Given a search string I need to select every record where (in the field the search is performed on) there is at least one word that begins with the given text.
For example:
'John Doe'
Have to be be selected with search strings like:
'joh'
'do'
'JOHN doe'
Have not to be selected with
'ohn'
'oe'
I need (possibly) to avoid full text search.
What I've found to work is
$query = 'SELECT * FROM MYTABLE WHERE SEARCHFIELD LIKE "' . $searchText . '%"'
. 'OR SEARCHFIELD LIKE "% ' . $searchText . '%"'
I'm asking if there is a better way to do that.
(for "better way" I mean better performance or same performance but more elegant)
Also, as the query will be built up with a prepared statement, how should I unescape LIKE metacharacters in the search string?
As already stated in the question the query
works for matching records where the
SEARCHFIELD
contains a word that begins with (or is equal to)$searchText
Regarding performance I've made a test on my development machine
MBP 2,2 GHz i7 quad core
:Searching for a word on 4.000 records takes around 40 milliseconds.
Records are normally indexed (no fulltext).
I have few thousands records and the query doesn't run very often so for me is good.
The solution may not be suitable for other contexts.
To build a prepared statement with the above query I used the technique described here:
Escaping MySQL wild cards
The resulting code is as follows:
Use this:
You don't need the
OR
condition -- if a field matchessearch%
, it will also match%search%
.