Like Operator With $variable
Solution 1:
Ultimately, what I want, is to be able to make a search for which the search-term is the current page's url, with wildcards before and after.
The SQL wildcard character is a percent sign. Therefore:
$variable = curPageURL();
$variable = mysql_real_escape_string($variable);
$query = "SELECT * FROM `tablename` WHERE `columnname` LIKE '%{$variable}%'";
Note: I've added in an extra bit of code. mysql_real_escape_string() will protect you from users deliberately or accidentally putting characters that will break your SQL statement. You're better off using parameterised queries, but that's a more involved topic than this simple fix.
Also note: I've fixed your string quoting, too. You can only use a variable in a string directly if that string is double quoted, and you were missing a quote at the end of $query.
edit 17 Jan 2015: Just got an upvote, so with that in mind, please don't use the mysql_* functions anymore.
Solution 2:
Use:
$query = "SELECT * FROM `tablename` WHERE `columnname` LIKE '{$variable}'" ;
To get an idea of why to prevent SQL injection attacks, like the above would be vulnerable to, I submit "Exploits of a Mom":

Solution 3:
Please don't do this, it is vulnerable to SQL injection (this is a list of 138 StackOverflow questions you should read, absorb and understand prior to returning to your application). Use parametrized queries or stored procedures.
Solution 4:
Use double quotes if you need to substitute variable values:
## this code isopenfor SQL injection attacks
$query = "SELECT * FROM `tablename` WHERE `columnname` LIKE '$variable'";
Or concat string manually:
## this code is open for SQL injection attacks$query = 'SELECT * FROM `tablename` WHERE `columnname` LIKE "' . $variable . '"';
Solution 5:
Your code is vulnerable to SQL injection attacks. User-supplied data should never be placed directly into a SQL query string. Instead, it must first be sanitized with a function such as mysql_real_escape_string().
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