[PHP-DB] What's wrong with this query?
I've just tried to do something I've done a thousand times. It does not work. I've checked all of the syntax, made sure the field and variable names are correct. No matter what I do it just doesn't work. The table remains empty. Here's the query: $logit = mysql_query(INSERT INTO log SET term='$search', returns='$arrayword', time=CURTIME(), date=CURDATE(), ip='$ip'); Now that doesn't look too difficult does it? Well, apparently it's impossible! I'm really hoping someone out there can see something that I missed. Nick -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] What's wrong with this query?
Ok guys, I found the problem. It's actually something I've run into before. If you take a look at the query you'll notice that one of the fields is named returns. When I created the test table with phpMyAdmin it created it and everything seemed fine. Then I decided to try something. I tweaked my install script to add creation of the new log table and ran it. It failed on that field. When I changed the name of the field it worked. I've seen this before when I tried to create a field for another database called group. Apparently these are MySQL reserved words and you can't use them as names for fields. Go figure. Anyway, that solves the problem. When I run it the data is put into the table, and the line where I echo the $logit var displays a 1. By the way, I changed the name of the returns field to found. Nick Micah Stevens wrote: Right.. a resource.. sorry. On Thursday 26 February 2004 12:55 pm, [EMAIL PROTECTED] wrote: An interesting thought. I tried this: echo Term: $search, Returns: $arrayword, UserIP: $ipbr; $logit = mysql_query(INSERT INTO log SET term='$search', returns='$arrayword', time=CURTIME(), date=CURDATE(), ip='$ip'); echo Query Value: $logit; And got this on the page: Term: skater, Returns: 312, UserIP: 192.168.1.234 Query Value: - mysql_query returns a Resource, which is not a printable 'string', You could print_r($logit), or var_dump($logit) and you would most likely see Resouce ID #3 (or some other number). So even if you're query excecutes properly printing $logit will always show as you've written above. The same would hold true of an array, when just using print. the returned value preceeds the function call in the manual. from php.net/mysql_query resource mysql_query ( string query [, resource link_identifier]) hth jeff - Notice that the variables are set with appropriate values, but the $logit variable is blank. This, I think is the problem. The question is why would it do this and not return any type of error? By the way, I tried this using the other syntax people where suggesting and I got the same results. I gotta tell you, this one is really kicking my ass. It's the last piece of an update that I can't release until it's finished. Nick Hutchins, Richard wrote: Been kind of following this thread off and on... If the syntax is acceptable by MySQL, which it appears to be, is it possible that the variables you are using within the query string are not set to anything? Or is it possible that there is something broken immediately before the query string is fired? Most times, when I run into query problems, immediately before I send the query to the database, I'll echo the statement out to the browser so I can see the exact string that's being sent to the db. So can you do this: $sql = INSERT INTO log SET term='$search', returns='$arrayword', time=CURTIME(), date=CURDATE(), ip='$ip'; echo $sql; $logit = mysql_query($sql) or die(mysql_error()); And check out what gets spit out to the browser when $sql is echoed? Maybe that'll point out something that's going wrong. If nothing is apparent, post the results of echo $sql back to the list and maybe one of us will find something. HTH. Rich -Original Message- From: Axel IS Main [mailto:[EMAIL PROTECTED] Sent: Thursday, February 26, 2004 2:37 PM To: PHP DB Subject: Re: [PHP-DB] What's wrong with this query? Ok, ok. I get the message about the syntax. Since I've used this syntax for a long time and so far there hasn't been a problem I'll assume that's not the problem. I will, however, review this and probably make some changes for the sake of compliance if nothing else. In any event, the syntax is NOT why it is failing, smart ass comments by people who think two years is a long time not withstanding. As to the useful questions that where asked. There is no error reported. Error reporting is set to E_ALL ~E_NOTICE. I removed the notice part so I would get everything and still nothing showed up. I'm also logging errors and nothing is showing up in the log either. Not all PHP/MySQL errors are reported. Sometimes what happens is not considered an error, even though it does not do what you would expect it to do. If the query syntax was wrong, I would get a syntax error. There is no error, it just doesn't write to the table. I can go into phpMyAdmin and with the same syntax insert into the table all day long. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] What's wrong with this query?
Actually, yes, a thousand times is an obvious exaggeration, but I have
done a whole lot of times, and the query syntax is the query syntax.
