如何在GWT中格式化我的字符串？
我做了一个方法
Formatter format = new Formatter();
int matches = 0;
Formatter formattedString = format.format("%d numbers(s, args) in correct position", matches);
return formattedString.toString();
但它抱怨道
Validating newly compiled units
[ERROR] Errors in 'file:/C:/Documents%20and%20Settings/kkshetri/workspace/MasterMind/MasterMind/src/com/kunjan/MasterMind/client/MasterMind.java'
[ERROR] Line 84: No source code is available for type java.util.Formatter; did you forget to inherit a required module?
不包括Formatter吗？
更新：在进一步查看此答案之前，请参阅（并投票通过）Joseph Lust的帖子。
看起来像this post不包括formatter。但是，他们提出了一些替代方案
作为替代方案，您可以使用类NumberFormat：
NumberFormat fmt = NumberFormat.getDecimalFormat();
double value = 12345.6789;
String formatted = fmt.format(value);
// Prints 1,2345.6789 in the default locale
java.text.MessageFormat.format（）的另一个非常简单的替代品：
public static String format(final String format, final Object... args) {
StringBuilder sb = new StringBuilder();
int cur = 0;
int len = format.length();
while (cur < len) {
int fi = format.indexOf('{', cur);
if (fi != -1) {
sb.append(format.substring(cur, fi));
int si = format.indexOf('}', fi);
if (si != -1) {
String nStr = format.substring(fi + 1, si);
int i = Integer.parseInt(nStr);
sb.append(args[i]);
cur = si + 1;
} else {
sb.append(format.substring(fi));
break;
}
} else {
sb.append(format.substring(cur, len));
break;
}
}
return sb.toString();
}
如上所述，有数字和日期的GWT格式化程序：NumberFormat和DateTimeFormat。不过，我需要一个知名的String.format(...)案例的解决方案。我最终得到了这个解决方案，我不知道它是否对性能有误，但它看起来很干净。我很高兴听到任何评论，或其他解决方案。
我的String格式化程序：
public class Strings {
public static String format(final String format, final Object... args) {
String retVal = format;
for (final Object current : args) {
retVal = retVal.replaceFirst("[%][s]", current.toString());
}
return retVal;
}
}
和JUTest如果想要重复使用它：
public class StringsTest {
@Test
public final void testFormat() {
this.assertFormat("Some test here %s.", 54);
this.assertFormat("Some test here %s and there %s, and test [%s]. sfsfs !!!", 54, 59, "HAHA");
}
private void assertFormat(final String format, final Object... args) {
Assert.assertEquals("Formatting is not working", String.format(format, args), Strings.format(format, args));
}
}
请参阅GWT日期和数字格式的official page。
他们建议如下：
myNum decimal = 33.23232;
myString = NumberFormat.getFormat("#.00").format(decimal);
最好使用他们支持的优化方法，而不是制作自己的非最佳方法。他们的编译器最终会优化它们到几乎相同的东西。
GWT 2.1+中String.format（）的一个非常简单的替换：
import com.google.gwt.regexp.shared.RegExp;
import com.google.gwt.regexp.shared.SplitResult;
public static String format(final String format, final Object... args) {
final RegExp regex = RegExp.compile("%[a-z]");
final SplitResult split = regex.split(format);
final StringBuffer msg = new StringBuffer();
for (int pos = 0; pos < split.length() - 1; ++pos) {
msg.append(split.get(pos));
msg.append(args[pos].toString());
}
msg.append(split.get(split.length() - 1));
return msg.toString();
}
甚至更简单，不使用RegExp，只使用字符串：
public static String format(final String format, final String... args) {
String[] split = format.split("%s");
final StringBuffer msg = new StringBuffer();
for (int pos = 0; pos < split.length - 1; pos += 1) {
msg.append(split[pos]);
msg.append(args[pos]);
}
msg.append(split[split.length - 1]);
return msg.toString();
}
另一个使用JSNI的another post和一个很好的JavaScript格式函数的建议：
import com.google.gwt.core.client.JsArrayString;
public abstract class StringFormatter {
public static String format(final String format, final Object... args) {
if (null == args || 0 == args.length)
return format;
JsArrayString array = newArray();
for (Object arg : args) {
array.push(String.valueOf(arg)); // TODO: smarter conversion?
