Sheila,

You have restored my sanity;  it works as hoped for.   Thank you.

I have a tenuous grasp of the logic and will study this further.   I should
have been able to get closer than I did, but I see that I need to recognise
how much php separates the steps in a script.    I did  get three other
scripts working on this archiving db and must have got too confident with
this, most complex, one!

Again thank you and everyone who has helped step by step through my muddle.

Joseph

> Joseph Harris wrote:
> >
> > I always hesitate to make mails too long with code,
>
> If you have someplace online to post the .php file as .txt then your can
> just reference the file in you email.
>
>
> OK, you've got extra calls to mysql_query() and you are using the return
> value from mysql_query() incorrectly.
>
> Here's part of the code, rewritten and commented, starting from:
>
> echo "<p>$source</p>";
>
> // define a query to select all poem ids where
> // the title matches the input.
>
> $query = "SELECT  spw_poems_id  FROM  spw_poems
>      WHERE spw_poems_title = '{$_POST['poem_title']}' ";
>
> // submit the query to mysql
> $result = mysql_query($query);
>
> // get the result of the query
> if (!$result) {
>        // there was a problem
>        echo mysql_error();
>        echo " SQL: $query \n";
>
> } else {
>        // read the results
>        // since there could be more than one id returned,
>        // we use a while loop
>        while ($row = mysql_fetch_row($result)) {
>           $spw_poems_id[] = $row[0];
>        }
>   }
>
> // print all the id's returned, there could be more than one
> print_r($spw_poems_id);
>
>
> // setup a new query
> $query2 = "SELECT spw_type_id FROM spw_type
>         WHERE spw_type_description='{$_POST['poemtype']}'";
>
> // submit the query to mysql
> $result = mysql_query($query2);
>
> // get the result of the query
> if (!$result) {
>        // there was a problem
>        echo mysql_error();
>        echo " SQL: $query2 \n";
> } else {
>        // read the results
>        // since there could be more than one id returned,
>        // we use a while loop
>        while ($row = mysql_fetch_row($result)) {
>            $spw_type_id[] = $row[0];
>        }
> }
>
> // now we have an array of type ids
> print_r($spw_type_id);
>
> // at this point you have two arrays of id's
> // get the first item in each array
> $title = $spw_poems_id[0];
> $type = $spw_type_id[0];
>
> // setup SQL for insert
> $sql = "INSERT INTO spw_typelist (
>         spw_typelist_id ,
>         spw_poems_id,
>         spw_type_id) VALUES ('',
>         '$title',
>         '$type')
>          ";
>
> // run the query
> $result = mysql_query($sql);
> if(!$result){
>       echo "Error inserting your info into mysql:  ".mysql_error();
> }
>
> footer();
> exit();
>
>
> That should do it. I assume someone else will point out any errors I've
> made.
>
> Sheila
>


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