Trouble with Update Page

[Discord]

Hi guys, i'm working on an account update page and I'm getting some weird behaviors. When i submit click the submit button i get the error message :

Warning: Supplied argument is not a valid MySQL result resource in /var/www/html/editprofile2.php on line 65

But line 65 works on the first time i load the page. So why does it stop working when i submit?

Here is a snippet of the code i'm using.. I put a
// HERE IS LINE 65 on the line in question.

<center>
<table width=100% border=1 cellspacing=0 cellpadding=1 valign=top>
<form method=post action="<?php echo $PHP_SELF?>">


<?php
$UID=$_GET['UID'];
$Uname=$_GET['Uname'];
$PWD=$_GET['PWD'];
$EML=$_GET['EML'];


      printf ("<tr><td bgcolor=darkgray><p>IM Number:</p></td><td align=center><p>$UID</p></td></tr>");
      printf ("<tr><td bgcolor=darkgray><p>Username:</p></td><td align=center><p>$Uname</p></td></tr>");
      printf ("<tr><td bgcolor=darkgray><p>Email:</p></td><td align=center><input type=Text name=email value=$EML></td></tr>");
      printf ("<tr><td bgcolor=darkgray><p>New Password:</p></td><td align=center><input type=Password name=password1 value=$PWD></td></tr>");
      printf ("<tr><td bgcolor=darkgray><p>Confirm Password:</p></td><td align=center><input type=Password name=password2 value=$PWD></td></tr>");


// Connect to the database
$db = mysql_connect("localhost", "root");
mysql_select_db("basmentIM",$db);

	// Select Current Info and Display an update form
	$sql = "SELECT * FROM Profiles WHERE Account_ID=$UID";
	$result = mysql_query($sql);
        $number_results = mysql_num_rows($result); //      HERE IS LINE 65
	$myrow = mysql_fetch_row($result);
        if ($number_results==0) {

           printf ("<tr><td bgcolor=darkgray><p>Real Name:</p></td><td align=center><input type=Text name=realname></td></tr>");
           printf ("<tr><td bgcolor=darkgray><p>Age:</p></td><td align=center><input type=Text name=age></td></tr>");
           printf ("<tr><td bgcolor=darkgray><p>Location:</p></td><td align=center><input type=Text name=location></td></tr>");
           printf ("<tr><td bgcolor=darkgray><p>Marital Status:</p></td><td align=center><SELECT Name=mstat><Option value=single>Single<Option value=married>Married<Option value=divorced>Divorced</SELECT></td></tr>");
           printf ("<tr><td bgcolor=darkgray><p>Homepage:</p></td><td align=center><input type=Text name=homepage></td></tr>");
           printf ("<tr><td bgcolor=darkgray><p>About:</p></td><td align=center><textarea name=about rows=3 cols=40></textarea></td></tr>");
           printf ("</table>");

	} else {


           printf ("<tr><td bgcolor=darkgray><p>Real Name:</p></td><td align=center><input type=Text name=realname value=%s></td></tr>", $myrow[1]);
           printf ("<tr><td bgcolor=darkgray><p>Age:</p></td><td align=center><input type=Text name=age value=%s></td></tr>", $myrow[2]);
           printf ("<tr><td bgcolor=darkgray><p>Location:</p></td><td align=center><input type=Text name=location value=%s></td></tr>", $myrow[3]);
           printf ("<tr><td bgcolor=darkgray><p>Marital Status:</p></td><td align=center><SELECT Name=mstat value=%s><Option value=single>Single<Option value=married>Married<Option value=divorced>Divorced</SELECT></td></tr>", $myrow[4]);
           printf ("<tr><td bgcolor=darkgray><p>Homepage:</p></td><td align=center><input type=Text name=homepage value=%s></td></tr>", $myrow[6]);
           printf ("<tr><td bgcolor=darkgray><p>About:</p></td><td align=center><textarea name=about rows=8 cols=55>%s</textarea></td></tr>", htmlentities($myrow[5]));
           printf ("</table>");
	}

?>

<br><input type="Submit" name="submit" value="Update">
</form>



<?php
if ($submit) {
echo "$UID";
}

?>


</center>

any help you could give me would be greatly appreciated.

[bretticus]

The query...

// Select Current Info and Display an update form
$sql = "SELECT * FROM Profiles WHERE Account_ID=$UID";

Specifically, the variable $UID is your problem. It get's set in the first pageload form via GET, since your form is submitted via POST, on the second time around the query is sent to the database as 'SELECT * FROM Profiles WHERE Account_ID=' which will error out and return a resource result thus the error the first time you attempt to use $result

[Discord]

Thanks for the post. Ok i understand why its happening. Is it possible to have these maintained so that they will be either reposted to the page or maintain constant?

