问题
I am expecting
System.out.println(java.net.URLEncoder.encode(\"Hello World\", \"UTF-8\"));
to output:
Hello%20World
(20 is ASCII Hex code for space)
However, what I get is:
Hello+World
Am I using the wrong method? What is the correct method I should be using?
回答1:
This behaves as expected. The URLEncoder implements the HTML Specifications for how to encode URLs in HTML forms.
From the javadocs:
This class contains static methods for converting a String to the application/x-www-form-urlencoded MIME format.
and from the HTML Specification:
application/x-www-form-urlencoded
Forms submitted with this content type must be encoded as follows:
- Control names and values are escaped. Space characters are replaced by `+'
You will have to replace it, e.g.:
System.out.println(java.net.URLEncoder.encode("Hello World", "UTF-8").replace("+", "%20"));
回答2:
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:23.0'
// or, for Android:
compile 'com.google.guava:guava:23.0-android'
}
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.
回答3:
This class perform application/x-www-form-urlencoded-type encoding rather than percent encoding, therefore replacing with + is a correct behaviour.
From javadoc:
When encoding a String, the following rules apply:
- The alphanumeric characters "a" through "z", "A" through "Z" and "0" through "9" remain the same.
- The special characters ".", "-", "*", and "_" remain the same.
- The space character " " is converted into a plus sign "+".
- All other characters are unsafe and are first converted into one or more bytes using some encoding scheme. Then each byte is represented by the 3-character string "%xy", where xy is the two-digit hexadecimal representation of the byte. The recommended encoding scheme to use is UTF-8. However, for compatibility reasons, if an encoding is not specified, then the default encoding of the platform is used.
回答4:
Encode Query params
org.apache.commons.httpclient.util.URIUtil
URIUtil.encodeQuery(input);
OR if you want to escape chars within URI
public static String escapeURIPathParam(String input) {
StringBuilder resultStr = new StringBuilder();
for (char ch : input.toCharArray()) {
if (isUnsafe(ch)) {
resultStr.append('%');
resultStr.append(toHex(ch / 16));
resultStr.append(toHex(ch % 16));
} else{
resultStr.append(ch);
}
}
return resultStr.toString();
}
private static char toHex(int ch) {
return (char) (ch < 10 ? '0' + ch : 'A' + ch - 10);
}
private static boolean isUnsafe(char ch) {
if (ch > 128 || ch < 0)
return true;
return " %$&+,/:;=?@<>#%".indexOf(ch) >= 0;
}
回答5:
Hello+World is how a browser will encode form data (application/x-www-form-urlencoded) for a GET request and this is the generally accepted form for the query part of a URI.
http://host/path/?message=Hello+World
If you sent this request to a Java servlet, the servlet would correctly decode the parameter value. Usually the only time there are issues here is if the encoding doesn't match.
Strictly speaking, there is no requirement in the HTTP or URI specs that the query part to be encoded using application/x-www-form-urlencoded key-value pairs; the query part just needs to be in the form the web server accepts. In practice, this is unlikely to be an issue.
It would generally be incorrect to use this encoding for other parts of the URI (the path for example). In that case, you should use the encoding scheme as described in RFC 3986.
http://host/Hello%20World
More here.
回答6:
The other answers either present a manual string replacement, URLEncoder which actually encodes for HTML format, Apache's abandoned URIUtil, or using Guava's UrlEscapers. The last one is fine, except it doesn't provide a decoder.
Apache Commons Lang provides the URLCodec, which encodes and decodes according to URL format rfc3986.
String encoded = new URLCodec().encode(str);
String decoded = new URLCodec().decode(str);
If you are already using Spring, you can also opt to use its UriUtils class as well.
回答7:
"+" is correct. If you really need %20, then replace the Plusses yourself afterwards.
