This issue is about a HTTP PUT request. I'm passing my url, query parameters and body to a method called buildURI().
private URI buildURI(String url, Map<String, Object> queryParameters, T body) {
try {
String bodyString = null;
if (body != null) {
bodyString = objectMapper.writeValueAsString(body);
}
} catch (JsonProcessingException e) {
System.out.println(e);
}
UriComponentsBuilder uriComponentsBuilder = UriComponentsBuilder.fromUriString(url);
if ((queryParameters != null) && (!queryParameters.isEmpty())) {
queryParameters.forEach((a, b) -> uriComponentsBuilder.queryParam(a, b));
}
return uriComponentsBuilder.encode().buildAndExpand().toUri();
}
The presence of reserved characters(& = %) can prevent correct parsing of the URI string. To substitute those characters I used this UriComponentsBuilder.fromUriString(url). Problem is those character list does not contain "#".
My url is http://1.1.1.1:1111/xx/xx/save/abc#gmail.com.
I need to substitute # as well.
But if I encode # and then call buildURI() this gives the url as http://1.1.1.1:1111/xx/xx/save/abc%2540gmail.com. (because of double encoding)
Calling this buildURI() is a must since I'm using common code.
I need to anyhow encode this "#" and get the url as http://1.1.1.1:1111/xx/xx/save/abc%40gmail.com.
Please give me a solution.
I receive a XML file with a tag whose value is "97ò00430 ò" while this tag initially contains only numbers. The encoding use is "ISO-8859-1".
How to detect the bad characters (ò...) in java, please ?
LNA
I guess you could use a Regex to check the format of your tag (here, "\d+" if you want numbers only).
public static String encode(String chr) {
try {
byte[] bytes = chr.getBytes("ISO-8859-1");
if (!validUTF8(bytes))
return chr;
return new String(bytes, "UTF-8");
} catch (UnsupportedEncodingException e) {
throw new IllegalStateException("No char" + e.getMessage());
}
}
I have tried multiple solutions here but none seem to work. I am getting error at String imageJsonStr = sh2.makeServiceCall(imgUrl[i], ServiceHandler.GET);
There are no special characters in string and variable imageId is a string containing only numbers like '98546214265231'
imgUrl[i] = "https://graph.facebook.com/v2.2/"
+ imageId
+ "?access_token="
+ static_token;
try {
try {
imgUrl[i] = URLEncoder.encode(imgUrl[i],"UTF-8");
} catch (UnsupportedEncodingException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
String imageJsonStr = sh2.makeServiceCall(
imgUrl[i], ServiceHandler.GET);
Java URL encoding of query string parameters refer this.. you need not encode the entire url .. encode only the parmaters
I merge two url with the following code.
String strUrl1 = "http://www.domainname.com/path1/2012/04/25/file.php";
String arg = "?page=2";
URL url1;
try {
url1 = new URL(strUrl1);
URL reconUrl1 = new URL(url1,arg);
System.out.println(" url : " + reconUrl1.toString());
} catch (MalformedURLException ex) {
ex.printStackTrace();
}
I'm surprise by the result : http://www.domainname.com/path1/2012/04/25/?page=2
I expect it to be (what browser do) : http://www.domainname.com/path1/2012/04/25/file.php?page=2
Tha javadoc about the constructor URL(URL context, String spec) explain it should respect the RFC.
I'm doing something wrong ?
Thanks
UPDATE :
This is the only problem I encountered with the fonction.
The code already works in all others cases, like browser do
"domain.com/folder/sub" + "/test" -> "domain.com/test"
"domain.com/folder/sub/" + "test" -> "domain.com/folder/sub/test"
"domain.com/folder/sub/" + "../test" -> "domain.com/folder/test"
...
You can always merge the String first and then created the URL based on the merged String.
