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Can anybody help me with the oauth 1.0 oauth_signature generation from the client secret and consumer key? I have got client secret and consumer key along with the oauth token. Just need helping hand to generate the header and oauth_signature.

I found a post on StackOverflow really helpful but complete code is not there.

Please help me to understand from my client secret and consumer key how will I encrypt and generate oauth_signature?

1 Answer 1

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There is complete OAuth 1.0 code at my fork of sfdc-oauth-playground (originally by Jesper Joergensen - I fixed a couple of bugs in Jesper's code and added the 'two-legged' OAuth flow). Here is the header/oauth_signature code from OAuth.cls:

private void refreshParameters() {
    parameters.clear();
    parameters.put('oauth_consumer_key',consumerKey);
    if(token!=null) {
        parameters.put('oauth_token',token);
    }
    if(verifier!=null) {
        parameters.put('oauth_verifier',verifier);
    }
    parameters.put('oauth_signature_method','HMAC-SHA1');
    parameters.put('oauth_timestamp',timestamp);
    parameters.put('oauth_nonce',nonce);
    parameters.put('oauth_callback',callbackUrl);
}

private Map<String,String> getUrlParams(String value) {

    Map<String,String> res = new Map<String,String>();
    if(value==null || value=='') {
        return res;
    }
    for(String s : value.split('&')) {
        System.debug('getUrlParams: '+s);
        List<String> kv = s.split('=');
        if(kv.size()>1) {
          // RFC 5849 section 3.4.1.3.1 and 3.4.1.3.2 specify that parameter names 
          // and values are decoded then encoded before being sorted and concatenated
          // Section 3.6 specifies that space must be encoded as %20 and not +
          String encName = EncodingUtil.urlEncode(EncodingUtil.urlDecode(kv[0], 'UTF-8'), 'UTF-8').replace('+','%20');
          String encValue = EncodingUtil.urlEncode(EncodingUtil.urlDecode(kv[1], 'UTF-8'), 'UTF-8').replace('+','%20');
          System.debug('getUrlParams:  -> '+encName+','+encValue);
          res.put(encName,encValue);
        }
    }
    return res;
}

private String createBaseString(Map<String,String> oauthParams, HttpRequest req) {
    Map<String,String> p = oauthParams.clone();
    if(req.getMethod().equalsIgnoreCase('post') && req.getBody()!=null && 
       req.getHeader('Content-Type')=='application/x-www-form-urlencoded') {
        p.putAll(getUrlParams(req.getBody()));
    }
    String host = req.getEndpoint();
    Integer n = host.indexOf('?');
    if(n>-1) {
        p.putAll(getUrlParams(host.substring(n+1)));
        host = host.substring(0,n);
    }
    List<String> keys = new List<String>();
    keys.addAll(p.keySet());
    keys.sort();
    String s = keys.get(0)+'='+p.get(keys.get(0));
    for(Integer i=1;i<keys.size();i++) {
        s = s + '&' + keys.get(i)+'='+p.get(keys.get(i));
    }

    // According to OAuth spec, host string should be lowercased, but Google and LinkedIn
    // both expect that case is preserved.
    return req.getMethod().toUpperCase()+ '&' + 
        EncodingUtil.urlEncode(host, 'UTF-8') + '&' +
        EncodingUtil.urlEncode(s, 'UTF-8');
}

public void sign(HttpRequest req) {
    nonce = String.valueOf(Crypto.getRandomLong());
    timestamp = String.valueOf(DateTime.now().getTime()/1000);

    refreshParameters();

    String s = createBaseString(parameters, req);

    System.debug('Signature base string: '+s);

    Blob sig = Crypto.generateMac('HmacSHA1', Blob.valueOf(s), 
                   Blob.valueOf(consumerSecret+'&'+
                                (tokenSecret!=null ? tokenSecret : '')));
    signature = EncodingUtil.urlEncode(EncodingUtil.base64encode(sig), 'UTF-8');
    System.debug('Signature: '+signature);

    String header = 'OAuth ';
    for (String key : parameters.keySet()) {
        header = header + key + '="'+parameters.get(key)+'", ';
    }
    header = header + 'oauth_signature="'+signature+'"';
    System.debug('Authorization: '+header);
    req.setHeader('Authorization',header);
}   
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  • Awesome !!!No words for this .I will try this Commented Dec 19, 2012 at 17:21

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