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MissingFormatArgumentException with square brackets on android

I have this string which I'm trying to format:
String url = "http://api/doSomething.json?params%5Bemail%5D=%s"
String.format(url,email).
The idea is that it ends up looking like this:
http://api/doSomething.json?params[email]=aValue;
I'm currently getting a MissingFormatArgumentException, Format specifier: 5D exception.
Has anyone had issues with this before?

String.format() doesn't like the %5D placeholder - %5D has to be %5d.
Reference: http://developer.android.com/reference/java/util/Formatter.html
... if it was about placeholders.
Anyway, it seems you just want the square brackets.
Therefore, change this
String url = "http://api/doSomething.json?params%5Bemail%5D=%s"
to
String url = "http://api/doSomething.json?params[email]=%s"

In the end i was able to resolve this using a URLEncoder.
This post was particularly helpful -> URL encoding in Android
String queryPart = String.format(PARAM_STRING,
email);
return baseUrl + URLEncoder.encode(queryPart, "utf-8");

Android webview does not take bad url

WebView does not take url if they are bad formated ??
For example good url format http://www.youtube.com/embed/KeSzOIUJ4xY?rel=0 and
bad url format //player.vimeo.com/video/142545652?badge=0
Is there a way to accept the following bad url as browsers do ??

No android web view didnt take a bad formatted url..
Uri.Builder builder = new Uri.Builder();
builder.scheme("https")
.authority("www."+YOUR_STRING+".com")
.appendPath("turtles")
.appendPath("types")
.appendQueryParameter("type", "1")
.appendQueryParameter("sort", "relevance")
.fragment("section-name");
String myUrl = builder.build().toString();
But you can play with above code.

How to get Hyperlink from whole string

I create an application in which i get the response from web service .
The response is
"I might be in danger. I have triggered my panic alarm which is connected to you. Call me now. If i'm not answering, contact the police. My position is:http://maps.google.com/maps?q=21.183783,72.823548"
3.I store the string in text view.and i want to open HTTP URL in browser,on the click of text.but how can i get HTTP URL in whole string plese give me idea.

You can do this easily with php...
If you are able to run php, this should do it.
$string = $_GET['string'];
OR
$string = $_POST['string'];
this may change depending on how you get the responce from the website, feel free to send me the form which you get the responce and ill change it accordingly.
$string_chunks = explode('http://',$string,2);
$url = 'http://'.$string_chunk['1'];
Basically, this will take the string, find the "http://" and create 2 strings out of it. one with the content before the "http://" and one with the content after, which is the url. so it would return $string_chunk['0'] and $string_chunk['1']

var response = "ur response string";
var indexofHttp = response.indexOf('http://');
var url = response.substring(indexofHttp);

Opening URL on Android web browser causes Google search

I'm having a slight problem opening a certain URL in the browser. First of all I use the following code to launch the browser:
Intent browserIntent = new Intent(Intent.ACTION_VIEW, Uri.parse(Globals.currentChatURL));
startActivity(Intent.createChooser(browserIntent, "Select browser:"));
Now if I set Globals.currentChatURL to something like http://www.google.com then it opens that site just fine. But my URL is a little more complicated as it contains multiple parameters which are all base64 encoded. Here is an example of how my URL looks:
http://webportal.mysite.com/ChatProgram/chat.php? intgroup=UFYyMA==&intid=UFYyMEZN&hg=Pw__&pref=user&en=U0NPVFQgTUlMTEFS&ee=cGF1bGdAbWFnbmF0ZWNoLmNvbQ==&eq=UFRWRkVI&ec=TUFHTkFURUNI
Now if I use my above code to try and launch this URL it brings me to the Google search page with the following message:
"Your search - http://URLabove ... did not match any documents"
Yet if I copy the URL and paste it into the address box it brings me to the right place. How can I fix this?? The whole point of this is to have the user click the button and the site to launch, not for the user to have to copy and paste the URL manually.
Any suggestions would be greatly appreciated.
Thanks a lot

There is unwanted equal signs in the query part of your http URI. Such signs have a specific meaning as delimiters in the form &parameter=value.
This equal signs represents padding values (0, 1 or 2) from your base64 encoding.
You can either
remove them because your base64 server decoder won't bother reconstructing them, or
percent encode them (with all other reserved characters).
In android you can use percent encode this way:
String value = URLEncoder.encode("annoying values with reserved chars &=#", "utf-8");
String url = "http://stackoverflow.com/search?q=" + value;
The RFC 2396 is now deprecated but that is what URI.parse() is based on as stated by the documentation:
uriString an RFC 2396-compliant, encoded URI

Size limit of a String while posting to server

I am posting a string to server. If string size is up to 6000KB then its posted successfully. But when size exceeded more than this its showing response -1.
I have tried method of posting: syn_data1 is string . records fetch from data base and then appending to A string builder and finally i create synData1 string from String builder
URL url = new URL(syn_data1);
URLConnection urlc = url.openConnection();
HttpURLConnection huc = (HttpURLConnection)urlc;
huc.setRequestMethod("POST");
huc.setConnectTimeout(3000);
huc.connect();
int response = huc.getResponseCode();
I do care about each special character and remove.But I did not get success

In theory, the URI in an HTTP request can be of any length, but the practical limit is on the order of 2k. Please read here for more info on that.
I am assuming the length is coming from the query string parameters (those name=value pairs that come after the ?). You should be putting these in the POST data, leaving the path part of the URI only. Of course, the server will have to be looking for those parameters in the POST data as well.

Are you passing the NameValue pairs properly . This is one successful way which i use .
List<NameValuePair> loginParams = new ArrayList<NameValuePair>(1);
loginParams.add(new BasicNameValuePair("ColumnName In DB",YourString));
then you do
httppost.setEntity(new UrlEncodedFormEntity(loginParams));
and proceed to execute

It's not clear exactly what you're trying to achieve, but this definitely looks wrong:
URL url = new URL(syn_data1.toString());
URLEncoder.encode(syn_data1.toString(),"UTF-16BE");
If syn_data1 is already a string, you don't need to call toString on it.. and calling URLEncoder.encode doesn't have any side-effects, so the second statement is pointless. Perhaps you want:
URL url = new URL(URLEncoder.encode(syn_data1, "UTF-16BE"));
That's just on the encoding side though - you still shouldn't be trying to use enormous URLs. If you have a lot of data, that should be in the body of the request rather than the URL.