[eriksmoen]

your form uses the "post" method, whereas you are trying to get information from a "get" method... thus making $uid, etc null. you need to change each instance of

$_GET['VAR_NAME'];

to

$_POST['var_name'];

hope that helps.

[Discord]

Thanks for the suggestion.
I changed:

$UID=$_GET['UID'];
$Uname=$_GET['Uname'];
$PWD=$_GET['PWD'];
$EML=$_GET['EML'];

to

$_POST['UID'];
$_POST['Uname'];
$_POST['PWD'];
$_POST['EML'];

But i am getting the exact same result. The pages loads fine and pulls down all the neccesary information. But when i click submit i get errors.

[bretticus]

You did add hidden fields to store the information right? Example:

<input type="hidden" name="UID" value="<?=$_POST['UID']?>">

[eriksmoen]

what kind of errors? what line numbers? copy and paste all the error messages you get... unless its a huge list :p then just paste the relevant errors...

[Discord]

yeah i have added

<input type="hidden" name="UID" value="<?=$_POST['UID']?>">
<input type="hidden" name="Uname" value="<?=$_POST['Uname']?>">
<input type="hidden" name="PWD" value="<?=$_POST['PWD']?>">
<input type="hidden" name="PWD" value="<?=$_POST['EML']?>">

but the same thing is till happening...
The Error message is:
"Warning: Supplied argument is not a valid MySQL result resource in" line ect..

[cgraz]

put single quotes around $UID

$sql = "SELECT * FROM Profiles WHERE Account_ID='$UID'";

Cgraz

[Discord]

Thanks that stopped the error message from showing up! But my code on submit is

<?php
if ($submit) {

printf("<table width=300>");
printf("<tr><td><p>$UID</p></td></tr>");
printf("</table>");

}

?>

and it doesn't appear that the values are retained.

[cgraz]

It's late and I just skimmed over your post. Can you explain what you're trying to do and what's actually happening? What values don't appear to be retained?

Sorry if I missed it. I'm running on little sleep.

Cgraz

[bretticus]

Please resubmit the code you have now. I'm going to bed but I know the update is gonna be useful to whomever takes it from here.

[cgraz]

I'm going to bed too. I'll check the post tomorrow.

Cgraz

[Discord]

This page is suppose to be a profile editor. I'm passing the page the ID which i then use to select all the current profile information down from the database. On this page i have an update button at the bottom that I'm try to get to then run an update statement to the database and update the database from the users info on the form. This is what is happening though...
1. The page loads and Pulls down all the correct information
2. I click update and the UID, Uname ect.. values no longer contain any information.

I would like to be able to keep the values so that i can run use them for an update statement to the database.

[Discord]

here is the current code i'm using

<center>
<table width=100% border=1 cellspacing=0 cellpadding=1 valign=top>
<form method=post action="<?php echo $PHP_SELF?>">


<?php

$_POST['UID'];
$_POST['Uname'];
$_POST['PWD'];
$_POST['EML'];




      printf ("<tr><td bgcolor=darkgray><p>IM Number:</p></td><td align=center><p>$UID</p></td></tr>");
      printf ("<tr><td bgcolor=darkgray><p>Username:</p></td><td align=center><p>$Uname</p></td></tr>");
      printf ("<tr><td bgcolor=darkgray><p>Email:</p></td><td align=center><input type=Text name=email value=$EML></td></tr>");
      printf ("<tr><td bgcolor=darkgray><p>New Password:</p></td><td align=center><input type=Password name=password1 value=$PWD></td></tr>");
      printf ("<tr><td bgcolor=darkgray><p>Confirm Password:</p></td><td align=center><input type=Password name=password2 value=$PWD></td></tr>");


// Connect to the database
$db = mysql_connect("localhost", "root");
mysql_select_db("basmentIM",$db);

