The call Character.isLetter(c)
returns true
if the character is a letter. But is there a way to quickly find if a String
only contains the base characters of ASCII?
This question is related to
java
string
character-encoding
ascii
It was possible. Pretty problem.
import java.io.UnsupportedEncodingException;
import java.nio.charset.Charset;
import java.nio.charset.CharsetEncoder;
public class EncodingTest {
static CharsetEncoder asciiEncoder = Charset.forName("US-ASCII")
.newEncoder();
public static void main(String[] args) {
String testStr = "¤EÀsÆW°ê»Ú®i¶T¤¤¤ß3¼Ó®i¶TÆU2~~KITEC 3/F Rotunda 2";
String[] strArr = testStr.split("~~", 2);
int count = 0;
boolean encodeFlag = false;
do {
encodeFlag = asciiEncoderTest(strArr[count]);
System.out.println(encodeFlag);
count++;
} while (count < strArr.length);
}
public static boolean asciiEncoderTest(String test) {
boolean encodeFlag = false;
try {
encodeFlag = asciiEncoder.canEncode(new String(test
.getBytes("ISO8859_1"), "BIG5"));
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
}
return encodeFlag;
}
}
commons-lang3 from Apache contains valuable utility/convenience methods for all kinds of 'problems', including this one.
System.out.println(StringUtils.isAsciiPrintable("!@£$%^&!@£$%^"));
This will return true if String only contains ASCII characters and false when it does not
Charset.forName("US-ASCII").newEncoder().canEncode(str)
If You want to remove non ASCII , here is the snippet:
if(!Charset.forName("US-ASCII").newEncoder().canEncode(str)) {
str = str.replaceAll("[^\\p{ASCII}]", "");
}
Here is another way not depending on a library but using a regex.
You can use this single line:
text.matches("\\A\\p{ASCII}*\\z")
Whole example program:
public class Main {
public static void main(String[] args) {
char nonAscii = 0x00FF;
String asciiText = "Hello";
String nonAsciiText = "Buy: " + nonAscii;
System.out.println(asciiText.matches("\\A\\p{ASCII}*\\z"));
System.out.println(nonAsciiText.matches("\\A\\p{ASCII}*\\z"));
}
}
//return is uppercase or lowercase
public boolean isASCIILetter(char c) {
return (c > 64 && c < 91) || (c > 96 && c < 123);
}
Iterate through the string and make sure all the characters have a value less than 128.
Java Strings are conceptually encoded as UTF-16. In UTF-16, the ASCII character set is encoded as the values 0 - 127 and the encoding for any non ASCII character (which may consist of more than one Java char) is guaranteed not to include the numbers 0 - 127
Or you copy the code from the IDN class.
// to check if a string only contains US-ASCII code point
//
private static boolean isAllASCII(String input) {
boolean isASCII = true;
for (int i = 0; i < input.length(); i++) {
int c = input.charAt(i);
if (c > 0x7F) {
isASCII = false;
break;
}
}
return isASCII;
}
Iterate through the string, and use charAt() to get the char. Then treat it as an int, and see if it has a unicode value (a superset of ASCII) which you like.
Break at the first you don't like.
You can do it with java.nio.charset.Charset.
import java.nio.charset.Charset;
public class StringUtils {
public static boolean isPureAscii(String v) {
return Charset.forName("US-ASCII").newEncoder().canEncode(v);
// or "ISO-8859-1" for ISO Latin 1
// or StandardCharsets.US_ASCII with JDK1.7+
}
public static void main (String args[])
throws Exception {
String test = "Réal";
System.out.println(test + " isPureAscii() : " + StringUtils.isPureAscii(test));
test = "Real";
System.out.println(test + " isPureAscii() : " + StringUtils.isPureAscii(test));
/*
* output :
* Réal isPureAscii() : false
* Real isPureAscii() : true
*/
}
}
private static boolean isASCII(String s)
{
for (int i = 0; i < s.length(); i++)
if (s.charAt(i) > 127)
return false;
return true;
}
try this:
for (char c: string.toCharArray()){
if (((int)c)>127){
return false;
}
}
return true;
Source: Stackoverflow.com