ralphschindler wrote
>
> Based off of what is in master, I can confirm that this works:
>
> https://github.com/ralphschindler/Zend_Db-Examples/blob/master/example-15.php
>
> What is the error/exception you are getting?
>
> -ralph
>
> On 5/13/12 9:51 PM, cmple wrote:
>> Hi,
>>
>> It seems like the LIKE expression cannot accept the '%' symbol when using
>> mysqli,
>> it just breaks the query:
>>
>> this works:
>> LIKE 'cm'
>>
>> this doesn't:
>> LIKE 'cm%'
>>
>> Is there a work around for this?
>>
>> Thanks!
>>
>> --
>> View this message in context:
>> http://zend-framework-community.634137.n4.nabble.com/ZF2-Db-Sql-Select-WHERE-name-LIKE-cmp-breaks-the-query-tp4631367.html
>> Sent from the Zend Framework mailing list archive at Nabble.com.
>>
>
>
> --
> List: [email protected]
> Info: http://framework.zend.com/archives
> Unsubscribe: [email protected]
>
instead of ->where->like('artist.name', '%Brit%');
try the following:
*->where("artist.name = '%Brit%");*
throws exception:
Zend\Db\Adapter\Exception\InvalidQueryException
File:
/***/vendor/ZendFramework/library/Zend/Db/Adapter/Driver/Mysqli/Statement.php:174
Message:
Statement couldn't be produced with sql: "SELECT `user`.* FROM `user` WHERE
"
Also, the join statement doesn't work when there is no space in between two
tables:
this works:
->join('album', '*artist.id = album.artist_id*', array('title',
'release_date'))
this doesn't:
->join('album', '*artist.id=album.artist_id*', array('title',
'release_date'))
--
View this message in context:
http://zend-framework-community.634137.n4.nabble.com/ZF2-Db-Sql-Select-WHERE-name-LIKE-cmp-breaks-the-query-tp4631367p4634871.html
Sent from the Zend Framework mailing list archive at Nabble.com.
--
List: [email protected]
Info: http://framework.zend.com/archives
Unsubscribe: [email protected]
