2

I would like to know how to perform pattern search using JDatabase

Here's my code:

$query->select('Common_name');
$query->from($db->quoteName('common_name'));
$query->where($db->quoteName('Common_name').' LIKE '. $db->quote('\'o%\''));
$db->setQuery($query);
$result = $db->loadResult();
1
  • The Joomla Documentation recommends using the 2nd parameter on both quote() and escape() when escape is nested inside of quote. This is to avoid performing double escaping. joomla.stackexchange.com/a/22712/12352 Apr 19, 2018 at 22:00

1 Answer 1

5

try this

$db = JFactory::getDbo();
$query = $db->getQuery(true);

$query
  ->select('Common_name')
  ->from($db->qn('your_table_name'))
  ->where($db->qn('Common_name').' LIKE '. $db->quote($db->escape('o%')));

$db->setQuery($query);
$results = $db->loadObjectList(); // use loadResult() for single value result

Documentation is your best friend: https://docs.joomla.org/Accessing_the_database_using_JDatabase

7
  • it produced an error You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE LIKE' at line 3 SQL=SELECT Common_name FROM WHERE LIKE
    – Joshua
    Aug 9, 2016 at 5:43
  • It works fine to me. I added the $db, $query initialization code also above, please check.
    – Nagarjun
    Aug 9, 2016 at 5:46
  • actually it still has the same error even if i already initialized the $db and $query
    – Joshua
    Aug 9, 2016 at 5:56
  • can you post what is the code you write for ->from($db->qn('your_table_name'))
    – Nagarjun
    Aug 9, 2016 at 6:05
  • ->from($db->qouteName('common_name'))
    – Joshua
    Aug 9, 2016 at 6:07

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