File Exists? Display image else display default image

Question

I can't seem to get the following code to work?

$playerfileLocation = "../wp-content/gallery/playerphotos/' . $playerid . '.jpg";
if(file_exits($playerfileLocation)) 
echo "The file File exists";
else
echo "The file File does not exist";

The "playerid" field is a number that gets passed through.

I just can't seem to get it working. Urgh!

No...the page stops once it gets to the if statement. I have text below this "test" that is not getting displayed. — DoubleA
– DoubleA, Commented Dec 24, 2011 at 4:21

Virendra · Accepted Answer · 2011-12-24 07:33:39Z

5

You have mismatch of quotes. Try this code.

$playerfileLocation = "../wp-content/gallery/playerphotos/" . $playerid . ".jpg";
if(file_exists($playerfileLocation)) 
echo "The file File exists";
else
echo "The file File does not exist";

Update: Actually I would suggest using below code as whenever PHP see double-quotes, it tries to parse anything in between it, which is not required in this case. This is a small optimization in performance.

$playerfileLocation = '../wp-content/gallery/playerphotos/' . $playerid . '.jpg';
if(file_exists($playerfileLocation)) 
echo "The file File exists";
else
echo "The file File does not exist";

Also, to just check if file exists and if not display default image use the following code:

$playerfileLocation = '../wp-content/gallery/playerphotos/' . $playerid . '.jpg';
$defaultfileLocation = '../wp-content/gallery/playerphotos/default.jpg';
if (!file_exists($playerfileLocation)) {
    $playerfileLocation = $defaultfileLocation;
}

edited Dec 24, 2011 at 7:33

answered Dec 24, 2011 at 4:16

Virendra

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3 Comments

Somnath Muluk Over a year ago

What's problem for double quotes in whenever PHP see double-quotes, it tries to parse anything in between it, which is not required in this case. This is a small optimization in performance.

Virendra Over a year ago

When PHP sees double quotes it would try to parse it which takes some time (although it is very less, but still takes some time). If you use single quotes then PHP does not need to parse it saving the time required to parse it. In this case there are no variables in the string and hence its not required to parse it, unless you do $playerfileLocation = "../wp-content/gallery/playerphotos/$playerid.jpg"; This will require PHP to parse the entire string. Also, the code is now not very readable.

DoubleA Over a year ago

I finally figure out the problem It was related to the path I was using. I need to define an absolute path: $absolutepath = $_SERVER{'DOCUMENT_ROOT'}.'wp-content/gallery/playerphotos/' . $photoname . '.jpg'; The file_exists function works fine now. ;)

Alireza · Accepted Answer · 2011-12-29 03:31:35Z

1

Your code is misspelled: (exist not exits)

$playerfileLocation = "../wp-content/gallery/playerphotos/$playerid.jpg";

$default_player = "../wp-content/gallery/playerphotos/default.jpg";

if(file_exits($playerfileLocation)) 

{

}

else

{

$playerfileLocation=$default_player;

}

edited Dec 29, 2011 at 3:31

answered Dec 24, 2011 at 4:31

Alireza

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1 Comment

DoubleA Over a year ago

Well now I feel like an idiot. Thank you for solving my problem.

Somnath Muluk · Accepted Answer · 2011-12-24 07:15:54Z

0

In php, we can get variable value in double quotes also. It's good practice to use double quotes in path, so no much indentation is required.

$playerfileLocation = "../wp-content/gallery/playerphotos/$playerid.jpg";


$default_player = "../wp-content/gallery/playerphotos/default.jpg";
if(!file_exists($playerfileLocation)) 
{
$playerfileLocation=$default_player;
}

edited Dec 24, 2011 at 7:15

answered Dec 24, 2011 at 4:21

Somnath Muluk

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