k
/
legado
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

Update RuleAnalyzer.kt

Browse Source 
原来的规则还有点问题，有些情况下会跳过不分隔规则。

原因是拿一个变量记录了两个状态，有时忘记回退切换状态，导致队列位置处理混乱。

我重新梳理了一遍，更改了所有所有相关规则，新增了类成员变量保存另一种状态。

然后测试了我自己写的那几个书源，和其它用大量采用“||”、“&&”、“%%”分隔规则的书源，总算搞定。
pull/1106/head
bushixuanqi 3 years ago committed by GitHub
parent beb03a5cf2
commit 08d53a974c
No known key found for this signature in database
GPG Key ID: 4AEE18F83AFDEB23
1 changed files with 141 additions and 102 deletions
Show all changes Ignore whitespace when comparing lines Ignore changes in amount of whitespace Ignore changes in whitespace at EOL
Show Stats Download Patch File Download Diff File
  1. 181
      app/src/main/java/io/legado/app/model/analyzeRule/RuleAnalyzer.kt

181
app/src/main/java/io/legado/app/model/analyzeRule/RuleAnalyzer.kt
Unescape View File

@ -6,8 +6,10 @@ class RuleAnalyzer(data: String) {
private var queue: String = data //被处理字符串
private var pos = 0 //处理到的位置
private var rule = arrayOf<String>()
private var start = 0 //每次处理字段的开始
private var startX = 0 //规则的开始
private var step:Int = 0 //分割字符的长度
var elementsType = ""
@ -17,6 +19,11 @@ class RuleAnalyzer(data: String) {
return queue.substring(start+stepStart,pos-stepEnd) //当前平衡字段
}
fun trim(){ // 修剪当前规则之前的"@"或者空白符
while (queue[pos] == '@' || queue[pos] < '!') pos++
}
//将pos重置为0，方便复用
fun reSetPos() {
pos = 0
@ -34,21 +41,6 @@ class RuleAnalyzer(data: String) {
return queue.substring(start)
}
/**
* pos位置回退
*/
fun back(num :Int = 0) {
if(num == 0)pos = start //回退
else pos -= num
}
/**
* pos位置后移
*/
fun advance(num :Int = 1) {
pos+=num
}
/**
* 是否已无剩余字符?
* @return 若剩余字串中已无字符则返回true
@ -109,9 +101,8 @@ class RuleAnalyzer(data: String) {
while (pos != queue.length) {
for (s in seq) {
if (matches(s)) {
if (queue.regionMatches(pos, s, 0, s.length)) {
step = s.length //间隔数
start = this.pos //匹配成功, 设置规则下次起始位置
this.pos = pos //匹配成功, 同步处理位置到类
return true //匹配就返回 true
}
@ -163,7 +154,7 @@ class RuleAnalyzer(data: String) {
var commits = false //多行注释
do {
if (isEmpty) break
if (pos == queue.length) break
var c = queue[pos++]
if (c != '\\') { //非转义字符
if (c == '\'' && !commits && !commit && !regex && !inDoubleQuote && !inOtherQuote) inSingleQuote = !inSingleQuote //匹配具有语法功能的单引号
@ -192,22 +183,23 @@ class RuleAnalyzer(data: String) {
']' -> bracketsDepth-- //闭合一层嵌套[]
}
}
}else if(c == '\n') commit = false
if (commits || commit || regex || inSingleQuote || inDoubleQuote || inOtherQuote) continue //语法单元未匹配结束，直接进入下个循环
val fn = f(c) ?: continue
if (fn) depth++ else depth-- //嵌套或者闭合
}else { //转义字符
var next = queue[pos++] //拉出被转义字符
if(commit && next == 'n') commit = false //匹配单行注释终点。当前为\,下个为n，表示换行
else if (!commits && !commit && next == '\\') {
queue[pos++] //当前为\,下个为\，双重转义中"\\"表示转义字符本身，根据if条件"\\"字串不在注释中，则只能在字串或正则中
next = queue[pos++] //拉出下个字符，因为在双重转义的字串或正则中，类似于 \\/ 这样的结构才是转义结构
if(next == '\\')queue[pos++] //若为转义字符则继续拉出，因为双重转义中转义字符成对存在,即 \\\\
}
}
}else pos++
// { //转义字符
// var next = queue[pos++] //拉出被转义字符
