syntax error
This is a discussion on syntax error within the PHP General forums, part of the PHP Programming Forums category; On a system with php4 and mysql 4.x I had these lines: require("../db-config"); // includes $dbhost, $...
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07-21-2008
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syntax error
On a system with php4 and mysql 4.x I had these lines:
require("../db-config"); // includes $dbhost, $buname, $dbpass $db = mysql_connect($dbhost, $dbuname, $dbpass); mysql_select_db($dbname,$db); $sql = "SELECT * FROM CATEGORY WHERE ....."; $result = mysql_query($sql,$db); $num=mysql_num_rows($result); while ($myrow = mysql_fetch_array($result)) { ..... Moving the *.php to a php5 and mysql 5.x site I get these errors: PHP Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in ... PHP Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in ... Looking at the manual, I cannot see what I am doing wrong. bye R. 
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07-21-2008
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Re: [PHP] syntax error
Not a syntax error. It's not successfully connecting to the database. Check your settings. ~Ted On 21-Jul-08, at 4:24 AM, Ronald Wiplinger wrote: > On a system with php4 and mysql 4.x I had these lines: > > require("../db-config"); // includes $dbhost, $buname, $dbpass > $db = mysql_connect($dbhost, $dbuname, $dbpass); > mysql_select_db($dbname,$db); > > $sql = "SELECT * FROM CATEGORY WHERE ....."; > $result = mysql_query($sql,$db); > $num=mysql_num_rows($result); > while ($myrow = mysql_fetch_array($result)) { > .... > > Moving the *.php to a php5 and mysql 5.x site I get these errors: > > PHP Warning: mysql_num_rows(): supplied argument is not a valid MySQL > result resource in ... > PHP Warning: mysql_fetch_array(): supplied argument is not a valid > MySQL > result resource in ... > > > Looking at the manual, I cannot see what I am doing wrong. > > bye > > R.
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07-21-2008
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Re: [PHP] syntax error
On Mon, Jul 21, 2008 at 7:24 AM, Ronald Wiplinger <tm.ronald@gmail.com> wrote:
> On a system with php4 and mysql 4.x I had these lines: > > require("../db-config"); // includes $dbhost, $buname, $dbpass > $db = mysql_connect($dbhost, $dbuname, $dbpass); > mysql_select_db($dbname,$db); > > $sql = "SELECT * FROM CATEGORY WHERE ....."; > $result = mysql_query($sql,$db); > $num=mysql_num_rows($result); > while ($myrow = mysql_fetch_array($result)) { > .... > > Moving the *.php to a php5 and mysql 5.x site I get these errors: > > PHP Warning: mysql_num_rows(): supplied argument is not a valid MySQL > result resource in ... > PHP Warning: mysql_fetch_array(): supplied argument is not a valid MySQL > result resource in ... > > > Looking at the manual, I cannot see what I am doing wrong. > > bye > > R. > Your mysql_connect is probably failing. Look at your error log or look at mysql_error() for a reason why.
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07-21-2008
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Re: [PHP] syntax error
Eric Butera wrote:
> On Mon, Jul 21, 2008 at 7:24 AM, Ronald Wiplinger <tm.ronald@gmail.com> wrote: > >> On a system with php4 and mysql 4.x I had these lines: >> >> require("../db-config"); // includes $dbhost, $buname, $dbpass >> $db = mysql_connect($dbhost, $dbuname, $dbpass); >> mysql_select_db($dbname,$db); >> >> $sql = "SELECT * FROM CATEGORY WHERE ....."; >> $result = mysql_query($sql,$db); >> $num=mysql_num_rows($result); >> while ($myrow = mysql_fetch_array($result)) { >> .... >> >> Moving the *.php to a php5 and mysql 5.x site I get these errors: >> >> PHP Warning: mysql_num_rows(): supplied argument is not a valid MySQL >> result resource in ... >> PHP Warning: mysql_fetch_array(): supplied argument is not a valid MySQL >> result resource in ... >> >> >> Looking at the manual, I cannot see what I am doing wrong. >> >> bye >> >> R. >> >> > > Your mysql_connect is probably failing. Look at your error log or > look at mysql_error() for a reason why. Probably the mysql extension is not found or not loaded (due to not being compiled with the right path or the default path in PHP is not the right one). Happened to me a couple of times..... -- Aschwin Wesselius /'What you would like to be done to you, do that to the other....'/
