[SOLVED] Checking if an image exists using fsock?

RickyF · July 13, 2007

Hello,

Is there a way to check if an image exists, using fsock or similar?

Thanks for any help!

jitesh · July 13, 2007

<?php

$filename = '/path/to/foo.jpg';

if (file_exists($filename)) {

echo "The file $filename exists";

} else {

echo "The file $filename does not exist";

}

?>

chocopi · July 13, 2007

why dont you just use file_exists() ???

<?php
$filename = '/path/to/foo.txt';

if (file_exists($filename)) {
    echo "The file $filename exists";
} else {
    echo "The file $filename does not exist";
}
?>

or i think this might work (this is untested):

<?php
$fp = fsockopen("www.example.com");
if (!$fp) {
    echo "File does not exist<br />\n";
} else {
    echo "File exist<br />\n";
    fclose($fp);
}
?>

Those are both off of php.net

~ Chocopi

infid3l · July 13, 2007

<?php
$fp = fsockopen("www.example.com", 80, $errno, $errstr, 30);
if (!$fp) {
    echo "$errstr ($errno)<br />\n";
} else {
    $out = "HEAD /pics/image.jpg HTTP/1.1\r\n";
    $out .= "Host: www.example.com\r\n";
    $out .= "Connection: Close\r\n\r\n";

    fwrite($fp, $out);
    while (!feof($fp)) {
        $data .= fgets($fp, 128);
    }
    fclose($fp);
}

if(eregi("404 Not Found\r\n", $data)) {
    $exists = false;
} else {
    $exists = true;
}
?>

To see if a off-server file exists (chocopi's code only checks if a server is online, missing arguments).

RickyF · July 13, 2007

<?php
$fp = fsockopen("www.example.com", 80, $errno, $errstr, 30);
if (!$fp) {
    echo "$errstr ($errno)<br />\n";
} else {
    $out = "HEAD /pics/image.jpg HTTP/1.1\r\n";
    $out .= "Host: www.example.com\r\n";
    $out .= "Connection: Close\r\n\r\n";

    fwrite($fp, $out);
    while (!feof($fp)) {
        $data .= fgets($fp, 128);
    }
    fclose($fp);
}

if(eregi("404 Not Found\r\n", $data)) {
    $exists = false;
} else {
    $exists = true;
}
?>

To see if a off-server file exists (chocopi's code only checks if a server is online, missing arguments).

Hello, sorry i didnt make my request clear enough!! I required off server image checking, thank you very much!

RickyF · July 13, 2007

The image string is $imageurl

The script seems to try and use www.example.com and then the image location on the domain seperately, can you make it so that $imageurl is the image thats checked to see if it exists?

Thanks!

infid3l · July 13, 2007

The image string is $imageurl

The script seems to try and use www.example.com and then the image location on the domain seperately, can you make it so that $imageurl is the image thats checked to see if it exists?

Thanks!

Change www.example.com to the host, ie. www.photobucket.com.

If you want a variable for the image, do:

$out = "HEAD /pics/$imageurl HTTP/1.1\r\n";

or something similar

RickyF · July 13, 2007

There are two strings, one is imageurl the other is weburl

An example of a users input would be:

imageurl - http://www.test.com/image.jpg

weburl - http://www.test.com

http:// is posted - i think this may cause problems, is there a way to send whatever comes after http://? so that only www.test.com is sent to the image checker

infid3l · July 13, 2007

There are two strings, one is imageurl the other is weburl

An example of a users input would be:

imageurl - http://www.test.com/image.jpg

weburl - http://www.test.com

http:// is posted - i think this may cause problems, is there a way to send whatever comes after http://? so that only www.test.com is sent to the image checker

$url = "http://www.example.com/pic.jpg";
$url = str_ireplace("http://", null, $url); // www.example.com/pic.jpg
$page = substr($url, strpos($url, "/"), strlen($url)); // /pic.jpg
$url = str_replace($page, null, $url); // www.example.com

edit: You can also do (recommended):

$url = "http://www.example.com/pic.jpg"; 
$parts = explode("/",  str_ireplace("http://", null, $url), 2);
$url = $parts[0]; // www.example.com
$page = $parts[1]; // pic.jpg

RickyF · July 13, 2007

Would this be correct?


