sending databese info to new page

locopoco

while ($Row = mysql_fetch_array ($Result))
{
print "$Row[firstname]\n";
print "$Row[lastname]\n";
print "$Row[username]\n";
print "$Row[password]\n";
}

$Result is the information I have received from a database using SELECT * FROM 'table'.

The above while loop retreives information one row/array at a time.

I want to use the information on the next page.
I have tried registering $Result with a session (ie session_register("Result") ). and then tried to fetch the array on the next page. this did not work.

Can you help?

mourisjj

Hello,

I encounter the same database-problem (mysql + php) and I am unable to figure out, where the error is located.

The job is to list the content of a DB-table with a link for every row; when clicking the link, you should get a form displaying the data and enabling to change it. Although I perform a mysql_pconnect to the DB, I am unable to access the result set in the second window.

Here are some code-snippets, to show what I want to do:

FILE 1:
/ Opening the connection /
$resrc = mysql_pconnect("localhost", "username","");
mysql_select_db("anasig");
/ Selecting the content of the table /
$tbl = "select from menu";
$lst_tbl = mysql_query($tbl);
/ Display some data /
(...)
/ Preparing the data to hand it over to the next script /
$res_set = rawurlencode($lst_tbl); // without the urlencode, I have trouble with blanks in $lst_tbl
echo "<td><a href=\"auto_change.php?row_num=$j&res_set=$res_set\">$j</a></td>\n";
(... Displaying more data ...)
/ Up to this point, everything works like I imagined */

FILE 2:
/ Opening the connection /
$resrc = mysql_pconnect("localhost", "username","");
mysql_select_db("anasig");
/ I want to set the row-pointer (in $row) for the previousely used result set (in $res_set) /
$lst_tbl = rawurldecode($res_set);
$check = mysql_data_seek($lst_tbl, $row_num);
/ Here is where the trouble starts: I get the following error:
Warning: Supplied argument is not a valid MySQL result resource in /home/my_path/auto_change.php on line 10
Apparentely, my result-set is not recognized (or available?) anymore ??
/

Am I thinking wrong about how to do this, or is there an error in my code?

Thanks for your help.

jjm

Cinquecento

I think the problem is you are trying to pass an array to the next page - which can't work...

You next to 'explode' the array to turn it into a string - and then 'implode' it on the next page to turn it back into an array.

Why not just pass the key to the next page and run a new query there - or put the edit fields in a conditional, include them on the same page and submit the relevent fields to itself with a trigger 'editmode' variable?

Helpful? Doubt it!

Ian

mourisjj

I don't think I am converting an array into a string and back again in the next script. This is simply because the value of my $lst_tbl variable is a value like "Resource id #2", which needs to be urlencoded to have something like "...auto_change.php?row_num=1&res_set=Resource%20id%20%232" in the URL.

The trouble with the key is, that I want this script to be applicable to any database table, regardless of its structure: whithout any knowledge about the primary key(s) of the table, I wanted to easily get back to the already selected data by using the row_num in the result_set. This way, I would have not to worry about multiple-colum primary keys, data changed by other users while goiing from one script to the next, etc.

Anyway, I can ever be sure to get the same row-order from the database when executing the same select twice, specially if other people are working on te DB simoultanousely.

Now, I'm afraid the only way to do the thing I want to do is only possible if I use a separate column in every table with some kind of row_id inside. This way, at least, I could be sure to get to the table-row I want to change, although this is not the way I'd like this to work. It is not really "pretty", and there is another query to run.

On the other hand, I could figure out a way to pass the values to a form and submit them for change, but this brings the trouble with the key right back.

I am really surprised that there seems to be no way to work in script 2 on a result-set created in script 1! When reading the PHP-documentation of mysql-functions, I thought I had found exactely what I was looking for when reading about mysql_pconnect:

"mysql_pconnect() acts very much like mysql_connect() with two major differences.

First, when connecting, the function would first try to find a (persistent) link that's already open with the same host, username and password. If one is found, an identifier for it will
be returned instead of opening a new connection.

Second, the connection to the SQL server will not be closed when the execution of the script ends. Instead, the link will remain open for future use (mysql_close() will not close
links established by mysql_pconnect()).

This type of links is therefore called 'persistent'. "

Either there is a serious misunderstanding of the documentation of mysql_pconnect, or there is something wrong about my connections.

Any help for solving my problem would be greatly appreciated. How do other people do this kind of job? I imagine, I am not the first to to this kind of thigs, am I?

Best regards,

Jean-Jacques