Why wont $img work?

[hoponhiggo]

Hello

 

I am using the code below, which is fine, but in the part of the code that follows the 'while' statement near the bottom, i want to use the $img variable for a second time in between the <div id='mempic'> tags, to display a picture but its not working. Can anybody tel me why?

 

<?php
echo "<center>";
if(isset($_GET['user'])) { //if there trying to view a profile
//gets the user name and makes it safe
$username = $_GET[user];
//querys the db to find the username
$getuser = mysql_query("SELECT * FROM `users` WHERE `username` = '$username'");
//checks see if the username exists in the db 

$usernum = mysql_num_rows($getuser);
//if it don't exist 
if($usernum == 0) 


{ 
//don't exist

echo ("User Not Found"); 

} 

//if it does exist then show there profile
else{
$user = mysql_fetch_array($getuser);

//to display image from source
$dir = "prof_pics";

$sql = "SELECT prof_pic FROM users WHERE username = '$username'";
$res = mysql_query($sql) or die(mysql_error());
if(mysql_num_rows($res) == 0) die("Username not found in database.");

$row = mysql_fetch_array($res);
$pic="$dir/".$row['prof_pic'];
$img="<img src=\"$pic\" width=\"88\" height=\"88\" align=\"center\"><br>";
echo "
<b>$user[username]'s Profile</b><br><br>
$img <br>
Email: $user[email]<br>
<a href='friendrequest.php?user=$user[username]'>Add as Friend</a>
";

}
}else{
//gets all the members from the database
$getusers = mysql_query("SELECT * FROM `users` ORDER BY `uid` ASC") or die(mysql_error());

//loops there name out
while (

$user = mysql_fetch_array($getusers)) 
{ 
echo "<div id='memcontainer'>
<div id='mempic'>$img</div>
<div id='memname'><a href='members.php?user=$user[username]'>$user[username]</a></div></div><br>
"; 
}
}
echo "<center>";
?>

 

The first instance of the $img works fine, but not the second one?

[djlee]

compare the code, your defining $img in the first one but not the second block of code for output.

[AyKay47]

you defined the $img variable inside of the if else statement, and only there can it be called without redefining it

[Zane]

$img probably doesn't work because you only create it at a condition.  you should create the $img before any if statements if you want to use it multiple times.  And I'm sure there's another easier fix for it, but I really don't want to look through all that code.  You should post snippets of what you're talking about below all that.

[AdRock]

you defined the $img variable inside of the if else statement, and only there can it be called without redefining it

 

I was just about to say that.

 

The image hasn't been called in the first place if the else is taken.

[hoponhiggo]

ah right. didnt realise i had to redefine the variable.

 

Does that mean i would also have to redefine the $username variable too?

 

//to display image from source again
$dir = "prof_pics";

$sql = "SELECT prof_pic FROM users WHERE username = '$username'";
$res = mysql_query($sql) or die(mysql_error());
if(mysql_num_rows($res) == 0) die("Username not found in database.");

$row = mysql_fetch_array($res);
$pic="$dir/".$row['prof_pic'];
$img="<img src=\"$pic\" width=\"88\" height=\"88\" align=\"center\"><br>";

[AyKay47]

yes, but I would do that before the condition anyway.

 

$username = $_GET['user'];
if(isset($username)) { 

this way you won't have to redefine

[hoponhiggo]

$username = $_GET['user'];

if(isset($username)) {

 

i cant get this to work. based on my code above, where exactly would you put this?

[AyKay47]

<?php
echo "<center>";
$username = $_GET['user']; //very top 
if(isset($username) { //if there trying to view a profile
//gets the user name and makes it safe
//querys the db to find the username
$getuser = mysql_query("SELECT * FROM `users` WHERE `username` = '$username'");
//checks see if the username exists in the db 

$usernum = mysql_num_rows($getuser);
//if it don't exist 
if($usernum == 0) 


{ 
//don't exist

echo ("User Not Found"); 

} 

//if it does exist then show there profile
else{
$user = mysql_fetch_array($getuser);

//to display image from source
$dir = "prof_pics";

$sql = "SELECT prof_pic FROM users WHERE username = '$username'";
$res = mysql_query($sql) or die(mysql_error());
if(mysql_num_rows($res) == 0) die("Username not found in database.");

$row = mysql_fetch_array($res);
$pic="$dir/".$row['prof_pic'];
$img="<img src=\"$pic\" width=\"88\" height=\"88\" align=\"center\"><br>";
echo "
<b>$user[username]'s Profile</b><br><br>
$img <br>
Email: $user[email]<br>
<a href='friendrequest.php?user=$user[username]'>Add as Friend</a>
";

