Re: [PHP-DB] Resource link errors
Thanks Rick, Now it works. On Thursday 07 February 2002 03:04 pm, Rick Emery wrote: > The following means you have not opened a link to your database. > Concerntrate your efforts there. > > Warning: Supplied argument is not a valid MySQL-Link > resource in /var/www/html/list.php3 on line 26 > > -Original Message- > From: Ken Thompson [mailto:[EMAIL PROTECTED]] > Sent: Thursday, February 07, 2002 3:23 PM > To: [EMAIL PROTECTED] > Subject: [PHP-DB] Resource link errors > > > Hello all, > I'm very new to php and have been beating my head on the wall over this for > 2 > days. I've searched through the php documentation but I guess I either > don't > > know where or what to look for or don't understand what I'm seeing. > I think this has to be an array but the original doesn't seem to think > so.(see below) > I've tried ($myrow = mysql_fetch_array($result)) and still get the same > errors. > The original that I used as an example works OK showing the results of > Name, > > Position. They didn't include a data cell for $myrow[3] ... > = > THIS WORKS: > > $title = "Welcome to the Web-Site of"; /*I commented this out and finaly > removed it cuz I don't want the title echoed in the final page.*/ > > include("header.inc"); > > $result = mysql_query("SELECT * FROM employees",$db); > > echo "\n"; > > echo "NamePosition\n"; > > while ($myrow = mysql_fetch_row($result)) { > > echo("%s %s%s\n", $myrow[1], $myrow[2], > $myrow[3]); > > } > > echo "\n"; > include("footer.inc"); > > ?> > === > What I am trying to do is to return the results of the query into a table > with these items: > +---+-+--+-+-++ > > | Field | Type| Null | Key | Default | Extra | > > +---+-+--+-+-++ > > | id| tinyint(4) | | PRI | NULL| auto_increment | > | year | int(4) | YES | | NULL|| > | make | varchar(20) | YES | | NULL|| > | model | varchar(20) | YES | | NULL|| > | engine| varchar(20) | YES | | NULL|| > | fuel | varchar(8) | YES | | NULL|| > | trans | varchar(10) | YES | | NULL|| > | condition | varchar(20) | YES | | NULL|| > | price | varchar(10) | YES | | 0 || > | location | varchar(20) | YES | | NULL|| > > +---+-+--+-+-++ > These are the errors that show up when I run the page in the browser. > The database is local and I'm on Linux-Mandrake 8.1 using Apache web > server. I can get the results I want from the command line, that is I can > make queries and get the proper info returned. > what have I done to cause it not to work? > > > Warning: Supplied argument is not a valid MySQL-Link > resource in /var/www/html/list.php3 on line 26 > > Warning: Supplied argument is not a valid MySQL result > resource in /var/www/html/list.php3 on line 32 > > THIS DOESN'T: > I get the table and the item name e.g. Year, Make etc. > Once I get the 3 cells returning the proper info, I want to add the rest of > the cells needed to display all the info in the web page. I don't think it > should be a problem, comments on this? > > > include("header.inc"); > > //(Line 26 below, everything above " $result = mysql_query("SELECT * FROM autos",$db); > > echo "\n"; > > echo "YearMakeModel\n"; > > //(Line 32 below) > while ($myrow = mysql_fetch_row($result)) { > > //What does the '%s' mean? I read it someplace in the tutorial > //but can't find it now. > > printf("%s %s%s\n", $myrow[1], $myrow[2], > $myrow[3]); > > } > > echo "\n"; > include("footer.inc"); > > ?> -- Ken Thompson, North West Antique Autos Payette, Idaho Email: [EMAIL PROTECTED] http://www.nwaa.com Sales and brokering of antique autos and parts. Linux- Coming Soon To A Desktop Near You Registered Linux User #183936 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Resource link errors
The following means you have not opened a link to your database. Concerntrate your efforts there. Warning: Supplied argument is not a valid MySQL-Link resource in /var/www/html/list.php3 on line 26 -Original Message- From: Ken Thompson [mailto:[EMAIL PROTECTED]] Sent: Thursday, February 07, 2002 3:23 PM To: [EMAIL PROTECTED] Subject: [PHP-DB] Resource link errors Hello all, I'm very new to php and have been beating my head on the wall over this for 2 days. I've searched through the php documentation but I guess I either don't know where or what to look for or don't understand what I'm seeing. I think this has to be an array but the original doesn't seem to think so.(see below) I've tried ($myrow = mysql_fetch_array($result)) and still get the same errors. The original that I used as an example works OK showing the results of Name, Position. They didn't include a data cell for $myrow[3] ... = THIS WORKS: \n"; echo "NamePosition\n"; while ($myrow = mysql_fetch_row($result)) { echo("%s %s%s\n", $myrow[1], $myrow[2], $myrow[3]); } echo "\n"; include("footer.inc"); ?