Array problem

pinehead18

Ok, well this is the last script for my site really. And i am having a huge problem.

Basically what i do for my pictures is i put them all into an array. Then i use a foreach loop to display them onto the page. However, if one of the pics in the array ie. pic[0] or pic[1] etc.. equals default.gif i want it to leave that array var out and post all the other ones that do not equal default.gif

Any suggestions on how to go about this?

Thank you
Anthony

laserlight

Compare the values, then.

e.g.

foreach ($pic as $value) {
	if ($value != 'default.gif') {
		echo '<img src="' . $value . '" alt="pic">';
	}
}

pinehead18

That does not seem to work. So i thought if i pasted you my code you might be able to help me out a little bit more. I've been pulling my hair out over this for the past couple of days.

Thank you for your help

As you can see i have commented out some other thigns i have tried as well.
Thank you again - Anthony

$sql = "SELECT * FROM users WHERE user='$name'";
$result = mysql_query($sql,$con);
$row = mysql_fetch_array($result) or die(mysql_error());

    $mainpic = $row['mainpic'];

    $pic1 = $row['pic1'];
    $pic2 = $row['pic2'];
    $pic3 = $row['pic3'];
    $pic4 = $row['pic4'];

    $pics = array($pic1,$pic2,$pic3,$pic4);


//foreach ($pics as $key => $value)
//{

//if ($value == "default.gif") { unset ($pics[$key]); }

//}


var_dump($pics);
foreach($pics as $picname)
{

 if ($picname != 'default.gif') {
$table .= "
<td align=center><a href=http://c.com/viewimage.php?image=$picname&name=$name> <span style='text-decoration: none'><img
src=http://lc.com/pics/$picname height=80 width=80></a></span></td>
";

}
}
echo $table;


?>

laserlight

What output do you get?

How does it not work?

I do not have access to your database and webserver (at least you better hope I dont), so I have no idea what your table looks like, and what your output looks like.

pinehead18

the page just gives me a white output.

If i do not use the if($picname != ) if i dont' use that and i allow it to show the array vars that equal default.gif it works great and has no problems.

What should happen is instead of not displaying anything if one var in the array == default.gif it should just ignore that var and display all the other ones.

but the script either wants to show them all. Or not show anything.

laserlight

You're saying that var_dump() doesnt give anything at times?
hmm...

anyway, try a simplified version first:

$sql = "SELECT * FROM users WHERE user='$name'";
$result = mysql_query($sql, $con)
	OR die(mysql_error());
$row = mysql_fetch_array($result);

$mainpic = $row['mainpic'];

$pics = array($row['pic1'], $row['pic2'], $row['pic3'], $row['pic4']);

foreach ($pics as $picname)
{
	if ($picname != 'default.gif') {
		echo '<td align=center><a href="http://c.com/viewimage.php?image=' . $picname . '&name=' . $name .'">
<span style="text-decoration: none">
<img src="http://lc.com/pics/' . $picname . '" height=80 width=80></a></span></td>';
	}
}
?>

pinehead18

That script worked great. I would like to say thank you for your responce and help and especially how fast you were able to help me and respond. You have helped me out a great deal. Time for me to go study what you did so maby I can learn from it.

Thank you again

Anthony