On 05/16/2012 02:59 AM, cmple wrote:
> cmple wrote
>>
>> 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.
>>>>
>>>
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>>>
>> 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'))
>>
>
> Edit:
> *->where("artist.name LIKE '%Brit%");*
>
> --
> View this message in context: 
> http://zend-framework-community.634137.n4.nabble.com/ZF2-Db-Sql-Select-WHERE-name-LIKE-cmp-breaks-the-query-tp4631367p4634875.html
> Sent from the Zend Framework mailing list archive at Nabble.com.
>

You're missing the closing '

Cheers,
David



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