theTaskmaster theTaskmaster - 1 year ago 60
PHP Question

How to display stackexchange data on my site?


I am developing my presentation site and I want to include my Stack Overflow profile info/posts/data (eg top tag, score and so on.)

I found to retrieve the desired data but I can't understand how can I show this data in my site.

In I found this prerequisites: which basically says that I must be a .NET programmer to be able to display this data but I am a PHP programmer, I work with


I know that there are lots of PHP MsSQL functions I can use but how can I connect to the Stack Exchange database (I think as a guest/limited user) with which username-password?

Even if this is not too much on-topic here, where can I find more info on how I can display Stack Overflow data on my site?

Answer Source

Even if CONFUS3D' answer is a good solution, any alteration to the User Interface may cause errors in your site.

I suggest you to use the Stack Exchange API set with which you can retrieve the most of data you probably need.

Any API query will return a JSON object. I use this PHP class to retrieve this object:

class ApiReader {
    public function getResponse($url) {
        $cH = curl_init();
        curl_setopt($cH, CURLOPT_URL, $url);
        curl_setopt($cH, CURLOPT_HEADER, 0);
        curl_setopt($cH, CURLOPT_RETURNTRANSFER, true);
        curl_setopt($cH, CURLOPT_TIMEOUT, 30);
        curl_setopt($cH, CURLOPT_USERAGENT, cURL);
        curl_setopt($cH, CURLOPT_FOLLOWLOCATION, true);
        curl_setopt($cH, CURLOPT_ENCODING, "gzip");

        $result = curl_exec($cH);

        if(curl_errno($cH)) {
            $retur = FALSE;
        else {
            $status = curl_getinfo($cH, CURLINFO_HTTP_CODE);
            if($status == 200) {
                $retur = $result;
             else {
                 $retur = FALSE;
         return $retur;

I use this little trick to test the site even if I am off-line.

In your host, save all the JSON objects you need to use, then declare two vars $UInfo_API containing the API query and $UInfo_Syn which gets the content of the saved JSON object

$UInfo_API = "";
$UInfo_Syn = file_get_contents("yourjsonobject.json");

Then save the result in a variable checking if the getResponse() method has failed or not. After that, you have the data on tap.

$sear = new ApiReader();
$uInfo = $sear->getResponse($UInfo_API);
$uInfo = ($uInfo !== FALSE)? json_decode($uInfo, TRUE): json_decode($UInfo_Syn, TRUE);
$rep = $uInfo["items"][0]["reputation"];