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Posted by Mαrio Gamito on 10/24/05 20:34
Hi,
Sorry, i do have the table name.
It just passed me while transcripting.
The code is:
---
$query = "SELECT COUNT (login) FROM formacao WHERE login = '$login'";
$result = mysql_query($query);
mysql_fetch_row($result);
---
It works perfectly on MySQL prompt.
Regards,
MΓ‘rio Gamito
Colin Shreffler wrote:
> It looks to me like you forgot to specify the table in your query:
>
> SELECT COUNT (login) FROM <TABLENAME GOES HERE> WHERE login = '$login'
>
>
>>Hi,
>>
>>Why this doesn't work ?
>>
>>---
>>$query = "SELECT COUNT (login) FROM WHERE login = '$login'";
>>$result = mysql_query($query);
>>mysql_fetch_row($result);
>>---
>>
>>It gives me
>>Warning: mysql_fetch_row(): supplied argument is not a valid MySQL
>>result resource in /var/www/html/registar_action.php on line 22
>>
>>Any help would be apreciated.
>>
>>Warm Regards,
>>MΓ‘rio Gamito
>>
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>>
>>
>>
>
>
>
> Thank you,
> Colin Shreffler
> Principal
> 303.349.9010 - cell
> 928.396.1099 - fax
> colin.shreffler@warp9software.com
>
> Warp 9 Software, LLC.
> 6791 Halifax Avenue
> Castle Rock, CO 80104
>
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