Php variable not showing up?

Question

Hi I'm trying to change the picture for facebook thumbnail by having php set the image in the meta tag depending on the page. I have this code but for some reason when I go to Facebook and debug it it actually shows the variable and not what the value of the variable is. Here is my code please help thank you!

<?php
if($ifpage == 'picture.php')
{
$metaimage = '<meta property="og:image" content="http://www.quickdailylaugh.com/photos/print $picture" />';
print($metaimage);
}
?>

Do you have PHP installed on the server? If you do, are you sure it is configured properly? View the page source — do you see raw `` tags? — Amal Murali, Apr 16 '14 at 17:46
yes i have php , my entire site is built with it and no raw php tags — user3495551, Apr 16 '14 at 17:48
Well, your code doesn't make much sense. You're trying to use the variable `$picture` on line 4, which is never defined. That would throw a `Notice` (if you have [error reporting enabled](http://stackoverflow.com/a/6575502/1438393)). — Amal Murali, Apr 16 '14 at 17:48
Maybe it's defined in `picture.php` (*who knows*) @AmalMurali OP will need to provide more details. — Funk Forty Niner, Apr 16 '14 at 17:49
@Fred-ii-: But it wouldn't still be available in this page. He's not including it here, he's only comparing if the value of `$ifpage` is equal to `picture.php`. (Oh, and by the way, `$ifpage` isn't defined either.) — Amal Murali, Apr 16 '14 at 17:51
$picture is defined in picture.php and $ifpage is defined as well in the header.php — user3495551, Apr 16 '14 at 17:55

score 1 · Answer 1 · answered Apr 16 '14 at 17:50

1

Your variable is inside of a string enclosed with single quotes. Single quotes will take the literal value of the string $varname and not translate the variable to it's value. You need to use double quotes. Example:

$var1 = 'test';
$foo = 'The value of var1 is: $var1'; // The value of var1 is: $var1
$bar = "The value of var1 is: $var1"; // The value of var1 is: test

answered Apr 16 '14 at 17:50

Jonathan Kuhn

15,279
3
32
43

1

The OP's code is full of undefined variables, so even if this is fixed, I doubt it will work. – Amal Murali Apr 16 '14 at 17:52
@AmalMurali I would say it is more likely that this isn't the full code and the variables are defined elsewhere. Or maybe they are defined in the query string and register globals = On. Or maybe they are defined in an auto_prepend file. Point is, it is still possible this is the complete code and no errors would be thrown. – Jonathan Kuhn Apr 16 '14 at 17:54

score 1 · Answer 2 · answered Apr 16 '14 at 17:52

1

To interpret the varible you must use double quotes.

<?php
if($ifpage == 'picture.php')
{
$metaimage = "<meta property=\"og:image\" content=\"http://www.quickdailylaugh.com/photos/print $picture\" />";
print($metaimage);
}
?>

answered Apr 16 '14 at 17:52

vgrdominik

91
4

okay i added that but for somereason print still shows ? – user3495551 Apr 16 '14 at 17:57
@user3495551 what are you expecting `print` to do? php can only do what you tell it and right there you are saying display `print`. `print` isn't a variable or anything. You have it as a literal part of that string. – Jonathan Kuhn Apr 16 '14 at 18:01
i want it to print the variable – user3495551 Apr 16 '14 at 18:04
nevermind i found out – user3495551 Apr 16 '14 at 18:08

score 0 · Answer 3 · answered Apr 16 '14 at 18:25

I had to do another query for it to show... Thank you all for the answers here is what i used...

<?php
if($ifpage == 'picture.php')
{
$photoid = $_GET['picture'];
$sql = "SELECT * FROM photos WHERE id=$photoid";
$pic = mysql_query($sql);
while($row = mysql_fetch_array($pic))
        {
$picture=$row['picture'];
}
$metaimage = "<meta property=\"og:image\" content=\"http://www.quickdailylaugh.com/photos/$picture\" />";
        print($metaimage);
}else{
?>

Php variable not showing up?

3 Answers3