问题
How to replace \0 (NUL) in the String?
String b = "2012yyyy06mm"; // sth what i want
String c = "2\0\0\0012yyyy06mm";
String d = c.replaceAll("\\\\0", ""); // not work
String e = d.replace("\0", ""); // er, the same
System.out.println(c+"\n"+d+"\n"+e);
String bb = "2012yyyy06mm";
System.out.println(b.length() + " > " +bb.length());
The above code will print 12 > 11 in console. Oops, What happened?
String e = c.replace("\0", "");
System.out.println(e); // just print 2(a bad character)2yyyy06mm
回答1:
Your string "2\0\0\0012yyyy06mm"
does not start 2 {NUL} {NUL} {NUL} 0 1 2
, but instead contains 2 {NUL} {NUL} {SOH} 2
.
The \001
is treated as a single ASCII 1 character (SOH
) and not as a NUL
followed by 1 2
.
The result is that only two characters are being removed, not three.
I don't think there's any way to represent digits following an abbreviated octal escape other than by breaking the string apart:
String c = "2" + "\0\0\0" + "012yyyy06mm";
or alternately, specify all three digits in the (last) octal escape such that the following digits are not interpreted as being part of the octal escape:
String c = "2\000\000\000012yyyy06mm";
Once you've done that, replacing "\0"
as per your line:
String e = c.replace("\0", "");
will work correctly.
来源:https://stackoverflow.com/questions/13857726/special-characters-0-nul-in-java