Re: [PHP-DB] Problems with mysql_num-rows()
The variables are not being expanded. As a minimum, you need to escape the single quotes. $query = "SELECT * from user where name=\'$userid\' and pass=password(\'$password\')" Alternatively, you can concatenate the expanded variables. $query = "SELECT * from user where name='".$userid."' and pass=password('".$password."')" Good luck, Doug Arne Essa Madsen wrote: I have the following problem: This one does not work: $query = "SELECT * from user where name='$userid' and pass=password('$password')"; $result = mysql_query($query, $link); $num_rows = mysql_num_rows($result); I get this error message: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\wamp\www\fs1812_new\authmain.php on line 18 This one works OK: $query = "SELECT * FROM user"; $result = mysql_query($query, $link); $num_rows = mysql_num_rows($result); What is wrong and how can I use mysql_num_rows when the query include "WHERE" A fast response will be very much appreciated. Arne Essa Madsen; [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Problems with mysql_num-rows()
Very simply, either your connection is messed up, you are not getting a result set back or there is an error in your sql try this: $result = mysql_query($query, $link) or die ("failed because ".mysql_error()); then start debugging bastien From: "Arne Essa Madsen" <[EMAIL PROTECTED]> Reply-To: "Arne Essa Madsen" <[EMAIL PROTECTED]> To: [EMAIL PROTECTED] Subject: [PHP-DB] Problems with mysql_num-rows() Date: Sun, 31 Oct 2004 19:32:09 +0100 I have the following problem: This one does not work: $query = "SELECT * from user where name='$userid' and pass=password('$password')"; $result = mysql_query($query, $link); $num_rows = mysql_num_rows($result); I get this error message: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\wamp\www\fs1812_new\authmain.php on line 18 This one works OK: $query = "SELECT * FROM user"; $result = mysql_query($query, $link); $num_rows = mysql_num_rows($result); What is wrong and how can I use mysql_num_rows when the query include "WHERE" A fast response will be very much appreciated. Arne Essa Madsen; [EMAIL PROTECTED] -- 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
Re: [PHP-DB] Something wierd with time() and curdate()
On Sunday 31 October 2004 22:45, Bobo Wieland wrote: > There is something wierd going on over here... > > I have a date field, datum, in mysql that I cast to a timestamp and check > against time(). If the datum value is more than one day old the table finds > a new random row and sets the datum field to CURDATE(). Now, this has > worked for two years now, but today the script suddenly finds a new random > row each time the script is run... Would this line have anything to with it? If not, post some concise code (see below). > mysql_data_seek($result, rand(0,$rows)); > I've notice this just after 11 pm today. > As I can see from some debugging time() allways returns a higher value than > the timestamp + 86400... Have I missed something in this script for the > last couple of years or is there a bug somewhere? Could it have something > to do with the fact that we turned the time back one hour last night? > > This is the code btw: Please *try* to isolate the problem and post CONCISE code which sufficiently illustrates your problem. For example (for this particular problem) if you suspect PHP is calculating the dates incorrectly then a few lines of code showing the date calculations would be sufficient. If you suspect MySQL is the culprit then a single query would be sufficient. -- Jason Wong -> Gremlins Associates -> www.gremlins.biz Open Source Software Systems Integrators * Web Design & Hosting * Internet & Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-db -- /* "There are things that are so serious that you can only joke about them" - Heisenberg */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Problems with mysql_num-rows()
On Sunday 31 October 2004 18:32, Arne Essa Madsen wrote: > I have the following problem: > > This one does not work: > $query = "SELECT * from user where name='$userid' and > pass=password('$password')"; print $query; > $result = mysql_query($query, $link); Use mysql_error(). -- Jason Wong -> Gremlins Associates -> www.gremlins.biz Open Source Software Systems Integrators * Web Design & Hosting * Internet & Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-db -- /* An honest tale speeds best being plainly told. -- William Shakespeare, "Henry VI" nly */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Problems with mysql_num-rows()
I have the following problem: This one does not work: $query = "SELECT * from user where name='$userid' and pass=password('$password')"; $result = mysql_query($query, $link); $num_rows = mysql_num_rows($result); I get this error message: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\wamp\www\fs1812_new\authmain.php on line 18 This one works OK: $query = "SELECT * FROM user"; $result = mysql_query($query, $link); $num_rows = mysql_num_rows($result); What is wrong and how can I use mysql_num_rows when the query include "WHERE" A fast response will be very much appreciated. Arne Essa Madsen; [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] selective query?
