Java中的特殊字符\0 {NUL} [英] Special characters \0 {NUL} in Java

问题描述

如何在字符串中替换\0（NUL）？

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());  

以上代码将在控制台中打印12> 11。哎呀，发生什么事了？

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

推荐答案

你的字符串2 \\\\0012yyyy06mm无法启动 2 {NUL} {NUL} {NUL} 0 1 2 ，而是包含 2 {NUL} {NUL} {SOH} 2 。

Your string "2\0\0\0012yyyy06mm" does not start 2 {NUL} {NUL} {NUL} 0 1 2, but instead contains 2 {NUL} {NUL} {SOH} 2.

\001 被视为单个ASCII 1字符（ SOH ）而不是 NUL 后跟 1 2 。

The \001 is treated as a single ASCII 1 character (SOH) and not as a NUL followed by 1 2.

结果是只有两个字符是被删除，而不是三个。

The result is that only two characters are being removed, not three.

我认为没有办法在缩写八进制转义，而不是将字符串分开：

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";

或者，在（最后一个）八进制转义中指定所有三位数，使得以下数字不是被解释为八进制转义的一部分：

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";

完成后，替换\0根据你的行：

Once you've done that, replacing "\0" as per your line:

String e = c.replace("\0", "");

将正常工作。

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