Different don't make it wrong. This is working in several other places
throughout this particular set of scripts.
Paul Fitz wrote:
You haven't done that a thousand times - the syntax is wrong :).
You are using a combo of INSERT and UPDATE syntax there.
Try
INSERT INTO log (term, returns, time, date, ip) VALUES ('$search',
'$arrayword',CURTIME(), CURDATE(), '$ip');
Cheers,
Paul
-Original Message-
From: Axel IS Main [mailto:[EMAIL PROTECTED]
Sent: Thursday, February 26, 2004 7:06 PM
To: PHP-DB
Subject: [PHP-DB] What's wrong with this query?
I've just tried to do something I've done a thousand times. It does not
work. I've checked all of the syntax, made sure the field and variable
names are correct. No matter what I do it just doesn't work. The table
remains empty. Here's the query:
$logit = mysql_query(INSERT INTO log SET term='$search',
returns='$arrayword', time=CURTIME(), date=CURDATE(), ip='$ip');
Now that doesn't look too difficult does it? Well, apparently it's
impossible! I'm really hoping someone out there can see something that I
missed.
Nick
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] What's wrong with this query?
An interesting thought. I tried this: echo Term: $search, Returns: $arrayword, UserIP: $ipbr; $logit = mysql_query(INSERT INTO log SET term='$search', returns='$arrayword', time=CURTIME(), date=CURDATE(), ip='$ip'); echo Query Value: $logit; And got this on the page: Term: skater, Returns: 312, UserIP: 192.168.1.234 Query Value: Notice that the variables are set with appropriate values, but the $logit variable is blank. This, I think is the problem. The question is why would it do this and not return any type of error? By the way, I tried this using the other syntax people where suggesting and I got the same results. I gotta tell you, this one is really kicking my ass. It's the last piece of an update that I can't release until it's finished. Nick Hutchins, Richard wrote: Been kind of following this thread off and on... If the syntax is acceptable by MySQL, which it appears to be, is it possible that the variables you are using within the query string are not set to anything? Or is it possible that there is something broken immediately before the query string is fired? Most times, when I run into query problems, immediately before I send the query to the database, I'll echo the statement out to the browser so I can see the exact string that's being sent to the db. So can you do this: $sql = INSERT INTO log SET term='$search', returns='$arrayword', time=CURTIME(), date=CURDATE(), ip='$ip'; echo $sql; $logit = mysql_query($sql) or die(mysql_error()); And check out what gets spit out to the browser when $sql is echoed? Maybe that'll point out something that's going wrong. If nothing is apparent, post the results of echo $sql back to the list and maybe one of us will find something. HTH. Rich -Original Message- From: Axel IS Main [mailto:[EMAIL PROTECTED] Sent: Thursday, February 26, 2004 2:37 PM To: PHP DB Subject: Re: [PHP-DB] What's wrong with this query? Ok, ok. I get the message about the syntax. Since I've used this syntax for a long time and so far there hasn't been a problem I'll assume that's not the problem. I will, however, review this and probably make some changes for the sake of compliance if nothing else. In any event, the syntax is NOT why it is failing, smart ass comments by people who think two years is a long time not withstanding. As to the useful questions that where asked. There is no error reported. Error reporting is set to E_ALL ~E_NOTICE. I removed the notice part so I would get everything and still nothing showed up. I'm also logging errors and nothing is showing up in the log either. Not all PHP/MySQL errors are reported. Sometimes what happens is not considered an error, even though it does not do what you would expect it to do. If the query syntax was wrong, I would get a syntax error. There is no error, it just doesn't write to the table. I can go into phpMyAdmin and with the same syntax insert into the table all day long. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] What's wrong with this query?
