[Resolved] function won't display images

brokeDUstudent

hey hows it goin,
i tried creating this function to insert 3 screenshots depending on who's logged in.

function InsertScreens(){

$getimages = mysql_query("SELECT image1, image2, image3 FROM screenshots WHERE username = '$username'");
$image = mysql_fetch_row($getimages);

$counter = 0;
$number = 0;

WHILE ($counter < 3 AND $number < 3) {

echo "<img src='../images/screenshots/";
echo $image[$counter];
echo $number;
echo ".gif'><br><br>";
$counter++;
$number++;
}
}
echo InsertScreens();

I'm not sure why, but it's only displaying 2 images and the image names are 1.gif and 2.gif.. I thought it would output username0.gif, username1.gif, username2.gif. anybody have any suggestions? I'll try changing some stuff around to see if it works, thanks for any help.

stolzyboy

what are the values for the images in the database, and do they always come from image1, image2, and image3 field from the table

brokeDUstudent

the setup probably isnt the best, but here's generally how the table is setup..

the values for image1, image2, and image3 are the same as the username.. (not very creative) and the images will always be there

its basically like this:

username image1 image2 image3

johndoe johndoe johndoe johndoe

and in the folder where the images are they're named:
johndoe0.gif
johndoe1.gif
johndoe2.gif

so I was thinking i could just use the mysql_fetch_row to get each image and then the $number to put 0, 1, and 2 on the end of the filename. ? I haven't been programming too long so I don't know if thats the best way, but it seemed simple enough. Thanks for your help again..

stolzyboy

well then you do it like this

$query = "blah, ";
$result = mysql_query($query);

$row = $row = mysql_fetch_array($result);

echo "Image One: " . $row['image1'] . "<br>";
echo "Image Two: " . $row['image2'] . "<br>";
echo "Image Three: " . $row['image3'] . "<br>";

that'll get the results, then you just simply put it into the <img> tag