StringBuffer buf = new StringBuffer();
buf.append(strURL1);
buf.append(arg);
URL url1 = new URL(buf.toString());
try
String k = url1+arg;
URL url1;
try {
url1 = new URL(k);
//URL reconUrl1 = new URL(url1,arg);
System.out.println(" url : " + url1.toString());
} catch (MalformedURLException ex) {
ex.printStackTrace();
}
I haven't read through the RFC, but the context (as mentioned in the Java Doc for URL) is presumably the directory of a URL, which means that the context of
"http://www.domainname.com/path1/2012/04/25/file.php"
is
"http://www.domainname.com/path1/2012/04/25/"
which is why
new URL(url1,arg);
yields
"http://www.domainname.com/path1/2012/04/25/?page=2"
The "workaround" is obviously to concatenate the parts yourself, using +.
you are using the constructor of URL here which takes paramter as URL(URL context, String spec). So you dont pass the php page with the URL but instead with the string. context needs to be the directory. the proper way to do this would be
String strUrl1 = "http://www.domainname.com/path1/2012/04/25";
String arg = "/file.php?page=2";
URL url1;
try {
url1 = new URL(strUrl1);
URL reconUrl1 = new URL(url1,arg);
System.out.println(" url : " + reconUrl1.toString());
} catch (MalformedURLException ex) {
ex.printStackTrace();
}
Try this
String strUrl1 = "http://www.domainname.com/path1/2012/04/25/";
String arg = "file.php?page=2";
URL url1;
try {
url1 = new URL(strUrl1);
URL reconUrl1 = new URL(url1,arg);
System.out.println(" url : " + reconUrl1.toString());
} catch (MalformedURLException ex) {
ex.printStackTrace();
}
When you read the java doc it mentions about the context of the specified URL
Which is the domain and the path:
"http://www.domainname.com" + "/path1/2012/04/25/"
Where "file.php" is considered the text where it belongs to the context mentioned above.
This two parameter overloaded constructor uses the context of a URL as base and adds the second param to create a complete URL, which is not what you need.
So it's better to String add the two parts and then create URL from them:
String contextURL = "http://www.domainname.com/path1/2012/04/25/";
String textURL = "file.php?page=2";
URL url;
try {
url = new URL(contextURL);
URL reconUrl = new URL(url, textURL);
System.out.println(" url : " + reconUrl.toString());
} catch (MalformedURLException murle) {
murle.printStackTrace();
}
I got this error message :
java.net.URISyntaxException: Illegal character in query at index 31: http://finance.yahoo.com/q/h?s=^IXIC
My_Url = http://finance.yahoo.com/q/h?s=^IXIC
When I copied it into a browser address field, it showed the correct page, it's a valid URL, but I can't parse it with this: new URI(My_Url)
I tried : My_Url=My_Url.replace("^","\\^"), but
It won't be the url I need
It doesn't work either
How to handle this ?
Frank
You need to encode the URI to replace illegal characters with legal encoded characters. If you first make a URL (so you don't have to do the parsing yourself) and then make a URI using the five-argument constructor, then the constructor will do the encoding for you.
import java.net.*;
public class Test {
public static void main(String[] args) {
String myURL = "http://finance.yahoo.com/q/h?s=^IXIC";
try {
URL url = new URL(myURL);
String nullFragment = null;
URI uri = new URI(url.getProtocol(), url.getHost(), url.getPath(), url.getQuery(), nullFragment);
System.out.println("URI " + uri.toString() + " is OK");
} catch (MalformedURLException e) {
System.out.println("URL " + myURL + " is a malformed URL");
} catch (URISyntaxException e) {
System.out.println("URI " + myURL + " is a malformed URL");
}
}
}
Use % encoding for the ^ character, viz. http://finance.yahoo.com/q/h?s=%5EIXIC
You have to encode your parameters.
Something like this will do:
import java.net.*;
import java.io.*;
public class EncodeParameter {
public static void main( String [] args ) throws URISyntaxException ,
UnsupportedEncodingException {
String myQuery = "^IXIC";
URI uri = new URI( String.format(
"http://finance.yahoo.com/q/h?s=%s",
URLEncoder.encode( myQuery , "UTF8" ) ) );
System.out.println( uri );
}
}
http://java.sun.com/javase/6/docs/api/java/net/URLEncoder.html
Rather than encoding the URL beforehand you can do the following
String link = "http://example.com";
URL url = null;
URI uri = null;
try {
url = new URL(link);
} catch(MalformedURLException e) {
e.printStackTrace();
}
try{
uri = new URI(url.toString())
} catch(URISyntaxException e {
try {
uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(),
url.getPort(), url.getPath(), url.getQuery(),
url.getRef());
} catch(URISyntaxException e1 {
e1.printStackTrace();
}
}
try {
url = uri.toURL()
} catch(MalfomedURLException e) {
e.printStackTrace();
}
String encodedLink = url.toString();
A general solution requires parsing the URL into a RFC 2396 compliant URI (note that this is an old version of the URI standard, which java.net.URI uses).
