> That the thing see, the form(using GET instead of POST) sends data with
> the new dates like this:-
>
> id=1&name=Mark&job_number=AA1&job_date=2002-5-19
> &id=2&name=Mark&job_number=AA2&job_date=2002-5-21
> &id=3&name=Mark&job_number=AA3&job_date=2002-5-25
>
> and the UPDATE query like this:-
>
> >   $sql = "UPDATE mytable set job_date='$job_date' where job_number
> >   ='$job_number'";
> >   $result = @mysql_query($sql, $connection) or die.....;
>
> just updates the last record.
>
> Can the query be written so that it updates each of the three records?
>
> cheers
>
> Mark
Hi Mark
Your problem is misused of form transfer data please notify that by GET
method the data transfer to server with name and equa sign (=) and then
value each data seperate by & so

id=1&name=Mark&job_number=AA1&job_date=2002-5-19

mean the data have 3 elecment id,name and job_date as your data you send 9
element but use the same name so finally it only have 3 element and the
value id the last assign so you will got at last

id=3&name=Mark&job_number=AA3&job_date=2002-5-25

the first and the second set has over write so when you update data you got
the last data (id=3) update only

how to solve this problem you shoul declare the name seperate like this

 id1=1&name1=Mark&job_number1=AA1&job_date1=2002-5-19
 &id2=2&name2=Mark&job_number2=AA2&job_date2=2002-5-21
 &id3=3&name3=Mark&job_number3=AA3&job_date3=2002-5-25

and update 3 time

$sql = "UPDATE mytable set job_date='$job_date1' where job_number
   ='$job_number1'";
@mysql_query($sql, $connection) or die.....;

$sql = "UPDATE mytable set job_date='$job_date2' where job_number
   ='$job_number2'";
@mysql_query($sql, $connection) or die.....;

$sql = "UPDATE mytable set job_date='$job_date3' where job_number
   ='$job_number3'";
@mysql_query($sql, $connection) or die.....;


May be your answer.
Kittiphum Worachat,M.T.
www.hatyailab.com


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