php / msql help displaying graphics

[Keith]

I am wanting to display a different graphic depending on the content of a field in a mysql database my code is below only problem is XX.gif displays regardless of content,


if ($row>X=="XX") {
$output .= "
<div><img src=XX.gif></div>";
}

i'm assuming its a basic error but i'm only a beginner when it comes to mysql

[OLBG]

Not sure I understand your sytax there but maybe its because I dont do the object oriented php

Hopefull this example will help

$mysql="SELECT link,image FROM table WHERE category='Lamps'";
$result=mysql_query($mysql);
while($row=mysql_fetch_array($result))
{
echo "<p><as href=\"".$row['link']."\"><image border=0 src=\"".$row['image']."\"></a></p>";
}

[morleymouse]

if ($row>X=="XX") {
$output .= "
<div><img src=XX.gif></div>";
}

$row>X=="XX"

should possibly be...


$row=="XX"

so...

if ($row=="XX") {
$output .= "
<div><img src=XX.gif></div>";
}


think might be about right, easier to see more code but the if greater than X == XX thing looked a bit off

[Keith]

thanks for the input guys, all sorted now, I was trying to over complicate things

[Marc Gear]

Originally Posted by Keith if ($row>X=="XX") { $output .= " <div><img src=XX.gif></div>"; }

$output = '';
$result = mysql_query('SELECT foo FROM bar');
$row = mysql_fetch_object($result);
if ($row->foo == 'xx' || $row == 'yy')
{
  $output .= '<div><img src="'.$row->foo.'.gif"></div>';
}