doesn't contain a period. Thanks for your help!
SELECT c1 FROM t1 WHERE c1 NOT LIKE '%.%'
Your regexp matches anything that contains a character that isn't a period. So if it contains foo.bar, the regexp matches the f and succeeds. You can do:
WHERE url REGEXP '^[^.]*$'
The anchors and repetition operator make this check that every character is not a period. Or you can do:
WHERE LOCATE(url, '.') = 0
BTW, you don't need to escape . when it's inside [] in a regexp.
Using regexp seems like an overkill here. A simple like operator would do the trick:
SELECT * FROM `links` WHERE url NOT LIKE '%.%
EDIT:
Having said that, if you really want to negate regexp, just use not regexp:
SELECT * FROM `links` WHERE url NOT REGEXP '[\\.]';
My table contains some columns with ;-separated numbers like this :
1;2;43;22;20;12
and so on. It's also possible there's only 1 number in this column like 110, or 2 numbers like this 110;143
I want to select the rows that contain a certain number in this column. The number is in a variable $search_var.
Let's say I need to search for the number 1 in my column. If I use a select with like statement like so :
"SELECT * FROM Table WHERE ids LIKE '%".$search_var."%'"
I get all results containing '1' and not only '1', so I also get 11, 14, 110, 1999 etc.
I think I need to use some sort of regex-statement but I'm lost here... Who can help ?
You might not need regex for this
Set #YourNUmber := 110;
SELECT *
FROM Table
WHERE ';' + ids + ';' LIKE '%;'+ #yourNumber + ';%'
This guarantees there are always ; surrounding all the numbers.
This is formatted for SQL Server. The variable syntax and wildcards might be different if you are using something else.
EDIT:
Thanks #FélixGagnon-Grenier for the suggestions. I think either of these two will work. See here for a SQL Fiddle example
SELECT *
FROM T
WHERE concat(';',ids,';') LIKE concat('%;', #YourNumber , ';%');
SELECT *
FROM T
WHERE LOCATE(concat(';', #YourNumber , ';'),concat(';',ids,';'))>0
Try this solution if you're using SQL Server. This searches for the number where adjcent characters are not numbers:
SELECT * FROM Table WHERE ids LIKE '%[^0-9]".$search_var."[^0-9]%'
I have a table with part numbers. sometimes people put xxx at the end of the part number when they want to include all of the different possible endings (a lot like the wildcard ### in Access). How do I write a query that will give me all of the part numbers and replace anything ending in xxx with ###?
So if my table has:
1234
1235-xxx
1236
How do I write a query that will give me:
1234
1235-###
1236
In Access, you can use the replace function to change some text into some other text: http://office.microsoft.com/en-ca/access-help/replace-function-HA001228898.aspx
Together with a select statement and the iif function to choose the exact rows that need the replacement, it's a fairly simple operation:
select
iif(
part_num like '*-xxx'
, replace(part_num, '-xxx', '-###')
, part_num
)
from my_table
Ok, so here is the issue.
I have a table with some columns and 'subject' is one of the columns.
I need to get the first 10 letters from the 'subject' field no matter the 'subject' field contains a string with 100 letters.
For example,
Table - tbl.
Columns - id, subject, value.
SQL Query:
SELECT subject FROM tbl WHERE id ='$id';
The result I am getting is, for example
Hello, this is my subject and how are you
I only require the first 10 characters
Hello, thi
I can understand that I can remove the rest of the characters using php substr() but that's not possible in my case. I need to get the excess characters removed by MySQL. How can this be done?
Using the below line
SELECT LEFT(subject , 10) FROM tbl
MySQL Doc.
SELECT SUBSTRING(subject, 1, 10) FROM tbl
Have a look at either Left or Substring if you need to chop it up even more.
Google and the MySQL docs are a good place to start - you'll usually not get such a warm response if you've not even tried to help yourself before asking a question.