Php linkedin oauth无效请求错误

Php linkedin oauth无效请求错误,php,curl,oauth,Php,Curl,Oauth,我正在测试一个简单的oauth curl php脚本,但它不断返回此错误 回调url上的代码: $code = $_GET['code']; $url = "https://www.linkedin.com/uas/oauth2/accessToken"; $redirect_uri = "http://127.0.0.1:8080/auth.php"; $api_key = "78v5tds9n6u10x"; $secret_key = "XXXXXXX"; $tokenArgume

我正在测试一个简单的oauth curl php脚本,但它不断返回此错误

回调url上的代码:

$code = $_GET['code'];

$url = "https://www.linkedin.com/uas/oauth2/accessToken";

$redirect_uri = "http://127.0.0.1:8080/auth.php"; 

$api_key = "78v5tds9n6u10x";
$secret_key = "XXXXXXX";

$tokenArguments = array("grant_type" => "authorization_code",
                        "code" => $code,
                        "redirect_uri" => $redirect_uri,
                        "client_secret" => $secret_key,
                        "client_id" => $api_key);



$ch = curl_init();

curl_setopt($ch, CURLOPT_URL, $url);

curl_setopt($ch, CURLOPT_POST, true);

curl_setopt($ch, CURLOPT_POSTFIELDS, $tokenArguments);

curl_setopt($ch, CURLOPT_HTTPHEADER, array('Content-Type: application/x-www-form-urlencoded'));   

curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); 

$result = curl_exec($ch);

curl_close($ch);

print_r($result);   
在添加内容类型标题之前,我遇到了错误,即使在添加内容类型标题之后,我仍然会遇到相同的错误

{"error_description":"missing required parameters, includes an invalid parameter value, parameter more than once. : client_id","error":"invalid_request"}

试着这样做,我在创建SHA的过程中得到了我的解决方案

try {
    PackageInfo info = context.getPackageManager().getPackageInfo(
            packageName, PackageManager.GET_SIGNATURES);
    for (Signature signature : info.signatures) {
        MessageDigest md = MessageDigest.getInstance("SHA");
        md.update(signature.toByteArray());

        // writtenToFile("FB_KEY_HASH.txt",
        // Base64.encodeToString(md.digest(),
        // Base64.DEFAULT).toString(), false);

        if (AppUtills.showLogs)
            Log.v(pageName,
                    "KeyHash:"
                            + Base64.encodeToString(md.digest(),
                                    Base64.DEFAULT));
    }
} catch (Exception e) {
    e.printStackTrace();
}

嘿@maximl337你解决这个问题了吗?