> === What I am trying to do is to return the results of the query into a table with these items: +---+-+--+-+-++ | Field | Type| Null | Key | Default | Extra | +---+-+--+-+-++ | id| tinyint(4) | | PRI | NULL| auto_increment | | year | int(4) | YES | | NULL|| | make | varchar(20) | YES | | NULL|| | model | varchar(20) | YES | | NULL|| | engine| varchar(20) | YES | | NULL|| | fuel | varchar(8) | YES | | NULL|| | trans | varchar(10) | YES | | NULL|| | condition | varchar(20) | YES | | NULL|| | price | varchar(10) | YES | | 0 || | location | varchar(20) | YES | | NULL|| +---+-+--+-+-++ These are the errors that show up when I run the page in the browser. The database is local and I'm on Linux-Mandrake 8.1 using Apache web server. I can get the results I want from the command line, that is I can make queries and get the proper info returned. what have I done to cause it not to work? Warning: Supplied argument is not a valid MySQL-Link resource in /var/www/html/list.php3 on line 26 Warning: Supplied argument is not a valid MySQL result resource in /var/www/html/list.php3 on line 32 THIS DOESN'T: I get the table and the item name e.g. Year, Make etc. Once I get the 3 cells returning the proper info, I want to add the rest of the cells needed to display all the info in the web page. I don't think it should be a problem, comments on this? \n"; echo "YearMakeModel\n"; //(Line 32 below) while ($myrow = mysql_fetch_row($result)) { //What does the '%s' mean? I read it someplace in the tutorial //but can't find it now. printf("%s %s%s\n", $myrow[1], $myrow[2], $myrow[3]); } echo "\n"; include("footer.inc"); ?> -- Ken Thompson, North West Antique Autos Payette, Idaho Email: [EMAIL PROTECTED] http://www.nwaa.com Sales and brokering of antique autos and parts. Linux- Coming Soon To A Desktop Near You Registered Linux User #183936 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Resource link errors
Hello all, I'm very new to php and have been beating my head on the wall over this for 2 days. I've searched through the php documentation but I guess I either don't know where or what to look for or don't understand what I'm seeing. I think this has to be an array but the original doesn't seem to think so.(see below) I've tried ($myrow = mysql_fetch_array($result)) and still get the same errors. The original that I used as an example works OK showing the results of Name, Position. They didn't include a data cell for $myrow[3] ... = THIS WORKS: \n"; echo "NamePosition\n"; while ($myrow = mysql_fetch_row($result)) { echo("%s %s%s\n", $myrow[1], $myrow[2], $myrow[3]); } echo "\n"; include("footer.inc"); ?> === What I am trying to do is to return the results of the query into a table with these items: +---+-+--+-+-++ | Field | Type| Null | Key | Default | Extra | +---+-+--+-+-++ | id| tinyint(4) | | PRI | NULL| auto_increment | | year | int(4) | YES | | NULL|| | make | varchar(20) | YES | | NULL|| | model | varchar(20) | YES | | NULL|| | engine| varchar(20) | YES | | NULL|| | fuel | varchar(8) | YES | | NULL|| | trans | varchar(10) | YES | | NULL|| | condition | varchar(20) | YES | | NULL|| | price | varchar(10) | YES | | 0 || | location | varchar(20) | YES | | NULL|| +---+-+--+-+-++ These are the errors that show up when I run the page in the browser. The database is local and I'm on Linux-Mandrake 8.1 using Apache web server. I can get the results I want from the command line, that is I can make queries and get the proper info returned. what have I done to cause it not to work? Warning: Supplied argument is not a valid MySQL-Link resource in /var/www/html/list.php3 on line 26 Warning: Supplied argument is not a valid MySQL result resource in /var/www/html/list.php3 on line 32 THIS DOESN'T: I get the table and the item name e.g. Year, Make etc. Once I get the 3 cells returning the proper info, I want to add the rest of the cells needed to display all the info in the web page. I don't think it should be a problem, comments on this? \n"; echo "YearMakeModel\n"; //(Line 32 below) while ($myrow = mysql_fetch_row($result)) { //What does the '%s' mean? I read it someplace in the tutorial //but can't find it now. printf("%s %s%s\n", $myrow[1], $myrow[2], $myrow[3]); } echo "\n"; include("footer.inc"); ?> -- Ken Thompson, North West Antique Autos Payette, Idaho Email: [EMAIL PROTECTED] http://www.nwaa.com Sales and brokering of antique autos and parts. Linux- Coming Soon To A Desktop Near You Registered Linux User #183936 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php