("INSERT INTO agents (agent_name)
>VALUES ('$agentname')
>
> but I need to find out the ID value it created, how can I do this easily?
>
> Thanks
>
> Chris
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gt;
> @ $db = mysql_connect("localhost");
> mysql_select_db("me2resh00");
> $query = "insert into recipients (recipient_name, recipient_email) values
> ('"friend"', '".$mails[$i]."')";
> $result = mysql_query($query);
> if ($result)
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ted was to use an ODBC driver to interface
> between PHP and MySQL but indicated that this may be slower.
>
> My question, therefore, is - is there a way to upgrade the MySQL
> client in PHP 4.3.3? If not, what are the alternatives (apart from
> reverting back to MySQL 4.0.x
he following query
>> SELECT Question.Text_Long, AVG( Response ) FROM `Response` INNER JOIN
>> Question ON Question.Question_Key = Response.Question_Key WHERE
>> Question.Question_Key LIKE '2003%' GROUP BY Response.Question_Key ORDER
> BY
>> Question.Question_Key ASC
>
> What does EXPLAIN tell you for this query? Is it using your indexes?
>
> ---John Holmes...
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p by grade having cnt>1
>> and then
>> select * from studs where sno in ('result of previous query')
>>
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ow can i update every row so that
> all of the times for Booking_Start_Date are 09.00 and all of the times for
> Booking_End_Date are 17.30, without affecting any of the dates?
>
> Thanks for your help
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to update Projects with one query?
>
> Thanks for your help
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as a work type associated with it, currently the Work Type
> column of the Project table holds the work type name, however i have now
> added a Work_Type_ID column to the Work_Type table and i would like Projects
> to hold the Work_Type_ID as opposed to the Work Type Name.
>
> Thanks for
once it has done this, I need to get the id it has generated in
> the DB and then update that record with the id it generate, so how
> can I get the id from the above insert?
>
> Any help would be greatly appreciated :-)
>
> Regards
>
> Chris
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