undefine variable ive tried everything

[minus84]

i dont knw why its keep telling me

 

Notice: Undefined variable: lowdownDisplay in C:\wamp\www\bullshit.php on line 112

 

some1 please tell me wot i can do to fix this

 

 

$sql = mysql_query("SELECT id, mem_id, the_lowdown, post_time FROM lowdown_db ORDER BY post_time DESC LIMI 20");
//$lowdownDisplayList = '';

while ($row = mysql_fetch_array($sql)) {
//$lowdownDisplay = '<table width="0%"  border="0" align="center" cellpadding="5"> 
					//<tr>';
$lowdownid = $row["id"];
$uid = $row["mem_id"];
$the_lowdown = $row["the_lowdown"];
$the_date = $row["post_time"];
$the_date = strftime("%b %d, %Y", strtotime($post_time));
///// inner SQL QUERY
$sql_mem_data = mysql_query("SELECT id, firstname, lastname FROM mymembers WHERE id='$uid' LIMIT 1");
$lowdownDisplayList.= '<td><div style="height:80px; overflow:hidden;" > ' .$user_pic. ' </div><a href="profile.php?id=' .$id. '"><br>'.$firstname.' '.$lastname.'</a></td> ';
//$lowdownDisplayList = ''<td><div style="height:80px; overflow:hidden;" > ' .$user_pic. ' </div><a href="profile.php?id=' .$id. '"><br>'.$firstname.' '.$lastname.'</a></td> ';
	while ($row = mysql_fetch_array($sql_mem_data)){
	$uid = $row["id"];
	$ufirstname = $row["firstname"];
	$ufirstname = $row["lastname"];
	$ufirstname = substr($ufirstname,0,10);
	///////display lowdown pic//////////
	$ucheck_pic = "memberFiles/$uid/image01.jpg";
	$udefault_pic = "memberFiles/0/image01.jpg";
	if (file_exists($ucheck_pic)) {
    	$lowdown_pic = "<img src=\"$check_pic\" width=\"40px\" />"; // forces picture to be 100px wide and no more
	} else {
	$lowdown_pic = "<img src=\"$default_pic\" width=\"40px\" />"; // forces default picture to be 100px wide and no more
}

	$lowdownDisplayList.= '
	<table width="95%" align="center" cellpadding="4" bgcolor="#ffffff">
	<tr>
	<td width="7%" height="161" bgcolor="#996600"><a href="profile.php?id=' .$id .'">' .$lowdown_pic . '</a><br/>		</td>
	 <td width="93%"   bgcolor="#CCCCCC" ><a href="profile.php?id=' .$id .'">' .$ufirstname . ' ' .$ulastname . ' </a>
	 • <span style="font-size:10px;"> ' .$post_time . ' </span><br/>
	 ' .$the_lowdown. ' </td>
	</table>';
	}

}
?>

[code/]

[pornophobic]

before you try

$lowdownDisplayList.=

Put in something like this:

$lowdownDisplayList = '';

That'll fix your problem.

[fortnox007]

$lowdownDisplayList.= '  <----- is bad

 

Do you have a program that shows line numbers? I use NetBeans, easier to detect errors

[pornophobic]

$lowdownDisplayList.= '  <----- is bad

 

Why?

[objnoob]

To avoid these notices, make sure to define variables before trying to to use them.

 

Remember, using a self acting concatenation or operand such as .= or +=  will rely on the said variable's value prior to reassignment. So if said variables value is unset, an undefined variable notice is thrown.

 

.=  <-- is not bad. Just declare the variable first using $myVar = null;

[minus84]

thnx i got that problem solved but the table styll wont loop to take the information out ot the database. i dont knw why im using the same sciprt on another loop and it works fine

 

  <?
   
   $sql_lowdown = mysql_query("SELECT id, mem_id, the_lowdown, post_time FROM lowdown_db WhHERE mem_id='$id' ORDER BY post_time DESC LIMI 20");

while($row = mysql_fetch_array($sql_lowdown)) {

$lowid = $row["id"];
$uid = $row["mem_id"];
$the_lowdown = $row["the_lowdown"];
$the_date = $row["post_time"];
$the_date = strftime("%b %d, %Y", strtotime($post_time));

$LowdownDisplayList.= '
	<table width="95%" align="center" cellpadding="4" bgcolor="#ffffff">
	<tr>
	<td width="7%" height="161" bgcolor="#996600"><a href="profile.php?id=' .$id .'">' .$lowdown_pic . '</a><br/>		</td>
	 <td width="93%"   bgcolor="#CCCCCC" ><a href="profile.php?id=' .$id .'">' .$firstname . ' ' .$lastname . ' </a>
	 • <span style="font-size:10px;"> ' .$post_time . ' </span><br/>
	 ' . $the_lowdown . ' </td>
	</table>';
   }
   
   ?>

[mikosiko]

in the code that you were showing (you just changed the code in your post) the variable $id is undefined, that will more likely cause that your query

$sql_lowdown = mysql_query("SELECT id, mem_id, the_lowdown, post_time FROM lowdown_db WhHERE mem_id='$id' ORDER BY post_time DESC LIMI 20");

will either fail or not return any value.

 

you must validate that your $id variable effectively has some value and then  rewrite that code in this alternative way:

$sql_lowdown = mysql_query("SELECT id, mem_id, the_lowdown, post_time FROM lowdown_db WhHERE mem_id='$id' ORDER BY post_time DESC LIMI 20") or die("Query Error: " . mysql_error());
if (mysql_num_rows($sql_lowdown) > 0) {
   // The query returned values... therefore here you canb continue the rest of your code
} else { echo "..No records found.."; };