Ok, ok. I get the message about the syntax. Since I've used this syntax for a long time and so far there hasn't been a problem I'll assume that's not the problem. I will, however, review this and probably make some changes for the sake of compliance if nothing else. In any event, the syntax is NOT why it is failing, smart ass comments by people who think two years is a long time not withstanding. As to the useful questions that where asked. There is no error reported. Error reporting is set to E_ALL ~E_NOTICE. I removed the notice part so I would get everything and still nothing showed up. I'm also logging errors and nothing is showing up in the log either. Not all PHP/MySQL errors are reported. Sometimes what happens is not considered an error, even though it does not do what you would expect it to do. If the query syntax was wrong, I would get a syntax error. There is no error, it just doesn't write to the table. I can go into phpMyAdmin and with the same syntax insert into the table all day long. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Strange DB Error
When returning to the main page of my script sometimes I'll get the
following error:
Warning: mysql_fetch_array(): 8 is not a valid MySQL result resource in
/home/nick/http/search/includes/main.inc on line 13
This happens only AFTER the first element in the resource has been
processed. Here's the code that is causing the problem:
$s = mysql_query(SELECT * FROM categories WHERE length(cat)='2');
while($t = mysql_fetch_array($s)){
$cat = $t[cat];
$name = $t[name];
if($tstruc == $dirwidth){
echo /trtr;
$tstruc = 0;
}
echo td align=\left\ valign=\top\font face=\verdana,sans
serif,helvetica\ size=\3\ba
href=\index.php?index=directory.inccat=$catname=$namedepth=0\$name/a/b/font;
$subcat = mysql_unbuffered_query(SELECT * FROM categories WHERE
length(cat)='4' substring(cat,1,2)='$cat' ORDER BY cat);
while($ct = mysql_fetch_array($subcat)){
$scats = $ct[name];
$dcats = $dcats . $scats . , ;
}
$dcats = trim(substr($dcats, 0, 28)) . trim(...);
if($dcats != ...){
echo brfont face = \tahoma,sans serif,helvetica\
size=\1\ . $dcats . /font;
}
$dcats = ;
echo /td;
++$tstruc;
}
You can go to the site http://www.icesource.com to see this script in
action. This is an intermittent problem so you may not see it at the
site, but it does happen there sometimes. If you click on one of the
categories on the list there, and to two or three deep, then hit the
Home link at the top left of the listings there's a pretty good chance
you'll see the problem. I've been trying to figure this out for a long
time, and nothing seems to work. Any ideas out there would be greatly
appreciated.
Nick
Re: [PHP-DB] Strange DB Error
Error trapping is not really relevant here as the server is properly
configured so the errors are going to show up in either event. In fact,
using the die() function on my server just makes thing worse. In any
case, the var_dump was a good idea as I was able to find out that what
is happening is when this loops up after the first run through the array
is blank. It doesn't reinitialize the array with the new values. As far
as the $cat, or any of the other variables go, they're always the same
when returning to the index. Whether I initialize them from the query
string, or in the index itself as in the case when it is loaded without
any parameters. The question is, why wouldn't the array be redefined on
the second loop? Hell if I know.
Nick
[EMAIL PROTECTED] wrote:
When returning to the main page of my script sometimes I'll get the
following error:
Warning: mysql_fetch_array(): 8 is not a valid MySQL result resource in
/home/nick/http/search/includes/main.inc on line 13
This happens only AFTER the first element in the resource has been
processed. Here's the code that is causing the problem:
First, i'd place an... or die (mysql_error()) after your mysql call. so
you can see what mysql is complaining about. basically your sql is
failing. I also like to split the first two items, when debegging, to see
what the actual query was. i.e.
psedocode
$qry = SELECT * FROM categories WHERE length(cat)='2';
$s = mysql_query($qry) or die (mysql_error() . \nbr . $qry .
\nbr);
psedocode
The only time i saw the error on your page the 'cat' GET variable was
defined as 0. Is this really a static query like below, or is it based on
what is passed to the page in the query string? The URL when it died...
http://www.icesource.com/index.php?depth=0cat=0index=main.inc, which to
me looks like a bad querystring, but may indeed be valid values.
If that doesn't lend itself to any useful information, try var_dump($t); in
your while loop, to see what has been processed and what is failing.
HTH
Jeff
$s = mysql_query(SELECT * FROM categories WHERE length(cat)='2');
while($t = mysql_fetch_array($s)){$cat = $t[cat];
$name = $t[name];
if($tstruc == $dirwidth){echo /trtr;
$tstruc = 0;
}
echo td align=\left\ valign=\top\font face=\verdana,sans
serif,helvetica\ size=\3\ba
href=\index.php?index=directory.inccat=$catname=$namedepth=0\$name/a/b/font;
$subcat = mysql_unbuffered_query(SELECT * FROM categories WHERE
length(cat)='4' substring(cat,1,2)='$cat' ORDER BY cat);
while($ct = mysql_fetch_array($subcat)){$scats
= $ct[name];
$dcats = $dcats . $scats . , ;
}
$dcats = trim(substr($dcats, 0, 28)) . trim(...);
if($dcats != ...){echo brfont face = \tahoma,sans
serif,helvetica\
size=\1\ . $dcats . /font;
}
$dcats = ;
echo /td;
++$tstruc;
}
You can go to the site http://www.icesource.com to see this script in
action. This is an intermittent problem so you may not see it at the
site, but it does happen there sometimes. If you click on one of the
categories on the list there, and to two or three deep, then hit the
Home link at the top left of the listings there's a pretty good chance
you'll see the problem. I've been trying to figure this out for a long
time, and nothing seems to work. Any ideas out there would be greatly
appreciated.
