SugarQuery with a Where clause field LIKE '%58%'

Question

I have a field in my DB, it is type INT. Lets call it test_int so

<?php

$query = new SugarQuery();
$query->from(BeanFactory::getBean('Accounts'));
$query->select(array('id', 'name', 'test_int'));
$query->where()->contains('test_int', '58');
$returnArray = $query->execute();

the SQL I get is

SELECT  accounts.id id, accounts.name name, accounts.test_int test_int FROM accounts 
LEFT JOIN accounts_cstm ON accounts_cstm.id_c = accounts.id  
WHERE accounts.deleted = 0 AND accounts.test_int LIKE 58

Which pulls up no records, what am I doing wrong? I want '%58%' for the where clause.

Kenneth Brill · Accepted Answer

Hi Kenneth Brill Once this field is type INT so why don't you call $query->where()->equals() ? As far I understand "contains" is a string specific condition.