How to write a regular expression to match a string if at least 3 characters from the start are matching?
Here is how my SQL query looks right now -
SELECT * FROM tableName WHERE columnName REGEXP "^[a-zA-Z]{3}someString";
You cannot use CONCAT or alike with REGEX, it will fail. Easiest way to do it, is:
$query = 'SELECT * FROM Test WHERE colb REGEXP "^'.substr($mystring,0,3).'"');
Another is:
SELECT * FROM Test WHERE LEFT(colb, 3) LIKE "{$mystring}%"
Please use jQuery and jqSQL plugin. Note that symbol $ must be escaped in SQL query with this plugin.
Related
I am currently having trouble finding all rows that contain the following data in a table "%2" or "%20".
For Example: 'Susan rides%2bike across town'.
SELECT * FROM [my_table] WHERE [description] LIKE % %2 %
I know this example will not work, however is there a way to search for these values "%2" or "%20" using the SQL "LIKE" operator? As I am required to find and replace them with a white space.
SELECT * FROM my_table WHERE description LIKE '%\%2%' ESCAPE '\';
For the LIKE keyword you can specify the ESCAPE sequence afterwards. In this example the leading backwards slash escapes the % so it will be interpreted as literal value.
Read more about it in the documentation: https://learn.microsoft.com/en-gb/sql/t-sql/language-elements/like-transact-sql?view=sql-server-ver15
you can test it in this SQL Fiddle.
Thank you for this.
SELECT * FROM [my_table] WHERE [description] LIKE '%[%]2%'
The solution for the above stated probelem is
select * from my_table where description like'%[%]2%'
Is there a way to express repetition in SQL LIKE.
I have the following query that matches 4 or 5 digits, then a hyphen, and then 8 additional digits:
SELECT * FROM Table
WHERE Field LIKE '[0-9][0-9][0-9][0-9]-[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]'
OR Field LIKE '[0-9][0-9][0-9][0-9][0-9]-[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]'
In other languages I would write something like this: [0-9]{4-5}-[0-9]{8}.
Is there a way to simplify the above expression in SQL?
This looks like SQL Server. You could construct the strings:
select '[0-9][0-9][0-9][0-9]-[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]',
replicate('[0-9]', 4) + '-' + replicate('[0-9]', 8)
However, SQL Server doesn't have built-in regular expression support.
Below are the equivalent regular expressions,
Oracle
`REGEXP_LIKE(X, '^[[:digit:]]+$');`
PostgreSql
`like '%[1-9]%' `
I want to search a table which has file name(s) with a {Numerical Pattern String}.PDF.
Example: 1.PDF, 12.PDF, 123.PDF 1234.PDF etc.....
select * from web_pub_subfile where file_name like '[0-9]%[^a-z].pdf'
But above SQL Query is resulting even these kind of files
1801350 Ortho.pdf
699413.processing2.pdf
15-NOE-301.pdf
Could any one help me what I am missing here.
One way to do it is getting the substring before the file extension and checking if it is numeric. This solution only works well if there is only one . character in the file name.
select * from web_pub_subfile
where isnumeric(left(file_name,charindex('.',file_name)-1)) = 1
Note:
ISNUMERIC returns 1 for some characters that are not numbers, such as plus (+), minus (-), and valid currency symbols such as the dollar sign ($).
To handle file names with mutliple . characters and if there is always a .filetype extension, use
select * from web_pub_subfile
where isnumeric(left(file_name,len(file_name)-charindex('.',reverse(file_name)))) = 1
and charindex('.',file_name) > 0
Sample demo
As suggested by #Blorgbeard in the comments, to avoid the use of isnumeric, use
select * from web_pub_subfile
where left(file_name,len(file_name)-charindex('.',reverse(file_name))) NOT LIKE '%[^0-9]%'
and len(left(file_name,len(file_name)-charindex('.',reverse(file_name)))) > 0
You can't really do what you are trying to do using plain out of the box sql. The reason you are seeing those results is that the % character matches any character, any number of times. It's not like * in a regex which matches the pervious character 0 or more times.
Your best option would probably be to create some CLR functions that implement regex functionality on the SQL Server side. You can take a look at this link to find a good place to start.
Depending on your version if 2012+, you could use Try_Convert()
select * from web_pub_subfile where Try_Convert(int,replace(file_name,'.pdf',''))>0
Declare #web_pub_subfile table (file_name varchar(100))
Insert Into #web_pub_subfile values
('1801350 Ortho.pdf'),
('699413.processing2.pdf'),
('15-NOE-301.pdf'),
('1.pdf'),
('1234.pdf')
select * from #web_pub_subfile where Try_Convert(int,replace(file_name,'.pdf',''))>0
Returns
file_name
1.pdf
1234.pdf
I need to use a string query to make a DB search for a C# program that interacts with MySQL server. What I want to find is a name that is 'like' one of my other variables (nameVar)
I have the following query in a C# program
string q = "SELECT *
FROM TABLE
WHERE name is like %?nameVar%";
As soon as execute the query in my program I get a syntax error telling me that syntax near
'like' is incorrect. As soon as I remove the "%" sign, it works fine.
I am confused, is mandatory to remove the % sign while building a query string?
Your parameter is replacing the ?nameVar part, including quotes. If the param is "TEST", your query gets presented as
string q = "SELECT *
FROM TABLE
WHERE name is like %'TEST'%";
As you can see, the % signs are out of place. either include them from the C# program into namevar, or change the query to
string q = "SELECT *
FROM TABLE
WHERE name is like '%' + ?nameVar + '%'";
you need to quote the query:
string q = "SELECT * from table where name is like '%?nameVar%'";
Strings in SQL need to be enclosed in single quotes:
string q = "SELECT *
FROM TABLE
WHERE name LIKE '%?nameVar%' ";
Also, there's no IS operator when using LIKE.
I think the correct syntax is:
SELECT * FROM table WHERE fields LIKE '%phrase%'
I think you have to leave out 'is'.
MySQL Pattern Matching
How do I do this?
For example, if my column is "cats,dogs,birds" and I want to get any rows where column contains cats?
Using LIKE:
SELECT *
FROM TABLE
WHERE column LIKE '%cats%' --case-insensitive
While LIKE is suitable for this case, a more general purpose solution is to use instr, which doesn't require characters in the search string to be escaped. Note: instr is available starting from Sqlite 3.7.15.
SELECT *
FROM TABLE
WHERE instr(column, 'cats') > 0;
Also, keep in mind that LIKE is case-insensitive, whereas instr is case-sensitive.