// if(commit && next == 'n') commit = false //匹配单行注释终点。当前为\,下个为n，表示换行
// else if (!commits && !commit && next == '\\') {
// queue[pos++] //当前为\,下个为\，双重转义中"\\"表示转义字符本身，根据if条件"\\"字串不在注释中，则只能在字串或正则中
// next = queue[pos++] //拉出下个字符，因为在双重转义的字串或正则中，类似于 \\/ 这样的结构才是转义结构
// if(next == '\\')queue[pos++] //若为转义字符则继续拉出，因为双重转义中转义字符成对存在,即 \\\\
// }
// }
} while (depth > 0 || bracketsDepth >0) //拉出全部符合js语法的字段
return if(depth > 0 || bracketsDepth > 0) false else {
@ -230,7 +222,7 @@ class RuleAnalyzer(data: String) {
var inDoubleQuote = false //双引号
do {
if (isEmpty) break
if (pos == queue.length) break
val c = queue[pos++]
if (c != ESC) { //非转义字符
if (c == '\'' && !inDoubleQuote) inSingleQuote = !inSingleQuote //匹配具有语法功能的单引号
@ -245,14 +237,15 @@ class RuleAnalyzer(data: String) {
if (fn) otherDepth++ else otherDepth--
}
}else { //转义字符
var next = queue[pos++] //拉出被转义字符，匹配\/、\"、\'等
if (next == ESC) {
queue[pos++] //当前为\,下个为\，双重转义中"\\"表示转义字符本身，根据语法特征当前字段在字串或正则中
next = queue[pos++] //拉出下个字符，因为在双重转义的字串或正则中，类似于 \\/ 这样的结构才是转义结构
if(next == ESC)queue[pos++] //若为转义字符则继续拉出，因为双重转义中转义字符成对存在,即 \\\\
}
}
}else pos++
// { //转义字符
// var next = queue[pos++] //拉出被转义字符，匹配\/、\"、\'等
// if (next == ESC) {
// queue[pos++] //当前为\,下个为\，双重转义中"\\"表示转义字符本身，根据语法特征当前字段在字串或正则中
// next = queue[pos++] //拉出下个字符，因为在双重转义的字串或正则中，类似于 \\/ 这样的结构才是转义结构
// if(next == ESC)queue[pos++] //若为转义字符则继续拉出，因为双重转义中转义字符成对存在,即 \\\\
// }
// }
} while (depth > 0 || otherDepth > 0) //拉出一个平衡字串
return if(depth > 0 || otherDepth > 0) false else {
@ -269,20 +262,32 @@ class RuleAnalyzer(data: String) {
if(split.size == 1) {
elementsType = split[0] //设置分割字串
return if(!consumeTo(elementsType)) {
rule += queue.substring(startX)
rule
}else {
step = elementsType.length //设置分隔符长度
return splitRule(arrayOf()) //仅一个分隔字串时，直接二段解析更快
}else if (!consumeToAny(* split)) return arrayOf(queue) //未找到分隔符
splitRule()
} //递归匹配
}else if (!consumeToAny(* split)) { //未找到分隔符
rule += queue.substring(startX)
return rule
}
val st = findToAny( '[','(' ) //查找筛选器
val end = pos //记录分隔位置
pos = start //重回开始，启动另一种查找
do{
val st = findToAny('[', '(') //查找筛选器位置
if (st == -1) {
var rule = arrayOf(queue.substring(0, pos)) //压入分隔的首段规则到数组
rule = arrayOf(queue.substring(startX, end)) //压入分隔的首段规则到数组
elementsType = queue.substring(pos, pos + step) //设置组合类型
pos += step //跳过分隔符
elementsType = queue.substring(end, end + step) //设置组合类型
pos = end + step //跳过分隔符
while (consumeToAny(* split)) { //循环切分规则压入数组
while (consumeTo(elementsType)) { //循环切分规则压入数组
rule += queue.substring(start, pos)
pos += step //跳过分隔符
}
@ -292,73 +297,107 @@ class RuleAnalyzer(data: String) {
return rule
}
val rule = if(st >pos ){ //先匹配到st1pos，表明分隔字串不在选择器中，将选择器前分隔字串分隔的字段依次压入数组
if (st > end) { //先匹配到st1pos，表明分隔字串不在选择器中，将选择器前分隔字串分隔的字段依次压入数组
var rule = arrayOf(queue.substring(0, pos)) //压入分隔的首段规则到数组
rule = arrayOf(queue.substring(startX, end)) //压入分隔的首段规则到数组
elementsType = queue.substring(pos, pos + step) //设置组合类型
pos += step //跳过分隔符
elementsType = queue.substring(end, end + step) //设置组合类型
