也许有人可以提供帮助？ 如何修改此方法next（）下一个标记可以是：＆＃39; abc＆＃39;带引号的文字。 现在，如果文本包含引用，则抛出ExpressionException Unknown operator＆＃39;＆＃39;＆＃39;＆＃39;在位置......
@Override
public String next() {
StringBuilder token = new StringBuilder();
if (pos >= input.length()) {
return previousToken = null;
}
char ch = input.charAt(pos);
while (Character.isWhitespace(ch) && pos < input.length()) {
ch = input.charAt(++pos);
}
if (Character.isDigit(ch)) {
while ((Character.isDigit(ch) || ch == decimalSeparator)
&& (pos < input.length())) {
token.append(input.charAt(pos++));
ch = pos == input.length() ? 0 : input.charAt(pos);
}
} else if (ch == minusSign
&& Character.isDigit(peekNextChar())
&& ("(".equals(previousToken) || ",".equals(previousToken)
|| previousToken == null || operators
.containsKey(previousToken))) {
token.append(minusSign);
pos++;
token.append(next());
} else if (Character.isLetter(ch)) {
while ((Character.isLetter(ch) || Character.isDigit(ch) || (ch == '_')) && (pos < input.length())) {
token.append(input.charAt(pos++));
ch = pos == input.length() ? 0 : input.charAt(pos);
}
} else if (ch == '(' || ch == ')' || ch == ',') {
token.append(ch);
pos++;
//FIXME
else if (ch == '\''){
pos++;
String temp = "\'"+next()+"\'";
token.append(temp);
pos++;
}
//
} else {
while (!Character.isLetter(ch) && !Character.isDigit(ch)
&& !Character.isWhitespace(ch) && ch != '('
&& ch != ')' && ch != ',' && (pos < input.length())) {
token.append(input.charAt(pos));
pos++;
ch = pos == input.length() ? 0 : input.charAt(pos);
if (ch == minusSign) {
break;
}
}
if (!operators.containsKey(token.toString())) {
throw new ExpressionException("Unknown operator '" + token
+ "' at position " + (pos - token.length() + 1));
}
}
return previousToken = token.toString();
}
EVAL
public Object eval() {
Stack<Object> stack = new Stack<Object>();
for (String token : getRPN()) {
mylog.pl("Reverse polish notation TOKEN : " + token + " RPN size: " + getRPN().size() );
if (operators.containsKey(token)) {
Object v1 = stack.pop();
Object v2 = stack.pop();
stack.push(operators.get(token).eval(v2, v1));
} else if (variables.containsKey(token)) {
stack.push(variables.get(token).round(mc));
} else if (functions.containsKey(token.toUpperCase())) {
Function f = functions.get(token.toUpperCase());
ArrayList<Object> p = new ArrayList<Object>(f.getNumParams());
for (int i = 0; i < f.numParams; i++) {
p.add(0, stack.pop());
}
Object fResult = f.eval(p);
stack.push(fResult);
} else if (isDate(token)) {
Long date = null;
try {
date = SU.sdf.parse(token).getTime();
} catch (ParseException e) {/* IGNORE! */
}
stack.push(new BigDecimal(date, mc));
} else {
if (BusinessStrategy.PREFIX_X.equals(Character.toString(token.charAt(0)))) {
stack.push(token);
} else {
stack.push(new BigDecimal(token, mc));
}
}
}
return stack.pop();
}
反向表示法
private List<String> getRPN() {
if (rpn == null) {
rpn = shuntingYard(this.expression);
}
return rpn;
}
堆场
private List<String> shuntingYard(String expression) {
List<String> outputQueue = new ArrayList<String>();
Stack<String> stack = new Stack<String>();
Tokenizer tokenizer = new Tokenizer(expression);
String lastFunction = null;
while (tokenizer.hasNext()) {
String token = tokenizer.next();
if (isNumber(token)) {
outputQueue.add(token);
} else if (variables.containsKey(token)) {
outputQueue.add(token);
} else if (functions.containsKey(token.toUpperCase())) {
stack.push(token);
lastFunction = token;
} else if (Character.isLetter(token.charAt(0))) {
if ("\'".equals(Character.toString(token.charAt(0)))){
outputQueue.add(token);
} else {
stack.push(token);
}
} else if (",".equals(token)) {
while (!stack.isEmpty() && !"(".equals(stack.peek())) {
outputQueue.add(stack.pop());
}
if (stack.isEmpty()) {
throw new ExpressionException("Parse error for function '"
+ lastFunction + "'");
}
} else if (operators.containsKey(token)) {
Operator o1 = operators.get(token);
String token2 = stack.isEmpty() ? null : stack.peek();
while (operators.containsKey(token2)
&& ((o1.isLeftAssoc() && o1.getPrecedence() <= operators
.get(token2).getPrecedence()) || (o1
.getPrecedence() < operators.get(token2)
.getPrecedence()))) {
outputQueue.add(stack.pop());
token2 = stack.isEmpty() ? null : stack.peek();
}
stack.push(token);
} else if ("(".equals(token)) {
stack.push(token);
} else if (")".equals(token)) {
while (!stack.isEmpty() && !"(".equals(stack.peek())) {
outputQueue.add(stack.pop());
}
if (stack.isEmpty()) {
throw new RuntimeException("Mismatched parentheses");
}
stack.pop();
if (!stack.isEmpty()
&& functions.containsKey(stack.peek().toUpperCase())) {
outputQueue.add(stack.pop());
}
}
}
while (!stack.isEmpty()) {
String element = stack.pop();
if ("(".equals(element) || ")".equals(element)) {
throw new RuntimeException("Mismatched parentheses");
}
if (!operators.containsKey(element)) {
throw new RuntimeException("Unknown operator or function: "
+ element);
}
outputQueue.add(element);
}
return outputQueue;
}
错误
*java.util.EmptyStackException
at java.util.Stack.peek(Unknown Source)
at java.util.Stack.pop(Unknown Source)
at com.business.Expression.eval(Expression.java:1033)*
它在eval方法Object v1 = stack.pop();行。
谢谢！
答案 0 :(得分：1)
在方法next中，您可以在两个地方进行递归调用：
第一种情况是构造令牌，其中一个减号后面跟一个数字（即一个不带数字的号码） - 好的。 （虽然没有标志但是一元减号运算符值得考虑。）
第二种情况意味着麻烦。在超过最初的撇号后，预计会出现另一个下一个结果，就好像字符串文字只包含一个数字或一个标识符或单个运算符。无论如何，next（）执行，让它说它返回一个数字：然后将一个apostroph添加到令牌中，但是没有努力检查是否有一个关闭撇号，也没有跳过它。 / p>
else if (ch == '\''){
token.append( '\'' );
pos++;
while( pos < input.length() &&
(ch = input.charAt(pos++)) != '\'' ){
token.append( ch );
}
token.append( '\'' );
这不允许撇号成为字符串中的字符，并且它不会诊断未终止的字符串。但这可以很容易地添加。