}
return nativeFormat(format, array);
}
private static native JsArrayString newArray()/*-{
return [];
}-*/;
private static native String nativeFormat(final String format, final JsArrayString args)/*-{
return format.replace(/{(\d+)}/g, function(match, number) {
return typeof args[number] != 'undefined' ? args[number] : match;
});
}-*/;
}
然后可以像这样打个电话：
StringFormatter.format("Greetings {0}, it's {1} o'clock, which is a {2} statement", "Master", 8, false);
......结果是
问候大师，现在是8点钟，这是一个错误的陈述
TODO评论有可能进一步改善，例如：利用NumberFormat。欢迎提出建议。
Daniels解决方案的扩展：还支持使用'转义，如果无法解析数字则抛出（如JVM版本那样）：
private static final char OPEN = '{';
private static final char CLOSE = '}';
private static final char ESCAPE = '\'';
@Override
public String format(String pattern, Object... arguments) {
if (pattern == null || pattern.isEmpty())
return "";
// Approximate the result length: format string + 16 character args
StringBuilder sb = new StringBuilder(pattern.length() + (arguments.length * 16));
int cur = 0;
int len = pattern.length();
// if escaped, then its >= 0
int escapedAtIndex = -1;
while (cur < len) {
char currentChar = pattern.charAt(cur);
switch (currentChar) {
case OPEN:
if (escapedAtIndex >= 0) {
// currently escaped
sb.append(currentChar);
} else {
// find close
int close = pattern.indexOf(CLOSE, cur + 1);
switch (close) {
case -1:
// Missing close. Actually an error. But just ignore
sb.append(currentChar);
break;
default:
// Ok, we have a close
final String nStr = pattern.substring(cur + 1, close);
try {
// Append the corresponding argument value
sb.append(arguments[Integer.parseInt(nStr)]);
} catch (Exception e) {
if (e instanceof NumberFormatException) {
throw new IllegalArgumentException(nStr +
" is not a number.");
}
// Append the curlies and the original delimited value
sb.append(OPEN).append(nStr).append(CLOSE);
}
// Continue after the close
cur = close;
break;
}
}
cur++;
break;
case ESCAPE:
// Special case: two '' are just converted to '
boolean nextIsEscapeToo = (cur + 1 < len) && pattern.charAt(cur + 1) == ESCAPE;
if (nextIsEscapeToo) {
sb.append(ESCAPE);
cur = cur + 2;
} else {
if (escapedAtIndex >= 0) {
// Escape end.
escapedAtIndex = -1;
} else {
// Escape start.
escapedAtIndex = cur;
}
cur++;
}
break;
default:
// 90% case: Nothing special, just a normal character
sb.append(currentChar);
cur++;
break;
}
}
return sb.toString();
}
此实现和JVM-Version都通过了这些测试：
// Replace: 0 items
assertFormat("Nothing to replace", "Nothing to replace");
// Replace: 1 item
assertFormat("{0} apples", "15 apples", 15);
assertFormat("number of apples: {0}", "number of apples: zero", "zero");
assertFormat("you ate {0} apples", "you ate some apples", "some");
// Replace 2 items
assertFormat("{1} text {0}", "second text first", "first", "second");
assertFormat("X {1} text {0}", "X second text first", "first", "second");
assertFormat("{0} text {1} X", "first text second X", "first", "second");
逃离小测试：
// Escaping with no replacement
assertFormat("It's the world", "Its the world");
assertFormat("It''s the world", "It's the world");
assertFormat("Open ' and now a second ' (closes)", "Open and now a second (closes)");
assertFormat("It'''s the world", "It's the world");
assertFormat("'{0}' {1} {2}", "{0} one two", "zero", "one", "two");
// Stays escaped (if end escape is missing)
assertFormat("'{0} {1} {2}", "{0} {1} {2}", "zero", "one", "two");
assertFormat("'{0} {1}' {2}", "{0} {1} two", "zero", "one", "two");
// No matter where we escape, stays escaped
assertFormat("It's a {0} world", "Its a {0} world", "blue");
// But we can end escape everywhere
assertFormat("It's a {0} world, but not '{1}",
"Its a {0} world, but not always", "blue", "always");
// I think we want this
assertFormat("It''s a {0} world, but not {1}",
"It's a blue world, but not always", "blue", "always");
// Triple
assertFormat("' '' '", " ' ");
// From oracle docs
assertFormat("'{''}'", "{'}");
// Missing argument (just stays 0)
assertFormat("begin {0} end", "begin {0} end");
// Throws
try {
assertFormat("begin {not_a_number} end", "begin {not_a_number} end");
throw new AssertionError("Should not get here");
} catch (IllegalArgumentException iae) {
// OK
}
这个很快，忽略了错误的卷曲分隔值：
public static String format(final String format, final Object... args)
{
if (format == null || format.isEmpty()) return "";
// Approximate the result length: format string + 16 character args
StringBuilder sb = new StringBuilder(format.length() + (args.length*16));
final char openDelim = '{';
final char closeDelim = '}';
int cur = 0;
int len = format.length();
int open;
int close;
while (cur < len)
{
switch (open = format.indexOf(openDelim, cur))
{
case -1:
return sb.append(format.substring(cur, len)).toString();
default:
sb.append(format.substring(cur, open));
switch (close = format.indexOf(closeDelim, open))
{
case -1:
return sb.append(format.substring(open)).toString();
default:
String nStr = format.substring(open + 1, close);
try
{
// Append the corresponding argument value
sb.append(args[Integer.parseInt(nStr)]);
}
catch (Exception e)
{
// Append the curlies and the original delimited value
sb.append(openDelim).append(nStr).append(closeDelim);
}
cur = close + 1;
}
}
}
return sb.toString();
}
也许最简单的方法就是像String.format这样做，可以用String.replace来做，例如;
而不是做String.format("Hello %s", "Daniel"); ==> "Hello %s".replace("%s", "Daniel")，
两者都给我们相同的结果，但只是第二种方式适用于GWT客户端
我并不热衷于滥用字符串操作来完成regexps的工作，但是，基于bodrin的解决方案，你可以编写代码：
public static String format (String pattern, final Object ... args) {
for (Object arg : args) {
String part1 = pattern.substring(0,pattern.indexOf('{'));
String part2 = pattern.substring(pattern.indexOf('}') + 1);
pattern = part1 + arg + part2;
}
return pattern;
}