pos = end + step //跳过分隔符
while (consumeToAny( * split ) && pos < st ) { //循环切分规则压入数组
while (consumeTo(elementsType) && pos < st) { //循环切分规则压入数组
rule += queue.substring(start, pos)
pos += step //跳过分隔符
}
return if(pos > st) {
startX = start
splitRule() //首段已匹配,但当前段匹配未完成,调用二段匹配
}
else { //执行到此，证明后面再无分隔字符
rule += queue.substring(pos) //将剩余字段压入数组末尾
rule
}else null
}
}
pos = st //位置推移到筛选器处
val next = if (queue[pos] == '[') ']' else ')' //平衡组末尾字符
return if (rule == null) { //rule为空,首段未匹配完成
if (!chompRuleBalanced(queue[pos], next)) throw Error(
queue.substring(
0,
start
) + "后未平衡"
) //拉出一个筛选器,不平衡则报错
if(!chompRuleBalanced(queue[pos],next)) throw Error(queue.substring(0, start)+"后未平衡") //拉出一个筛选器,不平衡则报错
splitRule(* split) //递归调用首段匹配
}while( end > pos )
} else {
if(!chompRuleBalanced(queue[pos],next)) throw Error(queue.substring(0, start)+"后未平衡") //拉出一个筛选器,不平衡则报错
splitRule(rule) //首段已匹配,但当前段匹配未完成,调用二段匹配
}
start = pos //设置开始查找筛选器位置的起始位置
return splitRule(* split) //递归调用首段匹配
}
@JvmName("splitRuleNext")
private tailrec fun splitRule(rules:Array<String>): Array<String>{ //二段匹配被调用,elementsType非空(已在首段赋值),直接按elementsType查找,比首段采用的方式更快
private tailrec fun splitRule(): Array<String>{ //二段匹配被调用,elementsType非空(已在首段赋值),直接按elementsType查找,比首段采用的方式更快
if (!consumeTo(elementsType,false)) return rules + queue.substring(start) //此处consumeTo(...)开始位置不是规则的开始位置,start沿用上次设置
val end = pos //记录分隔位置
pos = start //重回开始，启动另一种查找
val st = findToAny( '[','(' ) //查找筛选器
do{
val st = findToAny('[', '(') //查找筛选器位置
if (st == -1) {
var rule = rules + queue.substring(start, pos) //压入本次分隔的首段规则到数组
pos += step //跳过分隔符
rule += arrayOf(queue.substring(startX, end)) //压入分隔的首段规则到数组
pos = end + step //跳过分隔符
while (consumeTo(elementsType)) { //循环切分规则压入数组
rule += queue.substring(start, pos)
pos += step //跳过分隔符
}
rule += queue.substring(pos) //将剩余字段压入数组末尾
return rule
}
val rule = if(st > pos ){//先匹配到st1pos，表明分隔字串不在选择器中，将选择器前分隔字串分隔的字段依次压入数组
var rule = rules + queue.substring(start, pos) //压入本次分隔的首段规则到数组
pos += step //跳过分隔符
if (st > end) { //先匹配到st1pos，表明分隔字串不在选择器中，将选择器前分隔字串分隔的字段依次压入数组
rule += arrayOf(queue.substring(startX, end)) //压入分隔的首段规则到数组
pos = end + step //跳过分隔符
while (consumeTo(elementsType) && pos < st) { //循环切分规则压入数组
rule += queue.substring(start, pos)
pos += step //跳过分隔符
}
return if(pos > st) {
startX = start
splitRule() //首段已匹配,但当前段匹配未完成,调用二段匹配
}
else { //执行到此，证明后面再无分隔字符
rule += queue.substring(pos) //将剩余字段压入数组末尾
rule
}else rules
}
}
pos = st //位置推移到筛选器处
val next = if (queue[pos] == '[') ']' else ')' //平衡组末尾字符
if(!chompRuleBalanced(queue[pos],next)) throw Error(queue.substring(0, start)+"后未平衡") //拉出一个筛选器,不平衡时返回true,表示未平衡
if (!chompRuleBalanced(queue[pos], next)) throw Error(
queue.substring(
0,
start
) + "后未平衡"
) //拉出一个筛选器,不平衡则报错
return splitRule(rule) //递归匹配
}while( end > pos )
start = pos //设置开始查找筛选器位置的起始位置
return if(!consumeTo(elementsType)) {
rule += queue.substring(startX)
rule
}else splitRule() //递归匹配
}

Write Preview
Loading…
Cancel
Save