回答8:
This worked for me
org.apache.catalina.util.URLEncoder ul = new org.apache.catalina.util.URLEncoder().encode("MY URL");
回答9:
Just been struggling with this too on Android, managed to stumble upon Uri.encode(String, String) while specific to android (android.net.Uri) might be useful to some.
static String encode(String s, String allow)
https://developer.android.com/reference/android/net/Uri.html#encode(java.lang.String, java.lang.String)
回答10:
Although quite old, nevertheless a quick response:
Spring provides UriUtils - with this you can specify how to encoded and which part is it related from an URI, e.g.
encodePathSegment
encodePort
encodeFragment
encodeUriVariables
....
I use them cause we already using Spring, i.e. no additonal library is required!
回答11:
Check out the java.net.URI class.
回答12:
Am I using the wrong method? What is the correct method I should be using?
Yes, this method java.net.URLEncoder.encode wasn't made for converting " " to "20%" according to spec (source).
The space character " " is converted into a plus sign "+".
Even this is not the correct method, you can modify this to: System.out.println(java.net.URLEncoder.encode("Hello World", "UTF-8").replaceAll("\\+", "%20"));have a nice day =).
回答13:
USE MyUrlEncode.URLencoding(String url , String enc) to handle the problem
public class MyUrlEncode {
static BitSet dontNeedEncoding = null;
static final int caseDiff = ('a' - 'A');
static {
dontNeedEncoding = new BitSet(256);
int i;
for (i = 'a'; i <= 'z'; i++) {
dontNeedEncoding.set(i);
}
for (i = 'A'; i <= 'Z'; i++) {
dontNeedEncoding.set(i);
}
for (i = '0'; i <= '9'; i++) {
dontNeedEncoding.set(i);
}
dontNeedEncoding.set('-');
dontNeedEncoding.set('_');
dontNeedEncoding.set('.');
dontNeedEncoding.set('*');
dontNeedEncoding.set('&');
dontNeedEncoding.set('=');
}
public static String char2Unicode(char c) {
if(dontNeedEncoding.get(c)) {
return String.valueOf(c);
}
StringBuffer resultBuffer = new StringBuffer();
resultBuffer.append("%");
char ch = Character.forDigit((c >> 4) & 0xF, 16);
if (Character.isLetter(ch)) {
ch -= caseDiff;
}
resultBuffer.append(ch);
ch = Character.forDigit(c & 0xF, 16);
if (Character.isLetter(ch)) {
ch -= caseDiff;
}
resultBuffer.append(ch);
return resultBuffer.toString();
}
private static String URLEncoding(String url,String enc) throws UnsupportedEncodingException {
StringBuffer stringBuffer = new StringBuffer();
if(!dontNeedEncoding.get('/')) {
dontNeedEncoding.set('/');
}
if(!dontNeedEncoding.get(':')) {
dontNeedEncoding.set(':');
}
byte [] buff = url.getBytes(enc);
for (int i = 0; i < buff.length; i++) {
stringBuffer.append(char2Unicode((char)buff[i]));
}
return stringBuffer.toString();
}
private static String URIEncoding(String uri , String enc) throws UnsupportedEncodingException { //对请求参数进行编码
StringBuffer stringBuffer = new StringBuffer();
if(dontNeedEncoding.get('/')) {
dontNeedEncoding.clear('/');
}
if(dontNeedEncoding.get(':')) {
dontNeedEncoding.clear(':');
}
byte [] buff = uri.getBytes(enc);
for (int i = 0; i < buff.length; i++) {
stringBuffer.append(char2Unicode((char)buff[i]));
}
return stringBuffer.toString();
}
public static String URLencoding(String url , String enc) throws UnsupportedEncodingException {
int index = url.indexOf('?');
StringBuffer result = new StringBuffer();
if(index == -1) {
result.append(URLEncoding(url, enc));
}else {
result.append(URLEncoding(url.substring(0 , index),enc));
result.append("?");
result.append(URIEncoding(url.substring(index+1),enc));
}
return result.toString();
}
}
回答14:
use character-set "ISO-8859-1" for URLEncoder
来源：https://stackoverflow.com/questions/4737841/urlencoder-not-able-to-translate-space-character