}
}else{
//gets all the members from the database
$getusers = mysql_query("SELECT * FROM `users` ORDER BY `uid` ASC") or die(mysql_error());

//loops there name out
while (

$user = mysql_fetch_array($getusers)) 
{ 
echo "<div id='memcontainer'>
<div id='mempic'>$img</div>
<div id='memname'><a href='members.php?user=$user[username]'>$user[username]</a></div></div><br>
"; 
}
}
echo "<center>";
?>

[hoponhiggo]

Sorry, still getting my 'username not found in database' message.

 

I believe the problem is to do with my $username in the mysql query?

 

Full code below if anybody can help?

 

<?php
echo "<center>";
$username = $_GET[user];
//gets the user name and makes it safe
if(isset($_GET['user'])) { //if there trying to view a profile
//querys the db to find the username
$getuser = mysql_query("SELECT * FROM `users` WHERE `username` = '$username'");
//checks see if the username exists in the db 

$usernum = mysql_num_rows($getuser);
//if it don't exist 
if($usernum == 0) 


{ 
//don't exist

echo ("User Not Found"); 

} 

//if it does exist then show there profile
else{
$user = mysql_fetch_array($getuser);

//to display image from source
$dir = "prof_pics";

$sql = "SELECT prof_pic FROM users Where username = '$username'";
$res = mysql_query($sql) or die(mysql_error());
if(mysql_num_rows($res) == 0) die("Username not found in database.");

$row = mysql_fetch_array($res);
$pic="$dir/".$row['prof_pic'];
$img="<img src=\"$pic\" width=\"88\" height=\"88\" align=\"center\"><br>";
echo "
<b>$user[username]'s Profile</b><br><br>
$img <br>
Email: $user[email]<br>
<a href='friendrequest.php?user=$user[username]'>Add as Friend</a>
";

}
}else{
//gets all the members from the database
$getusers = mysql_query("SELECT * FROM `users` ORDER BY `uid` ASC") or die(mysql_error());

//to display image from source again
$dir = "prof_pics";

$sql = "SELECT prof_pic FROM users Where username = '$username'";
$res = mysql_query($sql) or die(mysql_error());
if(mysql_num_rows($res) == 0) die("Username not found in database.");

$row = mysql_fetch_array($res);
$pic="$dir/".$row['prof_pic'];
$img="<img src=\"$pic\" width=\"88\" height=\"88\" align=\"center\"><br>";

//loops there name out
while (

$user = mysql_fetch_array($getusers) )

{ 
echo "<table><td><div id='memcontainer'>
<div id='mempic'>$img</div>
<div id='memname'><a href='members.php?user=$user[username]'>$user[username]</a></div></div><br></td>
</table>
"; 
}
}
echo "<center>";
?>

[hoponhiggo]

I have changed the later part of the code to:

 

else{
//gets all the members from the database
$getusers = mysql_query("SELECT * FROM `users` ORDER BY `uid` ASC") or die(mysql_error());

//loops there name out
while (

$user = mysql_fetch_array($getusers) )
$username = $user[username];
//to display image from source again
$dir = "prof_pics";

$sql = "SELECT prof_pic FROM users Where username = '$username'";
$res = mysql_query($sql) or die(mysql_error());
if(mysql_num_rows($res) == 0) die("Username not found in database.");

$row = mysql_fetch_array($res);
$pic="$dir/".$row['prof_pic'];
$img="<img src=\"$pic\" width=\"88\" height=\"88\" align=\"center\"><br>";

{ 
echo "<table><td><div id='memcontainer'>
<div id='mempic'>$img</div>
<div id='memname'><a href='members.php?user=$user[username]'>$user[username]</a></div></div><br></td>
</table>
"; 

 

Which is now only showing one record, and other info such as $user is missing?

[AyKay47]

you need to be wrapping your indices in quotes..

 

$username = $_GET[user];

should be

$username = $_GET['user']; //$_GET is associative and the index needs to be wrapped in quotes unless its an integer

[hoponhiggo]

Ok, wrapped them up. Still no change. Any other ideas?

[dcro2]

You're missing some braces to start and end the while() so only this line is being executed for each row:

$username = $user[username];

I can't figure out why this brace is here either:

$img="<img src=\"$pic\" width=\"88\" height=\"88\" align=\"center\"><br>";

{ 
echo "<table><td><div id='memcontainer'>