I am using following code to get URI but it gives exception:
s3.putObject(bucket, fileToPut, file.ref.file)
val s3Object = s3.getObject(bucket, fileToPut).orNull
if(s3Object != null){
val assetUrl = s3Object.getObjectContent.getHttpRequest.getURI
}
I am unable to get uploaded object URI
Please help
If you just want the url of the object you could use
s3Client.getResourceUrl("your-bucket", "path/file.txt");
Related
I am using the textlocal short url API with the signed URL that I have generated for a Cloud Storage blob. I am getting MalformedSecurityHeader error in the browser when I use the short URL. I tried the same API with other random links and it worked fine. Can I get some help on this.
def shorten_url(apikey, url):
data = urllib.parse.urlencode({'apikey': apikey, 'url': url})
data = data.encode('utf-8')
request = urllib.request.Request("https://api.textlocal.in/create_shorturl/")
f = urllib.request.urlopen(request, data,
context=ssl.create_default_context(cafile=certifi.where()))
fr = f.read()
return(fr)
I was able to resolve this. data = urllib.parse.urlencode({'apikey': apikey, 'url': urllib.parse.quote(url)})
I have some problem on getting parameter from query string. I'm trying using
import { URLSearchParams } from
'#angular/http';
this.params = new URLSearchParams(window.location.search);
this.params1 = this.params.getAll('param');
console.log('parameter: '+JSON.stringify(this.params1))
and when I'm try to access http://localhost:8100/param=1 it getting error Cannot GET /param=1
How can I get that param?
Okay, i'm try using this.platform.getQueryParam('param'); and it working fine.
I am using httpClient lib to do some REST API calls using scala. I am able to post the data.
I am using following code to read the content. However, When I run on Spark Databricks Cluster it give me error.
val entity = response.getEntity
var content = ""
if (entity != null) {
val inputStream = entity.getContent
content = io.Source.fromInputStream(inputStream).getLines.mkString
inputStream.close()
}
Error
error: object Source is not a member of package io
content = io.Source.fromInputStream(inputStream).getLines.mkString
is there a way I can fix this error, or a different way to read HTTP response content.
Please try to import scala.io.Source
OR use the completed package name like this:
content = scala.io.Source.fromInputStream(inputStream).getLines.mkString
I succeeded to get every credentials(Oauth_token,Oauth_verifier).
With it, I tried to post a text to twitter account, but it always fail with error message "No authentication challenges found"
I found some solution like
"Check the time zone automatically",
"import latest twitter4j library" etc..
but after check it, still not work.
Is there anyone can show me the way.
code is like below
public static void updateStatus(final String pOauth_token,final String pOauth_verifier) {
new Thread() {
public void run() {
Looper.prepare();
try {
TwitterFactory factory = new TwitterFactory();
AccessToken accessToken = new AccessToken(pOauth_token,pOauth_verifier);
Twitter twitter = factory.getInstance();
twitter.setOAuthConsumer(Cdef.consumerKey, Cdef.consumerSecret);
twitter.setOAuthAccessToken(accessToken);
if (twitter.getAuthorization().isEnabled()) {
Log.e("btnTwSend","인증값을 셋팅하였고 API를 호출합니다.");
Status status = twitter.updateStatus(Cdef.sendText + " #" + String.valueOf(System.currentTimeMillis()));
Log.e("btnTwSend","status:" + status.getText());
}
} catch (Exception e) {
Log.e("btnTwSend",e.toString());
}
};
}.start();
}
"No authentication challenges found"
I think you are missing Access token secret in your code. That is why you are getting this exception.
Try following :
ConfigurationBuilder configurationBuilder;
Configuration configuration;
// Set the proper configuration parameters
configurationBuilder = new ConfigurationBuilder();
configurationBuilder
.setOAuthConsumerKey(TWITTER_CONSUMER_KEY);
configurationBuilder
.setOAuthConsumerSecret(TWITTER_CONSUMER_SECRET);
// Access token
configurationBuilder.setOAuthAccessToken(ACCESS_TOKEN);
// Access token secret
configurationBuilder
.setOAuthAccessTokenSecret(ACCESS_TOKEN_SECRET);
// Get the configuration object based on the params
configuration = configurationBuilder.build();
// Pass it to twitter factory to get the proprt twitter instance.
twitterFactory = new TwitterFactory(configuration);
twitter = twitterFactory.getInstance();
// use this instance to update
twitter.updateStatus("Your status");
I finally found the reason.
I thought parameter named 'oauth_token' , 'oauth_verifier' is member of accesstoken,
but it was not true.
I just had to pass one more way to get correct key.
and this way needs 'oauth_token' , 'oauth_verifier' to get accesstoken.
This code must add one more code below:
mAccessToken = mTwitter.getOAuthAccessToken(REQUEST_TOKEN,OAUTH_VERIFIER);
I've to make a new Component in JIRA
I found out the POST url /rest/api/2/component for making new component, but i'm unable to know what type of inputs to be given.
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost postRequest = new HttpPost("http://localhost:8080/rest/api/2/component/");
String authorization = JiraRequestResponseUtil.conversionForAuthorization();
postRequest.setHeader("Authorization", authorization);
StringEntity input = new StringEntity("\"name\":\"Component 1\",\"description\":\"This is a TEST JIRA component\" ,\"leadUserName\":\"fred\",\"assigneeType\":\"PROJECT_LEAD\",\"isAssigneeTypeValid\":\"false\",\"project\":\"TEST\"");
input.setContentType("application/json");
postRequest.setEntity(input);
HttpResponse response = httpClient.execute(postRequest);
this is the code i'm implementing.
Output i'm getting is Failed : HTTP error code : 400
Plz help.
we can't tell you. You need to find documentation on the service you are posting to.
The above code is correct, just add the { & } to the JSON string.