Fast "IF" Help

[erikhillis]

Hey, I wanted to limit my results to a certin thing... but dont know what to do...

I attempted to do the IF statement on my own, but couldnt get it to work... can someone please help? I want to only show "active" staff members.

<?php
$db = mysql_connect("localhost","jointops_site","pw");
mysql_select_db (jointops_site);
$query = "SELECT * FROM staff ORDER BY id asc";
if (!$status = active {
}
$result = mysql_query($query);
echo ("<table width=100%>");
while ($rows = mysql_fetch_row($result))
{
echo ("<tr>
<td width=25%><font face=Verdana size=1 color=000000><B>»</B> $rows[1]</td>
<td width=35%><font face=Verdana size=1 color=000000>$rows[6]</td>
<td width=20%><font face=Verdana size=1 color=000000><center>$rows[7]</td>
<td width=20%><font face=Verdana size=1 color=000000><center><a href=staff.php?id=$rows[0]>[Profile]</td></tr>");
}
echo "</table>";
?>

I tried this, dosnt work.

[dalecosp]

When you test a value you need to use "=="

if (!(status=="active")) {

this might work, but:

if ($status!="active") {

is standard usage...

HTH

[erikhillis]

This still just displays everyone...

           <?php
// Connect to the database
$db = mysql_connect("localhost","jointops_site","pass");
mysql_select_db (jointops_site);
// Ask the database for the information from the links table
$query = "SELECT * FROM staff ORDER BY id asc";
$result = mysql_query($query);
if (!(status=="inactive")) {
}
// Now we print out the results, starting by making a table
echo ("<table width=100%>");
while ($rows = mysql_fetch_row($result))
// Here we make the script keep making new rows until all the links in our database are shown, this is called a loop
{
echo ("<tr>
<td width=25%><font face=Verdana size=1 color=000000><B>»</B> $rows[1]</td>
<td width=35%><font face=Verdana size=1 color=000000>$rows[6]</td>
<td width=20%><font face=Verdana size=1 color=000000><center>$rows[7]</td>
<td width=20%><font face=Verdana size=1 color=000000><center><a href=staff.php?id=$rows[0]>[Profile]</td></tr>");
}
// Finally we close off the table
echo "</table>";
?>

Help with the code?

[dalecosp]

OK, so we're back to my original thought, which I didn't post, because another thought hit me at the time.

I was assuming that $active was some value whereby you determined whether or not you wanted to show the data. Is it, instead, that you want to show only a certain amount of data, i.e., staff members that are 'active'?

If that's the case, how do we know which of the staff are 'active'? Is this info in the db?

And, it would be better, (though it seems irrelevant if I'm right about the question now), to use

if ($status !="active")

[sarahk]

better still change your query to

$query = "SELECT * FROM staff where status != 'active' ORDER BY id asc";

your check for status happens before you have any database data to view

so the whole script should be

 <?php 
$db = mysql_connect("localhost","jointops_site","pw"); 
mysql_select_db (jointops_site); 
$query = "SELECT * FROM staff where `status` != 'active' ORDER BY id asc"; 

$result = mysql_query($query); 
echo "<table width='100%'>"; 
while ($rows = mysql_fetch_row($result)) 
{ 
  echo "<tr><td width='25%'><font face=Verdana size=1 color='000000'><B>»</B> {$rows[1]}</td> 
<td width=35%><font face=Verdana size=1 color='000000'>{$rows[6]}</td> 
<td width='20%'><font face=Verdana size=1 color=000000><center>{$rows[7]}</td> 
<td width=20%><font face=Verdana size=1 color=000000><center><a href=staff.php?id={$rows[0]}>[Profile]</a></td></tr>"; 
} //while
echo '</table>'; 
?> 

comments on your code
if you embed an array variable in a double quoted string it must have {} around it or it won't work
if you use % for a table cell width it must have quotes around it
* your profile hyperlink didn't have a closing </a>