[SOLVED] Using if () statements to display if field = 1 --- Please Help

[careym1989]

I am creating code for a PHP page to display Candidates if they have been approved. Currently, I'm using if () statements. There's a field in the table entitled "approved" that is automatically submitted as 0 on the form. I'm creating a vetting process page to change it to 1 to display after I've confirmed with the candidate's campaign. This code is displaying the else statement only after I upload it. I don't see any errors with the code and the page isn't giving me parse errors.

 

Help would be MUCH appreciated.

 

Thanks a lot, fellow coders!

 

			<?php 

		// BEGIN RECALL FROM DATABASE -- CANDIDATES

mysql_connect("XXX", "XXX", "XXX") or die(mysql_error()); 
mysql_select_db("XXX") or die(mysql_error()); 
$data = mysql_query("SELECT * FROM XXX ORDER BY id DESC") 
or die(mysql_error()); 

if ($approved == '1')
{	
	 	echo "<tr>";
		echo "<td><font color=#000000 face=Arial size=2>$info[n]</font></td>";
		echo "<td width=184><font color=#000000 face=Arial size=2>$info[cc]</font></td>";
		echo "<td width=68><font color=#000000 face=Arial size=2>$info[st]</font></td>";
		echo "<td><font color=#000000 face=Arial size=2>$info[ea]</font></td>";
		echo "<td><font color=#000000 face=Arial size=2>$info[cd]</font></td>";
		echo "<td><a href=$info[url]><font color=#000000 face=Arial size=2>$info[url]</font></a></td>";
		echo "</tr>";
} else {

echo "No candidates have taken the pledge.";

}

?>

[KevinM1]

When you run a query, you must fetch the results.  Right now, the value of your $approved variable is undefined because you didn't bind the results of your query to it.

 

In other words:

$data = mysql_query("SELECT * FROM XXX ORDER BY id DESC");

while($row = mysql_fetch_assoc($data))
{
   if($row['approved'] == 1) //more on this line below
   {
      //display candidate data
   }
}

 

Keep in mind that when using array notation (i.e. $row['approved']), the name within the quotes is supposed to be the column name in the DB you want to access.

[careym1989]

It doesn't display any records. I know there are several entries with $row['approved'] == 1.

 

Thanks a lot for your help though, I really appreciate it.

 

-Carey

[KevinM1]

It doesn't display any records. I know there are several entries with $row['approved'] == 1.

 

Thanks a lot for your help though, I really appreciate it.

 

-Carey

 

Did you fix the other variables too?  As in:

echo "<tr>";
echo "<td><font color='#000000' face='Arial' size='2'>{$row['n']}</font></td>";
echo "<td width='184'><font color='#000000' face='Arial' size='2'>{$row['cc']}</font></td>";
echo "<td width='68'><font color='#000000' face='Arial' size='2'>{$row['st']}</font></td>";
echo "<td><font color='#000000' face='Arial' size='2'>{$row['ea']}</font></td>";
echo "<td><font color='#000000' face='Arial' size='2'>{$row['cd']}</font></td>";
echo "<td><a href='{$row['url']}'><font color='#000000' face='Arial' size='2'>{$row['url']}</font></a></td>";
echo "</tr>";

 

??

 

I ask because your orginal code snippet is filled with variables that seemingly come out of nowhere and aren't related to the query results at all.  Remember: it's your job to bind query results to a variable.  That's what the following is for:

$row = mysql_fetch_assoc($data)

[careym1989]

Thanks a lot -- it worked.

 

I really appreciate it, Nightslyr.

 

Regards,

Carey