$url = "$imageurl"; 
$parts = explode("/",  str_ireplace("http://", null, $url), 2);
$url = $parts[0];
$page = $parts[1];


$fp = fsockopen("$parts[0]", 80, $errno, $errstr, 30);
if (!$fp) {
    echo "$errstr ($errno)<br />\n";
} else {
    $out = "HEAD $parts[0] HTTP/1.1\r\n";
    $out .= "Host: $parts[1]\r\n";
    $out .= "Connection: Close\r\n\r\n";

    fwrite($fp, $out);
    while (!feof($fp)) {
        $data .= fgets($fp, 128);
    }
    fclose($fp);
}

if(eregi("404 Not Found\r\n", $data)) {
    die ("image doesnt exist");
}

infid3l · July 13, 2007

Would this be correct?

$url = "$imageurl";
$url = str_ireplace("http://", null, $url); 
$page = strsub($url, strpos($url, "/"), strlen($url); 
$url = str_replace($page, null, $url); 

$fp = fsockopen("$page", 80, $errno, $errstr, 30);
if (!$fp) {
    echo "$errstr ($errno)<br />\n";
} else {
    $out = "HEAD $page HTTP/1.1\r\n";
    $out .= "Host: $weburl\r\n";
    $out .= "Connection: Close\r\n\r\n";

    fwrite($fp, $out);
    while (!feof($fp)) {
        $data .= fgets($fp, 128);
    }
    fclose($fp);
}

if(eregi("404 Not Found\r\n", $data)) {
    die ("image doesnt exist");
}

$parts = explode("/",  str_ireplace("http://", null, $imgurl), 2);
$url = $parts[0]; // www.example.com
$page = "/".$parts[1]; // pic.jpg

$fp = fsockopen($url, 80, $errno, $errstr, 30);
if (!$fp) {
    echo "$errstr ($errno)<br />\n";
} else {
    $out = "HEAD $page HTTP/1.1\r\n";
    $out .= "Host: $url\r\n";
    $out .= "Connection: Close\r\n\r\n";

    fwrite($fp, $out);
    while (!feof($fp)) {
        $data .= fgets($fp, 128);
    }
    fclose($fp);
}

if(eregi("404 Not Found\r\n", $data)) {
    die ("image doesnt exist");
}

RickyF · July 13, 2007

Hi,

Sorry i tried again with your second recommended code.

RickyF · July 13, 2007

Fatal error: Call to undefined function: str_ireplace() on line 25

Line 25: $parts = explode("/", str_ireplace("http://", null, $url), 2);

infid3l · July 13, 2007

Fatal error: Call to undefined function: str_ireplace() on line 25

Line 25: $parts = explode("/", str_ireplace("http://", null, $url), 2);

Ah, old PHP version. Use this instead:

$parts = explode("/",  str_replace("http://", null, strtolower($imgurl)), 2);

Replace it in the code I posted above.

RickyF · July 13, 2007

If the website doesnt exist it seems to say:

Warning: fsockopen() [function.fsockopen]: php_network_getaddresses: getaddrinfo failed: Name or service not known in /home/galaxy/public_html/plug/index.php on line 30

Warning: fsockopen() [function.fsockopen]: unable to connect to boasdasd.com:80 in /home/galaxy/public_html/plug/index.php on line 30

Permission denied (13)

But i have instructed the script to die and show an error?

Could you give the code a quick check to make sure the strings are in the right place etc

$url = "$imageurl"; 
$parts = explode("/",  str_replace("http://", null, strtolower($url)), 2);
$url = $parts[0];
$page = $parts[1];


$fp = fsockopen("$parts[0]", 80, $errno, $errstr, 30);
if (!$fp) {
    echo "$errstr ($errno)<br />\n";
} else {
    $out = "HEAD $parts[1] HTTP/1.1\r\n";
    $out .= "Host: $parts[0]\r\n";
    $out .= "Connection: Close\r\n\r\n";

    fwrite($fp, $out);
    while (!feof($fp)) {
        $data .= fgets($fp, 128);
    }
    fclose($fp);
}

if(eregi("404 Not Found\r\n", $data)) {
    die ("image doesnt exist");
}

infid3l · July 13, 2007

Fixed your errors, too tired to list them.

