[PHP-DB] mysql_fetch_array problem
Greetings, I'm experiencing the strangest problem and I was wondering if anyone else has had the same problem. I have a fairly simple script setup that queries a mySQL database and displays the records in a HTML table. Everything works fine except it keeps omitting the first record. Running the query directly on the database returns 3 records but only 2 are displayed in the table. I just upgraded to PHP 4.0.6 and I'm still having the problem. I've also tried using mysql_fetch_array and mysql_fetch_object, both produce the same results. The first record is left out every time. Any idea as to what the problem might be? Thanks. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] mysql_fetch_array problem
I was having the same problem for a while... although, I was using this: for($i=0;$imysql_num_rows($result);$i++) doStuffTo(mysql_result($result, $i, foo); If I remember correctly... it has to do with zero-based indexing versus 1-based indexing. Now... I think I fixed it by using = instead of just . But, it has been a while so I could be totally off = Good luck, Ben Quoting BrianSander [EMAIL PROTECTED]: Greetings, I'm experiencing the strangest problem and I was wondering if anyone else has had the same problem. I have a fairly simple script setup that queries a mySQL database and displays the records in a HTML table. Everything works fine except it keeps omitting the first record. Running the query directly on the database returns 3 records but only 2 are displayed in the table. I just upgraded to PHP 4.0.6 and I'm still having the problem. I've also tried using mysql_fetch_array and mysql_fetch_object, both produce the same results. The first record is left out every time. Any idea as to what the problem might be? Thanks. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP-DB] mysql_fetch_array problem...!
I want to run a query to my db, fetching different fields from three different tables. In order to recognise the individual fields I give them names: select portal.portal as portal, portal.portalid as id... etc etc How can I refer to one specific row in this query..? What I mean is, how can i refer to result row number 4...? If I only selected rows from one table I could do something like this: $i = mysql_fetch_array($sql) ; echo "$i[4]" ; -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] mysql_fetch_array problem...!
How can I refer to one specific row in this query..? What I mean is, how can i refer to result row number 4...? If I only selected rows from one table I could do something like this: $i = mysql_fetch_array($sql) ; echo "$i[4]" ; Actually, no. mysql_fetch_array return the _current row_ from the query (use this function in a loop to process each row of the resulting data). $i[4] in your case is the fourth field in the current row. If you really don't want to process the rows in order, look at the documentation for mysql_data_seek; it can be used to jump around the result set. - Darryl -- Darryl Friesen, B.Sc., Programmer/Analyst[EMAIL PROTECTED] Education Research Technology Services, http://gollum.usask.ca/ Department of Computing Services, University of Saskatchewan -- "Go not to the Elves for counsel, for they will say both no and yes" -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] mysql_fetch_array problem...!
You must use a loop to show each row of your query. "mysql_fetch_array($result)" only get the current row. while($row = mysql_fetch_array($result) ){ // do something } The "mysql_fetch_array" function returns an associative array where you can use the field names to have access to its value : $row = mysql_fetch_array($result) ; $v1 = $row["fieldname1"] ; $v2 = $row["fieldname2"] ; $v3 = $row["fieldname3"] ; You're going to use the alias names you are creating on the query. See more details on php manual at mysql functions. HTH. Jayme. -Mensagem Original- De: Trond Erling Hundal [EMAIL PROTECTED] Para: PHP-DB-LIST [EMAIL PROTECTED] Enviada em: segunda-feira, 5 de maro de 2001 09:56 Assunto: [PHP-DB] mysql_fetch_array problem...! I want to run a query to my db, fetching different fields from three different tables. In order to recognise the individual fields I give them names: select portal.portal as portal, portal.portalid as id... etc etc How can I refer to one specific row in this query..? What I mean is, how can i refer to result row number 4...? If I only selected rows from one table I could do something like this: $i = mysql_fetch_array($sql) ; echo "$i[4]" ; -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]