I have written a Java URL parsing library that makes this possible: galimatias. With this library, you can achieve your desired behaviour with this code:
String urlString = //...
URLParsingSettings settings = URLParsingSettings.create()
.withStandard(URLParsingSettings.Standard.RFC_2396);
URL url = URL.parse(settings, urlString);
Note that galimatias is in a very early stage and some features are experimental, but it is already quite solid for this use case.
A space is encoded to %20 in URLs, and to + in forms submitted data (content type application/x-www-form-urlencoded). You need the former.
Using Guava:
dependencies {
compile 'com.google.guava:guava:28.1-jre'
}
You can use UrlEscapers:
String encodedString = UrlEscapers.urlFragmentEscaper().escape(inputString);
Don't use String.replace, this would only encode the space. Use a library instead.
Coudn't imagine nothing better for
http://server.ru:8080/template/get?type=mail&format=html&key=ecm_task_assignment&label=Согласовать с контрагентом&descr=Описание&objectid=2231
that:
public static boolean checkForExternal(String str) {
int length = str.length();
for (int i = 0; i < length; i++) {
if (str.charAt(i) > 0x7F) {
return true;
}
}
return false;
}
private static final Pattern COLON = Pattern.compile("%3A", Pattern.LITERAL);
private static final Pattern SLASH = Pattern.compile("%2F", Pattern.LITERAL);
private static final Pattern QUEST_MARK = Pattern.compile("%3F", Pattern.LITERAL);
private static final Pattern EQUAL = Pattern.compile("%3D", Pattern.LITERAL);
private static final Pattern AMP = Pattern.compile("%26", Pattern.LITERAL);
public static String encodeUrl(String url) {
if (checkForExternal(url)) {
try {
String value = URLEncoder.encode(url, "UTF-8");
value = COLON.matcher(value).replaceAll(":");
value = SLASH.matcher(value).replaceAll("/");
value = QUEST_MARK.matcher(value).replaceAll("?");
value = EQUAL.matcher(value).replaceAll("=");
return AMP.matcher(value).replaceAll("&");
} catch (UnsupportedEncodingException e) {
throw LOGGER.getIllegalStateException(e);
}
} else {
return url;
}
}
I had this exception in the case of a test for checking some actual accessed URLs by users.
And the URLs are sometime contains an illegal-character and hang by this error.
So I make a function to encode only the characters in the URL string like this.
String encodeIllegalChar(String uriStr,String enc)
throws URISyntaxException,UnsupportedEncodingException {
String _uriStr = uriStr;
int retryCount = 17;
while(true){
try{
new URI(_uriStr);
break;
}catch(URISyntaxException e){
String reason = e.getReason();
if(reason == null ||
!(
reason.contains("in path") ||
reason.contains("in query") ||
reason.contains("in fragment")
)
){
throw e;
}
if(0 > retryCount--){
throw e;
}
String input = e.getInput();
int idx = e.getIndex();
String illChar = String.valueOf(input.charAt(idx));
_uriStr = input.replace(illChar,URLEncoder.encode(illChar,enc));
}
}
return _uriStr;
}
test:
String q = "\\'|&`^\"<>)(}{][";
String url = "http://test.com/?q=" + q + "#" + q;
String eic = encodeIllegalChar(url,'UTF-8');
System.out.println(String.format(" original:%s",url));
System.out.println(String.format(" encoded:%s",eic));
System.out.println(String.format(" uri-obj:%s",new URI(eic)));
System.out.println(String.format("re-decoded:%s",URLDecoder.decode(eic)));
If you're using RestangularV2 to post to a spring controller in java you can get this exception if you use RestangularV2.one() instead of RestangularV2.all()
Replace spaces in URL with + like If url contains dimension1=Incontinence Liners then replace it with dimension1=Incontinence+Liners.