Nick
[PHP-DB] Re:[PHP-DB] Strange DB Error
What you propose makes sense of course, but no matter what I do, where I
send the errors, even to the system log, the mysql_fetch_array () error
is the only one that shows up. It would appear that either the query is
failing is some way that PHP isn't recognizing, or the fetch isn't
working properly. It could also be that for whatever reason, the query
isn't finding all of the records that meet the criteria. I've tried
doing and unconditional SELECT from the database but that had no effect
on this error. I still got it. This is not the only place I look at this
database, but it is the only place I get this error.
Nick
[EMAIL PROTECTED] wrote:
Error trapping is not really relevant here as the server is properly
configured so the errors are going to show up in either event. In fact,
using the die() function on my server just makes thing worse.
---
it'll definitely make it worse, but hopefully only to find the problem and
make it better. you should be able to find the query on which this is not
a valid resource run that agains't mysql and find out why its failing,
if you can't see that from $qry being in die(). but yes, a bad idea in
production. send the output to a file on error and see what your
getting.the error that you want is from mysql_query, the error that you're
showing is from mysql_fetch_array(), so if the server is configured to show
them elsewhere then the answer should be right before the error below
hth
jeff
In any case, the var_dump was a good idea as I was able to find out that
what is happening is when this loops up after the first run through the
array is blank. It doesn't reinitialize the array with the new values. As
far as the $cat, or any of the other variables go, they're always the same
when returning to he index. Whether I initialize them from the query
string, or in the index itself as in the case when it is loaded without any
parameters. The question is, why wouldn't the array be redefined on the
second loop? Hell if I know.
Nick
[EMAIL PROTECTED] wrote:
When returning to the main page of my script sometimes I'll get
the
following error:
Warning: mysql_fetch_array(): 8 is not a valid MySQL result
resource in
/home/nick/http/search/includes/main.inc on line 13
This happens only AFTER the first element in the resource has
been
processed. Here's the code that is causing the problem:
First, i'd place an... or die (mysql_error()) after your mysql call.
so
you can see what mysql is complaining about. basically your sql is
failing. I also like to split the first two items, when debegging,
to see
what the actual query was. i.e.
psedocode
$qry = SELECT * FROM categories WHERE length(cat)='2';
$s = mysql_query($qry) or die (mysql_error() . \nbr . $qry .
\nbr);
psedocode
The only time i saw the error on your page the 'cat' GET variable was
defined as 0. Is this really a static query like below, or is it
based on
what is passed to the page in the query string? The URL when it
died...
http://www.icesource.com/index.php?depth=0cat=0index=main.inc,
which to
me looks like a bad querystring, but may indeed be valid values.
If that doesn't lend itself to any useful information, try
var_dump($t); in
your while loop, to see what has been processed and what is failing.
HTH
Jeff
$s = mysql_query(SELECT * FROM categories WHERE
length(cat)='2');
while($t = mysql_fetch_array($s)){$cat = $t[cat];
$name = $t[name];
if($tstruc == $dirwidth){echo /trtr;
$tstruc = 0;
}
echo td align=\left\ valign=\top\font
face=\verdana,sans
serif,helvetica\ size=\3\ba
href=\index.php?index=directory.inccat=$catname=$namedepth=0\$name/a/b/font;
$subcat = mysql_unbuffered_query(SELECT * FROM categories
WHERE
length(cat)='4' substring(cat,1,2)='$cat' ORDER BY cat);
while($ct = mysql_fetch_array($subcat)){$scats
= $ct[name];
$dcats = $dcats . $scats . , ;
}
$dcats = trim(substr($dcats, 0, 28)) . trim(...);
if($dcats != ...){echo brfont face
= \tahoma,sans
serif,helvetica\
size=\1\ . $dcats . /font;
}
$dcats = ;