$url = $imageurl; 
$parts = explode("/",  str_replace("http://", null, strtolower($url)), 2);

$fp = fsockopen($parts[0], 80, $errno, $errstr, 30);
if (!$fp) {
    echo "$errstr ($errno)<br />\n";
} else {
    $out = "HEAD ".$parts[1]." HTTP/1.1\r\n";
    $out .= "Host: ".$parts[0]."\r\n";
    $out .= "Connection: Close\r\n\r\n";

    fwrite($fp, $out);
    while (!feof($fp)) {
        $data .= fgets($fp, 128);
    }
    fclose($fp);
}

if(eregi("404 Not Found\r\n", $data)) {
    die ("image doesnt exist");
}

should work

RickyF · July 13, 2007

Posting a working image/website works, it posts as it should, but when the image/website is non existant, its showing:

Warning: fsockopen() [function.fsockopen]: php_network_getaddresses: getaddrinfo failed: Name or service not known in /home/galaxy/public_html/plug/index.php on line 28

Warning: fsockopen() [function.fsockopen]: unable to connect to gsdfsdfsdd.com:80 on line 28

Permission denied (13)

Line 28: $fp = fsockopen($parts[0], 80, $errno, $errstr, 30);

Can you not hide this output and make it show image doesnt exist?

MadTechie · July 13, 2007

Why not this

<?php

$url = "http://us2.php.net/images/php.gif";

if( !is_numeric(xfile_404($url)) )
{
echo "Found<br>";
echo "<img src='$url'>";
}

function xfile_404($file){
        $file = ereg_replace(' +', '%20', trim($file));
        if(substr($file, 0, 7) !== "http://")
	{
		$file = "http://" . $file;
	}

        $domain = substr($file, 7); 
        $domain_ext = strtoupper(array_pop(explode('.', substr($domain, 0, strpos($domain, '/'))))); 
        $file_headers = @get_headers($file);
        if(!$file_headers){
            return 3;
            break;
        }
        if($file_headers[0] == 'HTTP/1.1 404 Not Found') {
            return 404;
        }else{
	return ereg_replace('%20', ' ', $file);
        }
}
?>

RickyF · July 13, 2007

Is it not possible to modify the below to work as required?

$url = $imageurl; 
$parts = explode("/",  str_replace("http://", null, strtolower($url)), 2);

$fp = fsockopen($parts[0], 80, $errno, $errstr, 30);
if (!$fp) {
    echo "$errstr ($errno)<br />\n";
} else {
    $out = "HEAD ".$parts[1]." HTTP/1.1\r\n";
    $out .= "Host: ".$parts[0]."\r\n";
    $out .= "Connection: Close\r\n\r\n";

    fwrite($fp, $out);
    while (!feof($fp)) {
        $data .= fgets($fp, 128);
    }
    fclose($fp);
}

if(eregi("404 Not Found\r\n", $data)) {
    die ("image doesnt exist");
}

infid3l · July 13, 2007

MadTechie's alternative is better, but if you must:

$fp = @fsockopen($parts[0], 80, $errno, $errstr, 30);

RickyF · July 13, 2007

I am a php newbie, i dont know how to implement this code, i tried but it didnt work.

infid3l · July 13, 2007

I am a php newbie, i dont know how to implement this code, i tried but it didnt work.

Put @ before the fsockopen(). Does it work when the image does exist?

RickyF · July 13, 2007

It says Success (0) if the image is not existant, and says nothing but posts the data if its real.

(Your script)

MadTechie · July 13, 2007

To add your code

<?php
//add to code
$url = "http://us2.php.net/images/php.gif";
$found = rfile_exists($url);

//example
if($found)
{
echo "found";
}else{
echo "not found";
}

//Add to the very end of your file.

<?php
function rfile_exists($url)
{
if( !is_numeric(xfile_404($url)) )
{
	return true;
}else{
	return false;
}
}

function xfile_404($file){
        $file = ereg_replace(' +', '%20', trim($file));
        if(substr($file, 0, 7) !== "http://")
{
	$file = "http://" . $file;
}

        $domain = substr($file, 7); 
        $file_headers = @get_headers($file);
        if(!$file_headers){
            return 3;
            break;
        }
        if($file_headers[0] == 'HTTP/1.1 404 Not Found') {
            return 404;
        }else{
	return ereg_replace('%20', ' ', $file);
        }
}
?>

Posting your code may help us

RickyF · July 13, 2007

Fatal error: Call to undefined function: rfile_exists() on line 4

Line 4: $found = rfile_exists($url);

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[SOLVED] Checking if an image exists using fsock?

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