Load php image from string

Question

I have a text file called database.txt and it contains 3 URLs separated with a " - ".

<?
$dir = $_POST['path123'];
$percorso = file($dir."/database.txt");
while(list(,$value) = each($percorso)){
list($gp1, $gp2, $gp3) = split("[:]", $value);
#declaration trim()
$params["gp1"] = trim($gp1);
$params["gp2"] = trim($gp2);
$params["gp3"] = trim($gp3);
#print results
}
echo '<img src="$gp1" border=0>';
?>

As you can see, path123 is the name of the folder and $percorso is the path of the database.txt. With that code I should load the 3 URLs in 3 different variables (gp1, gp2 and gp3).

My problem is that when I use echo '<img src="$gp1" border=0>' mozilla gives me an error here that says "syntax error, unexpected $end". So I can't show on the screen the 1st URL on my database.txt file. Any help?

you didn't close your while

Lefsler
– Lefsler

2013-07-08 14:10:42 +00:00
Commented Jul 8, 2013 at 14:10 — Lefsler
– Lefsler, Commented Jul 8, 2013 at 14:10
fixed. By the way I still can't see the image

Emma Rossignoli
– Emma Rossignoli

2013-07-08 14:12:14 +00:00
Commented Jul 8, 2013 at 14:12 — Emma Rossignoli
– Emma Rossignoli, Commented Jul 8, 2013 at 14:12
Can you show us the resulting HTML?

user254875486
– user254875486

2013-07-08 14:12:39 +00:00
Commented Jul 8, 2013 at 14:12 — user254875486
– user254875486, Commented Jul 8, 2013 at 14:12

Amal · Accepted Answer · 2013-07-08 14:18:29Z

3

Change

list($gp1, $gp2, $gp3) = split("[:]", $value); //will output http://

to

list($gp1, $gp2, $gp3) = split("[-]", $value); //will output http://linkhere.tld

answered Jul 8, 2013 at 14:18

Amal

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Comments

Guerra · Accepted Answer · 2013-07-08 14:13:32Z

2

Change

echo '<img src="$gp1" border=0>';

To

echo '<img src="'.$gp1.'" style="border: none;" />';

answered Jul 8, 2013 at 14:13

Guerra

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