I'm tweaking my internet speed tester thingamabob (http://www.dibcomputers.com/bandwidthmeter/index.php) and am rather stumped. Currently, I'm grouping together all results from each domain (all results from, say, adelphia.net are averaged and displayed as one, etc) and to that end I stopped displaying IP address results. Is there a way I can do a selective group? For example, all results from 11.22.33.44 should be averaged and grouped together, and all results from adelphia.net should be averaged and grouped as well. Here's the code I've got that displays all results: echo "Average speeds of each domain tested:"; $mongo = @mysql_query ("SELECT substring_index(name,'.',-2),ROUND(MIN(speed),1),ROUND(AVG(speed),1),ROUND(M AX(speed),1) FROM `readings` WHERE name <> ip GROUP BY substring_index(name,'.',-2) ORDER BY substring_index(name,'.',-2) ASC"); echo "DomainLowest SpeedAverage SpeedHighest Speed\n"; while ($mongorow = mysql_fetch_array($mongo, MYSQL_NUM)) { if($bg=='#dd') $bg='#ff' ; else $bg='#dd' ; echo "http://www.$mongorow[0]\"; target=\"new\">$mongorow[0]$mongorow[1]$mongorow[2]$mongorow[3]\n"; } echo ""; Any ideas? TIA, Dan -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Something wierd with time() and curdate()
There is something wierd going on over here... I have a date field, datum, in mysql that I cast to a timestamp and check against time(). If the datum value is more than one day old the table finds a new random row and sets the datum field to CURDATE(). Now, this has worked for two years now, but today the script suddenly finds a new random row each time the script is run... I've notice this just after 11 pm today. As I can see from some debugging time() allways returns a higher value than the timestamp + 86400... Have I missed something in this script for the last couple of years or is there a bug somewhere? Could it have something to do with the fact that we turned the time back one hour last night? This is the code btw: function bibelvers($link_id, $color = "white") { $myBible = new Bible($link_id, "FB"); $myBible->set_delimiters("span","p","bibelversnum","text"); $vers_nr = ""; $vers_txt = ""; $reset = false; $q = "SELECT vers,textkod,UNIX_TIMESTAMP(datum) as datum FROM dagensvers WHERE visad = 'NU'"; $result = mysql_query($q, $link_id); if (mysql_affected_rows($link_id) == 1) { $v = mysql_fetch_object($result); if (time() > ($v->datum + ONEDAY)) { mysql_free_result($result); $q = "SELECT vers,textkod FROM dagensvers WHERE visad = 'NEJ'"; $result = mysql_query($q, $link_id); $rows = mysql_affected_rows($link_id)-1; if ($rows >= 0) { mysql_data_seek($result, rand(0,$rows)); $v = mysql_fetch_object($result); $q = "UPDATE dagensvers SET visad = 'JA' WHERE visad = 'NU'"; mysql_query($q, $link_id); $q = "UPDATE dagensvers SET datum = CURDATE(), visad = 'NU' WHERE textkod = '".$v->textkod."'"; mysql_query($q, $link_id); } else { $reset = true; } } } else { $reset = true; } if ($reset) { $q = "UPDATE dagensvers SET visad = 'NEJ' WHERE visad = 'JA'"; mysql_query($q, $link_id); $q = "SELECT vers,textkod FROM dagensvers WHERE visad = 'NEJ'"; $result = mysql_query($q, $link_id); $rows = mysql_affected_rows($link_id)-1; mysql_data_seek($result, rand(0,$rows)); $v = mysql_fetch_object($result); $q = "UPDATE dagensvers SET visad = 'NEJ' WHERE visad = 'NU'"; mysql_query($q, $link_id); $q = "UPDATE dagensvers SET datum = CURDATE(), visad = 'NU' WHERE textkod = '".$v->textkod."'"; mysql_query($q, $link_id); } $vers_nr = $v->vers; $vers_txt = $v->textkod; $myBible->str_vers_format($vers_txt, false, false); $str = "\n". "\t DAGENS BIBELVERS | ".$vers_nr."\n". "\t\n". "\t\t". $vers_txt. "\t\t\n". "\t\n". "\n"; return $str; }
RE: [PHP-DB] PHP & Javascript
Actaully No , as Javascript has to run at Web Browser end and PHP has to run at Server Side , And more at matter javascript memory ( variable values ) cannot be shared with PHP memory ( variable values ) , If you want want to share the variable values or result values then you can do away with PHP-PEAR Extension JavaScript. Nikhil. -Original Message- From: Jason Wong [mailto:[EMAIL PROTECTED] Sent: Sunday, October 31, 2004 6:35 AM To: [EMAIL PROTECTED] Subject: Re: [PHP-DB] PHP & Javascript On Saturday 30 October 2004 12:33, balwantsingh wrote: > can anybody guide me how to embed php script into javascript. You can't. > actually i am > using a function in javascript and want that when this fuctions is called > than some values to be stored in php session so that the values can be > displayed in next page. You can pass values from client-side using Javascript either via the URL ($_GET) or form ($_POST) or cookies ($_COOKIE). You can't manipulate the session variables directly. -- Jason Wong -> Gremlins Associates -> www.gremlins.biz Open Source Software Systems Integrators * Web Design & Hosting * Internet & Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-db -- /* Defeat is worse than death because you have to live with defeat. -- Bill Musselman */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php