📄 fullrepiterable.java
字号:
/*
* Created on 28/07/2005
*/
package org.python.pydev.core;
import java.util.ArrayList;
import java.util.Iterator;
/**
* iterates through a string so that parts of it are gotten each time in a progressive way based on dots
*
* (e.g.: a.b.c
*
* will return
*
* a
* a.b
* a.b.c
*
* @author Fabio
*/
public class FullRepIterable implements Iterable<String>{
private static final class ReverseFullRepIterator implements Iterator<String> {
private String fullRep;
public ReverseFullRepIterator(String fullRep) {
this.fullRep = fullRep;
}
public boolean hasNext() {
return fullRep.length() > 0;
}
public String next() {
if(fullRep.length() == 0){
throw new RuntimeException("no more items");
}
String ret = fullRep;
int l = fullRep.lastIndexOf('.');
if(l == -1){
fullRep = "";
}else{
fullRep = fullRep.substring(0, l);
}
return ret;
}
public void remove() {
throw new RuntimeException("Not supported");
}
}
private static final class FullRepIterator implements Iterator<String> {
private int i = -1;
private boolean lastStep; //even if there is no point, we should return the last string
private String fullRep;
public FullRepIterator(String fullRep) {
this.fullRep = fullRep;
lastStep = false;
}
public boolean hasNext() {
boolean ret = !lastStep || fullRep.indexOf('.', i) != -1;
return ret;
}
public String next() {
int j = fullRep.indexOf('.', i);
if(j == -1){
lastStep = true;
return fullRep;
}
i = j+1;
return fullRep.substring(0,j);
}
public void remove() {
throw new RuntimeException("not supported");
}
}
private String fullRep;
private boolean reverse;
public FullRepIterable(String fullRep) {
this(fullRep, false);
}
/**
* User is able to specify reverse mode.
*
* If it is the standard mode, 'aa.bb' would be 'aa', 'aa.bb'.
*
* In reverse mode 'aa.bb' would be 'aa.bb', 'aa'
*
* @param fullRep The dotted string we want to gather in parts
* @param reverse whether we want to get it in reverse order
*/
public FullRepIterable(String fullRep, boolean reverse) {
this.fullRep = fullRep;
this.reverse = reverse;
}
public Iterator<String> iterator() {
if(!reverse){
return new FullRepIterator(this.fullRep);
}else{
return new ReverseFullRepIterator(this.fullRep);
}
}
public static String[] headAndTail(String fullRep){
return headAndTail(fullRep, true);
}
public static final int HEAD = 1;
public static final int TAIL = 0;
/**
* on string 'aa.bb.cc', the head is 'cc' and the tail is 'aa.bb'
*
* head is pos 1 (cc)
* tail is pos 0 (aa.bb)
*
* if it does not have a ".", everything is part of the tail - this is a strange behaviour by the way... but it is what I need, because
* of the way to calculate the modules on some cases.
*/
public static String[] headAndTail(String fullRep, boolean emptyTailIfNoDot){
int i = fullRep.lastIndexOf('.');
if(i != -1){
return new String[]{
fullRep.substring(0, i),
fullRep.substring(i+1)
};
}else{
if(emptyTailIfNoDot){
return new String[]{
"",
fullRep
};
}else{
return new String[]{
fullRep,
""
};
}
}
}
/**
* @return the name of the parent module of the module represented by currentModuleName
*/
public static String getParentModule(String currentModuleName) {
return headAndTail(currentModuleName, true)[0];
}
/**
* @return All that is after the last dot (or the whole string if there is no dot)
*/
public static String getLastPart(String tokToCheck) {
int i = tokToCheck.lastIndexOf('.');
if(i == -1){
return tokToCheck;
}
return tokToCheck.substring(i+1);
}
/**
* @return All that is before the first dot (or the whole string if there is no dot)
*/
public static String getFirstPart(String tokToCheck) {
int i = tokToCheck.indexOf('.');
if(i == -1){
return tokToCheck;
}
return tokToCheck.substring(0, i);
}
/**
* @return All that is before the first dot (or the whole string if there is no dot)
*/
public static String getFirstPart(String tokToCheck, char[] toks) {
String ret = tokToCheck;
for(char c: toks){
int i = tokToCheck.indexOf(c);
if(i != -1){
String s = tokToCheck.substring(0, i);
if(s.length() < ret.length()){
ret = s;
}
}
}
return ret;
}
/**
* @return All that is before the last dot (or an empty string if there is no dot)
*/
public static String getWithoutLastPart(String currentModuleName) {
int i = currentModuleName.lastIndexOf('.');
if(i == -1){
return "";
}
return currentModuleName.substring(0, i);
}
public static String joinParts(String[] actToks, int parts) {
StringBuffer buffer = new StringBuffer();
for (int i = 0; i < parts; i++) {
if(i > 0){
buffer.append('.');
}
buffer.append(actToks[i]);
}
return buffer.toString();
}
public static String joinFirstParts(String[] actToks) {
return joinParts(actToks, actToks.length-1);
}
/**
* @return whether the foundRep contains some part with the nameToFind
*/
public static boolean containsPart(String foundRep, String nameToFind) {
String[] strings = dotSplit(foundRep);
for (String string : strings) {
if(string.equals(nameToFind)){
return true;
}
}
return false;
}
/**
* Splits some string given some char
*/
public static String[] split(String string, char toSplit) {
ArrayList<String> ret = new ArrayList<String>();
int len = string.length();
int last = 0;
char c = 0;
for (int i = 0; i < len; i++) {
c = string.charAt(i);
if(c == toSplit){
if(last != i){
ret.add(string.substring(last, i));
}
while(c == toSplit && i < len-1){
i++;
c = string.charAt(i);
}
last = i;
}
}
if(c != toSplit){
if(last == 0 && len > 0){
ret.add(string); //it is equal to the original (no dots)
}else if(last < len){
ret.add(string.substring(last, len));
}
}
return ret.toArray(new String[ret.size()]);
}
/**
* Splits the string as would string.split("\\."), but without yielding empty strings
*/
public static String[] dotSplit(String string) {
return split(string, '.');
}
}
⌨️ 快捷键说明
复制代码
Ctrl + C
搜索代码
Ctrl + F
全屏模式
F11
切换主题
Ctrl + Shift + D
显示快捷键
?
增大字号
Ctrl + =
减小字号
Ctrl + -