Need help with UPDATE command

MsMittens

I keep getting this weird glitch and I'm not sure what's causing it. When I change a team name I end up with "Newteamnameteam=Newteamname" as being entered in the database. So, for example if I change the team name The Fuzzy Wuzzles I get The Fuzzy Wuzzlesteam=The Fuzzy Wuzzles. What is frustrating me is I can't figure out where the "team=.. " comes from. Help?

I have the following code (newteam.php):

<? 
session_start();
// add the team
include_once 'dbconnect.php';
$username_use = $_SESSION['username'];
$teamname = $_POST['team'];
$sql_pre_query = mysql_query ("SELECT userid, username FROM id WHERE username='$username_use'") or die (mysql_error());
$row_pre = mysql_fetch_object($sql_pre_query);
$userid = $row_pre -> userid;

$sql_query_name = mysql_query ("SELECT * FROM users  WHERE userid='$userid'") or die (mysql_error());
$row = mysql_fetch_object($sql_query_name);
$teamname2 = $row -> team;

echo "Userid = $userid<BR><BR>";
echo "This is the entered name: $teamname and what exists: $teamname2<BR><BR>";

if (!$teamname) {
	echo "Uh oh. No team name was entered.You put in $teamname. <a href=login2.php>Try again!</a>";
	} else {
	$sql_query = mysql_query("UPDATE users SET team='$teamname' WHERE userid='$userid'") or die(mysql_error());
	echo "<BR><BR>The team entered is $teamname<BR><BR>";
	echo "<a href=login2.php>Back to Main</a>";
	}
?>

It grabs the new team name from options.php:

<?
// get database info
include_once 'dbconnect.php';
// get user info
$username_use = $_SESSION['username'];
$sql_query_team = mysql_query("SELECT * from id WHERE username='$username_use'") or die (mysql_error());
$row_team = mysql_fetch_object($sql_query_team);
$userid = $row_team -> userid;
// get present team name
$sql_query_team2 = mysql_query("SELECT * from users WHERE userid='$userid'") or die (mysql_error());
$row_team2 = @mysql_fetch_array($sql_query_team2);
$teamname1 = $row_team2['team'];
echo "<form method='post' action='newteam.php'>";
echo "Your present team name is: $teamname1<BR>";
echo "Enter a new team name to change: <input type=text name='team' size='25' maxsize='50'><BR>";
echo "<input type='submit' value='Change Team/Store Name'></form>";
?>

nemonoman

I'm going to guess that this isn't EXACTLY the code that is causing the problem you describe.

I would guess, in fact that the problem is actually found in some more complicated code that you have written, and that you have posted this example after cleaning up the actual code a bit.

My guess is that in that code you have an if statement that says something like

if($team=$teamname)

or some other similar conditional

when what you meant is

if ($team == $teamname)

In other words, in your code instead of CHECKING ($team=='FUZZIEWUZZIES")

You are assigning ($team='FUZZIEWUZZIES')

Anyway, that's my guess.

Another guess would be using a

.= (concat)

instead of an

= (assignment)

at some point in your code before the variable is used to update the database.

You may ask yourself why I'm guessing this? It's because I do this kind of thing all the time.

MsMittens

There isn't any code like that (using the equal value sign rather than the equal comparitor sign). The code I've posted is what I'm having issue with.

I think you might be misunderstanding what I'm asking. When I enter in the name "Fuzzy Wuzzies" the output is becoming "Fuzzy Wuzziesteam=Fuzzy Wuzzies".

There is nothing that has $team='=team$team', which is what might cause that.

Oddly enough, it's only happening in my team field. When I change the user's first name, last name, address, etc. I have no issues.

gtibok

where is the code that outputs NewTeamNameteam=NewTeamName ?

MsMittens

What you see is what I have. That's why I brought it here. To see if there is something obvious that I am missing.

nemonoman

MsMittens:

FIRST -- What you are being asked is: How do you KNOW that teamname is now 'Fuzzy Wuzziesteam=Fuzzy Wuzzies'. Where do you see those words -- do you telnet into your database? Do you see it in PHP output? That's the question. What we're REALLY asking is: Has that value been placed in the database that way, or are you seeing it in some PHP page that might have a bug or typo?