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07-21-2008
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Re: [PHP] syntax error
On Mon, Jul 21, 2008 at 7:55 AM, Aschwin Wesselius
<aschwin@illuminated.nl> wrote: > Eric Butera wrote: > > On Mon, Jul 21, 2008 at 7:24 AM, Ronald Wiplinger <tm.ronald@gmail.com> > wrote: > > > On a system with php4 and mysql 4.x I had these lines: > > require("../db-config"); // includes $dbhost, $buname, $dbpass > $db = mysql_connect($dbhost, $dbuname, $dbpass); > mysql_select_db($dbname,$db); > > $sql = "SELECT * FROM CATEGORY WHERE ....."; > $result = mysql_query($sql,$db); > $num=mysql_num_rows($result); > while ($myrow = mysql_fetch_array($result)) { > .... > > Moving the *.php to a php5 and mysql 5.x site I get these errors: > > PHP Warning: mysql_num_rows(): supplied argument is not a valid MySQL > result resource in ... > PHP Warning: mysql_fetch_array(): supplied argument is not a valid MySQL > result resource in ... > > > Looking at the manual, I cannot see what I am doing wrong. > > bye > > R. > > > > Your mysql_connect is probably failing. Look at your error log or > look at mysql_error() for a reason why. > > Probably the mysql extension is not found or not loaded (due to not being > compiled with the right path or the default path in PHP is not the right > one). Happened to me a couple of times..... > > > -- > > Aschwin Wesselius > > 'What you would like to be done to you, do that to the other....' > Wouldn't that be call to undefined function then?
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07-21-2008
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Re: [PHP] syntax error
Eric Butera wrote:
> On Mon, Jul 21, 2008 at 7:55 AM, Aschwin Wesselius > <aschwin@illuminated.nl> wrote: > >> Probably the mysql extension is not found or not loaded (due to not being >> compiled with the right path or the default path in PHP is not the right >> one). Happened to me a couple of times..... >> >> >> -- >> >> Aschwin Wesselius >> >> 'What you would like to be done to you, do that to the other....' >> >> > > Wouldn't that be call to undefined function then? Oh my..... I'm so sorry. You're absolutely right. I mixed these two. I'm a bit side-tracked today..... Thanks for pointing that out. -- Aschwin Wesselius /'What you would like to be done to you, do that to the other....'/
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07-21-2008
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Re: [PHP] syntax error
On Mon, Jul 21, 2008 at 8:32 AM, Aschwin Wesselius
<aschwin@illuminated.nl> wrote: > Oh my..... I'm so sorry. You're absolutely right. I mixed these two. I'm a > bit side-tracked today..... > > Thanks for pointing that out. > > > -- > > Aschwin Wesselius It's no problem, I got scared I was spreading propaganda! hehe :)
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07-21-2008
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Re: [PHP] syntax error
On Mon, Jul 21, 2008 at 7:24 AM, Ronald Wiplinger <tm.ronald@gmail.com> wrote:
Try this: > $result = mysql_query($sql,$db) or die(mysql_error()); -- </Daniel P. Brown> Better prices on dedicated servers: Intel 2.4GHz/60GB/512MB/2TB $49.99/mo. Intel 3.06GHz/80GB/1GB/2TB $59.99/mo. Dedicated servers, VPS, and hosting from $2.50/mo.
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07-21-2008
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Re: [PHP] syntax error
Please keep replies on list for all to benefit and be able to assist.
On Mon, Jul 21, 2008 at 5:47 PM, Ronald Wiplinger <tm.ronald@gmail.com> wrote: > > On Tue, Jul 22, 2008 at 1:51 AM, Daniel Brown <parasane@gmail.com> wrote: >> >> Try this: >> >> > $result = mysql_query($sql,$db) or die(mysql_error()); Did you try my suggestion above? If so, did you receive any errors? > echo "result=$result<br>"; > $num=mysql_num_rows($result); > echo "num=$num"; > > I get: > > result= > num= The most you'd get from $result in this case would be a resource identifier message, because that's all mysql_query() returns. And since the resource link doesn't seem to be correctly established, $num will be empty. -- </Daniel P. Brown> Better prices on dedicated servers: Intel 2.4GHz/60GB/512MB/2TB $49.99/mo. Intel 3.06GHz/80GB/1GB/2TB $59.99/mo. Dedicated servers, VPS, and hosting from $2.50/mo.
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