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Posted by Kimmo Laine on 12/18/18 11:50
<cdtsly@relayeur.com> wrote in message
news:1150444378.177655.41090@c74g2000cwc.googlegroups.com...
> Hi i have a table with all value at 4
> i select all lines in a fetch
> i update one with a value of 7
This is where it goes wrong. You update nothing, see explanation below.
> i update all the row in the fetch with a value 5
>
> the result is that all my row are at 5
> and the only line which should be at 7 is at 5 too.
>
> i see that the problem is at the first mysql_db_query, and i don t know
> how to solve it
>
> thanks
>
>
> $query="SELECT * FROM val WHERE str='4'";
> $result=mysql_db_query($datamysql,$query);
> $query7="UPDATE val SET str='7',updated='1' WHERE val='2'";
You update all rows whose val=2 (should this be 4 maybe?), that is none,
because all rows val=4. Nothing is updated. Even if it was, the old record
that you fetched earlier are intact, only the database is updated but the
updates do not reflect on already fetched records. So when you select
something, then update something, the old values are still inside $result,
so the updating means nothing again.
> $result7=mysql_db_query($datamysql,$query7);
>
> while ($row=mysql_fetch_array($result))
> {
>
> $str=$row["str"];
> $val=$row["val"];
> $query2="UPDATE val SET str='5' WHERE val='$val'";
> $result2=mysql_db_query($datamysql,$query2);
>
>
> }
>
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