Please help us by doing the following:

Change this line:

$sql_query = mysql_query("UPDATE users SET team='$teamname' WHERE userid='$userid'") or die(mysql_error());

to these three lines:

$updatesql="UPDATE users SET team='$teamname' WHERE userid='$userid'";

echo $updatesql;

$sql_query = mysql_query($updatesql) or die(mysql_error());

then send the echo'ed output. That will clarify the value being sent to the db.

THIRD:
This may seem stupid, but does this happen with ANY team name, or only with teams named 'fuzzywuzzies'? I'm assuming this is a generic problem, yes?

MsMittens

To answer your first question, yes, that's what appears in the database. I always check to see that UPDATE and INSERT and other modification commands work. So what I am seeing is in the database and what's being presented.

I did the changes as you suggested and got back as output:

Userid = 18

This is the entered name: asfsdfsdfsdateam=asfsdfsdfsda and what exists: very new teamteam=very new team

UPDATE users SET team='asfsdfsdfsdateam=asfsdfsdfsda' WHERE userid='18'

The team entered is asfsdfsdfsdateam=asfsdfsdfsda

And yes, it happens to anything I put it. It seems to add "team=whateverthe value is". I think it's something with the input line but haven't figured out where on there it is. I thought at first it might have been the maxsize option but no.

I just find it weird that this is the only one I have this error with.

Sigh. The head bruises are getting bigger. =P

nemonoman

Please, be calm! You are now much closer to your answer!

Please add the following line somewhere:

echo $_POST['team']." is posted team";

edited this line slightly.

MsMittens

Hrmm. I tried that and it would seem that the code for posting is where the problem might be. The response I got was:

The Fuzzy Wussysteam=The Fuzzy Wussys is posted team

Userid = 18

This is the entered name: The Fuzzy Wussysteam=The Fuzzy Wussys and what exists: asdf asfsdfadteam=asdf asfsdfadteam=asdf asfsdfadt

UPDATE users SET team='The Fuzzy Wussysteam=The Fuzzy Wussys' WHERE userid='18'

The team entered is The Fuzzy Wussysteam=The Fuzzy Wussys

I checked the options page where all this entry was going into and got this out put:

SELECT * from users WHERE userid='18'
Your present team name is: The Fuzzy Wussysteam=The Fuzzy Wussys
Enter a new team name to change:

I didn't want to put up the full options code but perhaps I should.. I'm gonna do that in another post.

As for being calm, I am. I was just trying to add a little levity to the situation as I've been searching all weekend for an answer to this weirdness.

MsMittens

Hrmm.. I think perhaps it's too long to paste here so I'm adding it as a file.

xblue

I'm not absolutely sure, but could that be a problem with the server rather than with php? just a thought, but would you mind posting your server software and version?

(hope I'm not adding to the mess)

MsMittens

No problem. As I was getting ready to go out for the afternoon and pondering another UPDATE issue (not updating when all the info is there) I've begun to wonder if it is the server. (The bold information is the acutal version info).

I have a feeling there are some major bugs in the 4.0.9 gamma as I've had problems using ACID for SNORT (creating tables with ACID) and originally, had attributed it to the PHP files being written for earlier versions of PHP.

mysql client: Ver 12.16 Distrib 4.0.9-gamma, for unknown-freebsd4.7 (i386)

mysql server: Ver 4.0.9-gamma-max for unknown-freebsd4.7 on i386

PHP: PHP 4.3.0 (cli) (built: Jan 27 2003 05:36:01)
Copyright (c) 1997-2002 The PHP Group
Zend Engine v.1.3.0, Copyright (c) 1998-2002 Zend Technologies

Apache:

Server version: Apache/2.0.44
Server built: Jan 26 2003 19:04:28

xblue

I'd recommend to downgrade to an older apache version.

this is a quote from another post by feldon23 (fairly recent):

PLEASE don't install Apache 2 and expect PHP to work flawlessly just yet unless you are a seasoned Apache/PHP operator. Apache 2 is still a CVS release and so it is a moving target for compatibility!

MsMittens

I finally figured out what was causing the problem. I had

if (!$teamname)

For whatever reason this was doing it. When I changed it to:

if (!isset($teamname))

Problem solved.

Amazing when you clear your head what pops out at 5 in the afternoon. 😃