Android url中文编码问题
最近项目遇见一个很奇葩问题,关于URL问题,项目中加载图片,图片的URL含有中文,但是,我的手机可以加载,没问题,同事也都可以,但是测试手机却不可以,加载失败,找到问题,就是URL含有中文问题。
解决方案:
把中文字符encode即可:
方法1:
public static String encodeUrl(String url) {
return Uri.encode(url, "-![.:/,%?&=]");
}
方法2:
public static String toUtf8String(String s) {
StringBuffer sb = new StringBuffer();
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c >= 0 && c <= 255) {
sb.append(c);
} else {
byte[] b;
try {
b = String.valueOf(c).getBytes("utf-8");
} catch (Exception ex) {
System.out.println(ex);
b = new byte[0];
}
for (int j = 0; j < b.length; j++) {
int k = b[j];
if (k < 0)
k += 256;
sb.append("%" + Integer.toHexString(k).toUpperCase());
}
}
}
return sb.toString();
}
或者
import java.io.CharArrayWriter;
import java.io.UnsupportedEncodingException;
import java.net.URLDecoder;
import java.nio.charset.Charset;
import java.nio.charset.IllegalCharsetNameException;
import java.nio.charset.UnsupportedCharsetException;
import java.util.BitSet; public class URLEncoderURI { static BitSet dontNeedEncoding;
static final int caseDiff = ('a' - 'A'); static { /*
* The list of characters that are not encoded has been determined as
* follows:
*
* RFC 2396 states: ----- Data characters that are allowed in a URI but
* do not have a reserved purpose are called unreserved. These include
* upper and lower case letters, decimal digits, and a limited set of
* punctuation marks and symbols.
*
* unreserved = alphanum | mark
*
* mark = "-" | "_" | "." | "!" | "~" | "*" | "'" | "(" | ")"
*
* Unreserved characters can be escaped without changing the semantics
* of the URI, but this should not be done unless the URI is being used
* in a context that does not allow the unescaped character to appear.
* -----
*
* It appears that both Netscape and Internet Explorer escape all
* special characters from this list with the exception of "-", "_",
* ".", "*". While it is not clear why they are escaping the other
* characters, perhaps it is safest to assume that there might be
* contexts in which the others are unsafe if not escaped. Therefore, we
* will use the same list. It is also noteworthy that this is consistent
* with O'Reilly's "HTML: The Definitive Guide" (page 164).
*
* As a last note, Intenet Explorer does not encode the "@" character
* which is clearly not unreserved according to the RFC. We are being
* consistent with the RFC in this matter, as is Netscape.
*/ 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(' '); /*
* encoding a space to a + is done in the
* encode() method
*/
dontNeedEncoding.set('-');
dontNeedEncoding.set('_');
dontNeedEncoding.set('.');
dontNeedEncoding.set('*');
dontNeedEncoding.set(':');
dontNeedEncoding.set('/');
dontNeedEncoding.set('?');
dontNeedEncoding.set(';');
dontNeedEncoding.set('&');
dontNeedEncoding.set('='); } /**
* You can't call the constructor.
*/
private URLEncoderURI() {
} /**
* Translates a string into <code>application/x-www-form-urlencoded</code>
* format using a specific encoding scheme. This method uses the supplied
* encoding scheme to obtain the bytes for unsafe characters.
* <p>
* <em><strong>Note:</strong> The <a href=
* "http://www.w3.org/TR/html40/appendix/notes.html#non-ascii-chars">
* World Wide Web Consortium Recommendation</a> states that
* UTF-8 should be used. Not doing so may introduce
* incompatibilites.</em>
*
* @param s
* <code>String</code> to be translated.
* @param enc
* The name of a supported <a
* href="../lang/package-summary.html#charenc">character
* encoding</a>.
* @return the translated <code>String</code>.
* @exception UnsupportedEncodingException
* If the named encoding is not supported
* @see URLDecoder#decode(java.lang.String, java.lang.String)
* @since 1.4
*/
public static String encode(String s, String enc) throws UnsupportedEncodingException { boolean needToChange = false;
StringBuffer out = new StringBuffer(s.length());
Charset charset;
CharArrayWriter charArrayWriter = new CharArrayWriter(); if (enc == null)
throw new NullPointerException("charsetName"); try {
charset = Charset.forName(enc);
} catch (IllegalCharsetNameException e) {
throw new UnsupportedEncodingException(enc);
} catch (UnsupportedCharsetException e) {
throw new UnsupportedEncodingException(enc);
} for (int i = 0; i < s.length();) {
int c = (int) s.charAt(i);
// System.out.println("Examining character: " + c);
if (dontNeedEncoding.get(c)) {
if (c == ' ') {
c = '+';
needToChange = true;
}
// System.out.println("Storing: " + c);
out.append((char) c);
i++;
} else {
// convert to external encoding before hex conversion
do {
charArrayWriter.write(c);
/*
* If this character represents the start of a Unicode
* surrogate pair, then pass in two characters. It's not
* clear what should be done if a bytes reserved in the
* surrogate pairs range occurs outside of a legal surrogate
* pair. For now, just treat it as if it were any other
* character.
*/
if (c >= 0xD800 && c <= 0xDBFF) {
/*
* System.out.println(Integer.toHexString(c) +
* " is high surrogate");
*/
if ((i + 1) < s.length()) {
int d = (int) s.charAt(i + 1);
/*
* System.out.println("\tExamining " +
* Integer.toHexString(d));
*/
if (d >= 0xDC00 && d <= 0xDFFF) {
/*
* System.out.println("\t" +
* Integer.toHexString(d) +
* " is low surrogate");
*/
charArrayWriter.write(d);
i++;
}
}
}
i++;
} while (i < s.length() && !dontNeedEncoding.get((c = (int) s.charAt(i)))); charArrayWriter.flush();
String str = new String(charArrayWriter.toCharArray());
byte[] ba = str.getBytes(charset);
for (int j = 0; j < ba.length; j++) {
out.append('%');
char ch = Character.forDigit((ba[j] >> 4) & 0xF, 16);
// converting to use uppercase letter as part of
// the hex value if ch is a letter.
if (Character.isLetter(ch)) {
ch -= caseDiff;
}
out.append(ch);
ch = Character.forDigit(ba[j] & 0xF, 16);
if (Character.isLetter(ch)) {
ch -= caseDiff;
}
out.append(ch);
}
charArrayWriter.reset();
needToChange = true;
}
} return (needToChange ? out.toString() : s);
}
}
参考:
原文链接:http://www.jianshu.com/p/9be694c8fee2
著作权归作者所有,转载请联系作者获得授权,并标注“简书作者”。
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