[SOLVED] no user with id echo

runnerjp · December 8, 2007

<?php

  

  session_start();

  require_once '../settings.php';



  $query = "SELECT * FROM users WHERE Username = '$username' LIMIT 1"; 

  if ($result = mysql_query($query)) {

    if (mysql_num_rows($result)) { 

      $array = mysql_fetch_assoc($result);

      $pemail = $array['Email'];

      $puser = $array['Username'];

  $pid = $array['ID'];

    } else {

      echo "No users found with id $id<br />";

    }

  } else {

    echo "Query failed<br />$sql<br />" . mysql_error();

  }

$sql = "SELECT `ext` FROM `user_images` WHERE `user_id`='$pid' LIMIT 1"; 

$q = mysql_query($sql) or die("Error running query:".mysql_error());

if($q && mysql_num_rows($q) > 0){

  $row = mysql_fetch_array($q);

  if(!empty($row)){

    echo "<img src='http://www.runningprofiles.com/members/images/".

         $pid . "." . $row['ext']

         . "' />";

  }

}else{

  echo '<img src="images/pic.jpg">';

  }

?>

where it says no users with id.. how can i make it so only that comes up as at the moment it shows no users with id but also shows the other information?

revraz · December 8, 2007

Change your logic so when the record is found you do the rest of your code, and if not, it returns the error and then their is no code after.

You only want to take action if it found what you want it to.

runnerjp · December 8, 2007

ok so could i have example to work with or link as i dont quite get you lol

revraz · December 8, 2007

Something like this

<?php
  session_start();
  require_once '../settings.php';
  $query = "SELECT * FROM users WHERE Username = '$username' LIMIT 1"; 

  if ($result = mysql_query($query)) {
    if (mysql_num_rows($result)) { 
      $array = mysql_fetch_assoc($result);
      $pemail = $array['Email'];
      $puser = $array['Username'];
  	$pid = $array['ID'];		
		$sql = "SELECT `ext` FROM `user_images` WHERE `user_id`='$pid' LIMIT 1"; 
		$q = mysql_query($sql) or die("Error running query:".mysql_error());

		if($q && mysql_num_rows($q) > 0){
  				$row = mysql_fetch_array($q);

							 if(!empty($row)){
    							echo "<img src='http://www.runningprofiles.com/members/images/". $pid . "." . $row['ext']  . "' />";
  						}
	}
else {

  echo '<img src="images/pic.jpg">';
  	}
} 
else {
      echo "No users found with id $id<br />";
    }
  } 
else {
    echo "Query failed<br />$sql<br />" . mysql_error();
  }
?>

runnerjp · December 8, 2007

ohhh so have the bumph 1st then add no user last ahh that makes sence

revraz · December 8, 2007

Yeah, I wasnt sure what you were actually doing, but you want it to flow logically. No point going through all the code if one part isn't valid right?

runnerjp · December 8, 2007

ah yes buuuuut what if you want to add a table into the mix

<?php
  session_start();
  require_once '../settings.php';
  $query = "SELECT * FROM users WHERE Username = '$username' LIMIT 1"; 

  if ($result = mysql_query($query)) {
    if (mysql_num_rows($result)) { 
      $array = mysql_fetch_assoc($result);
      $pemail = $array['Email'];
      $puser = $array['Username'];
  	$pid = $array['ID'];		
		$sql = "SELECT `ext` FROM `user_images` WHERE `user_id`='$pid' LIMIT 1"; 
		$q = mysql_query($sql) or die("Error running query:".mysql_error());

		if($q && mysql_num_rows($q) > 0){
  				$row = mysql_fetch_array($q);

							 if(!empty($row)){
    							echo "<img src='http://www.runningprofiles.com/members/images/". $pid . "." . $row['ext']  . "' />";
  						}
	}
else {

  echo '<img src="images/pic.jpg">';
  	}
} 
else {
      echo "No users found with id $id<br />";
    }
  } 
else {
    echo "Query failed<br />$sql<br />" . mysql_error();
  }
?>



<table width="100%">

  <tr>

    <td colspan="3"><div align="center">Welcome to <?echo "$puser"?>'s profile</div></td>

  </tr>

  <tr>

    <td width="162">your id is no <?echo "$pid"?></td>

    <td width="711"></td>

    <td width="100">

  </td>

  </tr>

</table>

revraz · December 8, 2007

You only want to add the table if there was an ID, so you'll have to mix it with your PHP.

You can either keep PHP on and use echo or you can exit PHP, do the HTML, then turn PHP back on.

runnerjp · December 8, 2007

what would be the best way?? how would i turn php off if their is no id then turn it back on again :S

revraz · December 8, 2007

No best way really, it's your preference.