	// Select Current Info and Display an update form
	$sql = "SELECT * FROM Profiles WHERE Account_ID='$UID'";
	$result = mysql_query($sql);
        $number_results = mysql_num_rows($result);
	$myrow = mysql_fetch_row($result);
        if ($number_results==0) {

           printf ("<tr><td bgcolor=darkgray><p>Real Name:</p></td><td align=center><input type=Text name=realname></td></tr>");
           printf ("<tr><td bgcolor=darkgray><p>Age:</p></td><td align=center><input type=Text name=age></td></tr>");
           printf ("<tr><td bgcolor=darkgray><p>Location:</p></td><td align=center><input type=Text name=location></td></tr>");
           printf ("<tr><td bgcolor=darkgray><p>Marital Status:</p></td><td align=center><SELECT Name=mstat><Option value=single>Single<Option value=married>Married<Option value=divorced>Divorced</SELECT></td></tr>");
           printf ("<tr><td bgcolor=darkgray><p>Homepage:</p></td><td align=center><input type=Text name=homepage></td></tr>");
           printf ("<tr><td bgcolor=darkgray><p>About:</p></td><td align=center><textarea name=about rows=3 cols=40></textarea></td></tr>");
           printf ("</table>");

	} else {


           printf ("<tr><td bgcolor=darkgray><p>Real Name:</p></td><td align=center><input type=Text name=realname value=%s></td></tr>", $myrow[1]);
           printf ("<tr><td bgcolor=darkgray><p>Age:</p></td><td align=center><input type=Text name=age value=%s></td></tr>", $myrow[2]);
           printf ("<tr><td bgcolor=darkgray><p>Location:</p></td><td align=center><input type=Text name=location value=%s></td></tr>", $myrow[3]);
           printf ("<tr><td bgcolor=darkgray><p>Marital Status:</p></td><td align=center><SELECT Name=mstat value=%s><Option value=single>Single<Option value=married>Married<Option value=divorced>Divorced</SELECT></td></tr>", $myrow[4]);
           printf ("<tr><td bgcolor=darkgray><p>Homepage:</p></td><td align=center><input type=Text name=homepage value=%s></td></tr>", $myrow[6]);
           printf ("<tr><td bgcolor=darkgray><p>About:</p></td><td align=center><textarea name=about rows=8 cols=55>%s</textarea></td></tr>", htmlentities($myrow[5]));
           printf ("</table>");
	}

?>

<input type="hidden" name="UID" value="<?=$_POST['UID']?>">
<input type="hidden" name="Uname" value="<?=$_POST['Uname']?>">
<input type="hidden" name="PWD" value="<?=$_POST['PWD']?>">
<input type="hidden" name="PWD" value="<?=$_POST['EML']?>">


<br><input type="Submit" name="submit" value="Update">
</form>



<?php
if ($submit) {

printf ("<table width=300>");
printf ("<tr><td><p>$UID</p></td></tr>");
printf ("</table>");

}

?>


</center>

[cgraz]

what does

$POST['UID'];
$POST['Uname'];
$POST['PWD'];
$POST['EML'];

do?

How i'd do this is pass the value of $UID (wherever it's coming from) in the url when you submit your form.

<form method="post" action="<?= $PHP_SELF?UID=$UID ?>">

replace $UID with however you're getting $UID to this page. I have no idea where it's coming from. if it's from a form
on the previous page, then use $_POST['UID']

Anyways, then when the form is submitted,

<?php
if (isset($_POST['submit'])) {

echo "<table width=300>";
echo "<tr><td><p>" . $_GET['UID'] . "</p></td></tr>"; // pull out of the url using $_GET
echo "</table>";

}

?>

It's bedtime for me. Let me know how that goes.

Cgraz

[eriksmoen]

$_POST['UID'];

should be

$UID = $_POST['UID'];

same with uname, eml, and pwd. the problem was that you didn't have $uid, $uname, $pwd, or $eml defined. i don;t know if this makes sense, but im tired and i have a paper to finish on iDen wireless... stupid 9am classes...

hope it helps.

[Discord]

Before my code is fired i load the page from the previous one, which is simply a login screen. This is the code i'm using on the login screen to send the basic information to the next page.

	    // Check for Account existence
	    $sql = "SELECT * FROM Accounts WHERE Username='$uname' And Password='$password'";
            $result = mysql_query($sql);
	    $number_results = mysql_num_rows($result);

        if ($number_results==0) {

	// Invalid Login!
	echo "<br><center><p><font color=red>Invalid Username or Password!</font></p></center>";
	exit;

	} else {

	$myrow = mysql_fetch_row($result);


	// Valid Login
	header("Location: editprofile2.php?UID=$myrow[0]&Uname=$myrow[1]&PWD=$myrow[2]&EML=$myrow[3]");
	}
}

[cgraz]

Well if the value for $UID is in the url on this script, then pass it to the next page through your form like this:

<form method="post" action="<?= $PHP_SELF?UID=$_GET['UID'] ?>">

Then access it as I mentioned above (with the if(isset($_POST['submit'])) code)

Cgraz

[Discord]

got it working... thanks for the help guys I appreciate it.