mysql_fetch_array problem

I set up phpmyadmin, and it works fine, but my code isn't working. id is
the correct column (it is the primary key), stock is the correct database.

CODE:
$qresult = mysql_query("SELECT id FROM 'stock'");
echo ("START<P><HR>");
while ($row = mysql_fetch_array($qresult)) {
echo ($row ["name"]);
echo ("<br>");
}
echo ("<P><HR><P>END");

OUTPUT:
START
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
result resource in /home/houseproudlancs_co_uk/index_to_be.php on line 14
END

does anybody have any idea's why?

just thought i would add that i am using uk2.net as my hosting, and mysql
and php are setup fine.

TIA

Matthew Robinson wrote:
> OUTPUT:
> START
> Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
> result resource in /home/houseproudlancs_co_uk/index_to_be.php on line 14
> END
>
> does anybody have any idea's why?

Use error checking!


Instead of

<?php
$qresult = mysql_query("select ...");
?>

do

<?php
$qresult = mysql_query("select ...") or die(mysql_error());
?>

or, easier to interpret the (eventual) error

<?php
$command = 'select ...';
$qresult = mysql_query($command) or die('Error ' . mysql_error() .
' in ' . $command);
?>
--
--= my mail box only accepts =--
--= Content-Type: text/plain =--
--= Size below 10001 bytes =--

Matthew Robinson wrote:
> I set up phpmyadmin, and it works fine, but my code isn't working. id is
> the correct column (it is the primary key), stock is the correct database.
>
> CODE:
> $qresult = mysql_query("SELECT id FROM 'stock'");
> echo ("START<P><HR>");
> while ($row = mysql_fetch_array($qresult)) {
> echo ($row ["name"]);
> echo ("<br>");
> }
> echo ("<P><HR><P>END");
>
> OUTPUT:
> START
> Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
> result resource in /home/houseproudlancs_co_uk/index_to_be.php on line 14
> END
>
> does anybody have any idea's why?
>


Omit the single quotes from your query, like so:
$qresult = mysql_query("SELECT id FROM stock");

Putting single quotes around the table name causes a SQL error.

Regards,

- Dan
dantripp.com

$command = "SELECT id FROM stock";
$qresult = mysql_query($command) or die('Error ' mysql_error() ' in '
$command); //$qresult = mysql_query("SELECT id FROM stock"); echo
("START<P><HR>");

while ($row = mysql_fetch_array($qresult)) {
echo ($row ["name"]);
echo ("<br>");
}
echo ("<P><HR><P>END");


gives this error: Parse error: parse error in
/home/houseproudlancs_co_uk/index_to_be.php on line 12

line 12 in the page is the second line on this post ( $qresult =
mysql_query($command) or die('Error ' mysql_error() ' in ' $command);)

Matthew Robinson wrote:
> $qresult = mysql_query($command) or die('Error ' mysql_error() ' in '
> $command); //$qresult = mysql_query("SELECT id FROM stock"); echo
> ("START<P><HR>");

> gives this error: Parse error: parse error in
> /home/houseproudlancs_co_uk/index_to_be.php on line 12


You lack the dots to concatenate the strings.

$qresult = mysql_query($command) or die('Error ' . mysql_error() . ' in ' . $command);
// ______________________________________________^___ ____________^________^___________

// sorry for the long lines :)
--
--= my mail box only accepts =--
--= Content-Type: text/plain =--
--= Size below 10001 bytes =--

removing the quotes from "SELECT id FROM stock"; didn't work either.

thanks - it works now, well that bit anyway. here's the entire page, and
the output below it. im trying to work out php/mysql, so im just trying to
get it to print out all the fields in the 'name' field in the 'stock'
table in the 'houseproudlancs_co_uk1' database.

<?php

$dbcnx = @mysql_connect("server", "username", "password");

$select = @mysql_select_db("houseproudlancs_co_uk1");


$command = "SELECT id FROM stock";
$qresult = mysql_query($command) or die('Error ' . mysql_error() . ' in ' . $command); echo ("START<P><HR>");

while ($row = mysql_fetch_array($qresult)) {
echo ($row ["name"]);
echo ("<br>");
}
echo ("<P><HR><P>END");

?>



START
--------------------------------


--------------------------------
END

Matthew Robinson wrote:
><?php
> $dbcnx = @mysql_connect("server", "username", "password");
> $select = @mysql_select_db("houseproudlancs_co_uk1");
>
> $command = "SELECT id FROM stock";
> $qresult = mysql_query($command) or die('Error ' . mysql_error() . ' in ' . $command);
> echo ("START<P><HR>");
>
> while ($row = mysql_fetch_array($qresult)) {
> echo ($row ["name"]);

Do you want the "name" or the "id"?

At this point $row['name'] -- I prefer single quotes :) -- does not exist.

Either also select the name: "SELECT id, name FROM stock"
or change id's identification: "SELECT id as name FROM stock"

> START
> --------------------------------
empty, of course :)
> --------------------------------
> END
--
--= my mail box only accepts =--
--= Content-Type: text/plain =--
--= Size below 10001 bytes =--

oh ye, just realised i didn't mention that the database has 2 rows, both
with the 'name' column not null, so the output should be different to what
it is.