And you just do this:

<?php
echo "PHP is on";
?>
HTML is on
<?php
echo "PHP is back on";
?>

runnerjp · December 8, 2007

ahh yer but i have php in my table... dam lol how would i mix it in the the php??

sorry lots of questions but you explain it well

revraz · December 8, 2007

The exact same way

<table>
<tr>
<td>HTML</td>
<td><?php echo "PHP" ?></td>
</tr>
</table>

runnerjp · December 8, 2007

oh no ok i can allready do that just i dont want the table to apear if no id is found

revraz · December 8, 2007

Right, so you put that part up in the code where the ID is found.

runnerjp · December 8, 2007

what would i just echo it?

revraz · December 8, 2007

Up to you. Either echo or turn off PHP, do your html, then turn php back on.

runnerjp · December 8, 2007

yes but i dont get how that works to echo a table

revraz · December 8, 2007

echo "<table><tr><td>Hi</td></tr></table>";

runnerjp · December 8, 2007

ok i tried

<?php
  session_start();
  require_once '../settings.php';
  $query = "SELECT * FROM users WHERE Username = '$username' LIMIT 1"; 

  if ($result = mysql_query($query)) {
    if (mysql_num_rows($result)) { 
      $array = mysql_fetch_assoc($result);
      $pemail = $array['Email'];
      $puser = $array['Username'];
  	$pid = $array['ID'];		
		$sql = "SELECT `ext` FROM `user_images` WHERE `user_id`='$pid' LIMIT 1"; 
		$q = mysql_query($sql) or die("Error running query:".mysql_error());

		if($q && mysql_num_rows($q) > 0){
  				$row = mysql_fetch_array($q);

							 if(!empty($row)){
    							echo "<img src='http://www.runningprofiles.com/members/images/". $pid . "." . $row['ext']  . "' />";
  						}
	}
else {

  echo '<img src="images/pic.jpg">';
  	}
} 
echo '<table width="100%">

  <tr>

    <td colspan="3"><div align="center">Welcome to <?echo "$puser"?>'s profile</div></td>

  </tr>

  <tr>

    <td width="162">your id is no <?echo "$pid"?></td>

    <td width="711"></td>

    <td width="100">

  </td>

  </tr>

</table>

'
else {
      echo "No users found with id $id<br />";
    }
  } 
else {
    echo "Query failed<br />$sql<br />" . mysql_error();
  }
  ?>

but get unexpected T_STRING, expecting ',' or ';' on line 31

revraz · December 8, 2007

You are already in PHP since you used Echo so this line isnt valid

<td colspan="3"><div align="center">Welcome to <?echo "$puser"?>'s profile</div></td>

you just need

<td colspan="3"><div align="center">Welcome to '.$puser.'\'s profile</div></td>

revraz · December 8, 2007

You also may have it placed wrong since I dont see a ending } and you dont have a semi colon at the end.

runnerjp · December 8, 2007

so hav ei put it wrong by just echoing it

runnerjp · December 8, 2007

<?php
  session_start();
  require_once '../settings.php';
  $query = "SELECT * FROM users WHERE Username = '$username' LIMIT 1"; 

  if ($result = mysql_query($query)) {
    if (mysql_num_rows($result)) { 
      $array = mysql_fetch_assoc($result);
      $pemail = $array['Email'];
      $puser = $array['Username'];
  	$pid = $array['ID'];		
		$sql = "SELECT `ext` FROM `user_images` WHERE `user_id`='$pid' LIMIT 1"; 
		$q = mysql_query($sql) or die("Error running query:".mysql_error());

		if($q && mysql_num_rows($q) > 0){
  				$row = mysql_fetch_array($q);

							 if(!empty($row)){
    							echo "<img src='http://www.runningprofiles.com/members/images/". $pid . "." . $row['ext']  . "' />";
  						}
	}
else {

  echo '<img src="images/pic.jpg">';
  	}
} 
echo '<table width="100%">

  <tr>

    <td colspan="3"><div align="center">Welcome to '.$puser.'\'s profile</div></td>

  </tr>

  <tr>

    <td width="162">your id is no '.$puser.'</td>

    <td width="711"></td>

    <td width="100">

  </td>

  </tr>

</table>


else {
      echo "No users found with id $id<br />";
    }
  } 
else {
    echo "Query failed<br />$sql<br />" . mysql_error();
  }
  ?>

getting Parse error: syntax error, unexpected T_ELSE, expecting ',' or ';' in /home/runningp/public_html/members/profile.php on line 50

revraz · December 8, 2007

You need a ending quote ' after your /table and you need a semicolon ; to end the statement as well as a } probably too.

runnerjp · December 8, 2007

  echo '<img src="images/pic.jpg">';
  	}
} 
{
echo '<table width="100%">

  <tr>

    <td colspan="3"><div align="center">Welcome to '.$puser.'\'s profile</div></td>

  </tr>

  <tr>

    <td width="162">your id is no '.$puser.'</td>

    <td width="711"></td>

    <td width="100">

  </td>

  </tr>

</table'>;

tried it and didnt respond well lol

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[SOLVED] no user with id echo

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