问题
I am coding a basic telnet server in Java and I would like to process the backspace terminal control character denoted by '\b'. The backspace character deletes/removes the preceding character in the string.
I am currently using the example method below to successfully achieve this but does anyone know of a cleaner/more efficient method?
Many thanks for any assistance you can provide.
/*
Example input:
"This is a dog\b\b\bcat"
"\b\b\bThis is x\b a cat"
"\b\b\bThis\b\b\bThis is a dog\b\b\bcat"
*/
private String processBackspace(String input)
{
StringBuilder output = new StringBuilder();
int backSpaceCount = 0, index = 0;
boolean isBackSpace = false;
for (int i = input.length() - 1; i >= 0; i--)
{
char c = input.charAt(i);
if (c == '\b')
{
isBackSpace = true;
backSpaceCount++;
}
else
isBackSpace = false;
if (!isBackSpace)
{
index = i - backSpaceCount;
if (index >= 0)
output.append(input.charAt(index));
}
}
output.reverse();
return output.toString();
}
回答1:
I would do it like this
private String processBackspace(String input) {
StringBuilder sb = new StringBuilder();
for (char c : input.toCharArray()) {
if (c == '\b') {
if (sb.length() > 0) {
sb.deleteCharAt(sb.length() - 1);
}
} else {
sb.append(c);
}
}
return sb.toString();
}
来源：https://stackoverflow.com/questions/16381879/how-to-process-the-backspace-terminal-control-character-in-java