我试图写一个Java例程来计算数学表达式从字符串值,如:
"5 + 3" "10-4 * 5" "(1 + 10) * 3"
我想避免很多如果-then-else语句。 我该怎么做呢?
我试图写一个Java例程来计算数学表达式从字符串值,如:
"5 + 3" "10-4 * 5" "(1 + 10) * 3"
我想避免很多如果-then-else语句。 我该怎么做呢?
当前回答
使用带有代码注入处理的JDK1.6 Javascript引擎尝试下面的示例代码。
import javax.script.ScriptEngine;
import javax.script.ScriptEngineManager;
public class EvalUtil {
private static ScriptEngine engine = new ScriptEngineManager().getEngineByName("JavaScript");
public static void main(String[] args) {
try {
System.out.println((new EvalUtil()).eval("(((5+5)/2) > 5) || 5 >3 "));
System.out.println((new EvalUtil()).eval("(((5+5)/2) > 5) || true"));
} catch (Exception e) {
e.printStackTrace();
}
}
public Object eval(String input) throws Exception{
try {
if(input.matches(".*[a-zA-Z;~`#$_{}\\[\\]:\\\\;\"',\\.\\?]+.*")) {
throw new Exception("Invalid expression : " + input );
}
return engine.eval(input);
} catch (Exception e) {
e.printStackTrace();
throw e;
}
}
}
其他回答
看来应该由JEP来做这项工作
如果我们要实现它,那么我们可以使用下面的算法
While there are still tokens to be read in, 1.1 Get the next token. 1.2 If the token is: 1.2.1 A number: push it onto the value stack. 1.2.2 A variable: get its value, and push onto the value stack. 1.2.3 A left parenthesis: push it onto the operator stack. 1.2.4 A right parenthesis: 1 While the thing on top of the operator stack is not a left parenthesis, 1 Pop the operator from the operator stack. 2 Pop the value stack twice, getting two operands. 3 Apply the operator to the operands, in the correct order. 4 Push the result onto the value stack. 2 Pop the left parenthesis from the operator stack, and discard it. 1.2.5 An operator (call it thisOp): 1 While the operator stack is not empty, and the top thing on the operator stack has the same or greater precedence as thisOp, 1 Pop the operator from the operator stack. 2 Pop the value stack twice, getting two operands. 3 Apply the operator to the operands, in the correct order. 4 Push the result onto the value stack. 2 Push thisOp onto the operator stack. While the operator stack is not empty, 1 Pop the operator from the operator stack. 2 Pop the value stack twice, getting two operands. 3 Apply the operator to the operands, in the correct order. 4 Push the result onto the value stack. At this point the operator stack should be empty, and the value stack should have only one value in it, which is the final result.
下面是GitHub上另一个名为EvalEx的开源库。
与JavaScript引擎不同,这个库只专注于计算数学表达式。此外,该库是可扩展的,支持使用布尔运算符和圆括号。
现在回答已经太晚了,但我也遇到过同样的情况,在java中计算表达式,这可能会帮助到一些人
MVEL对表达式进行运行时求值,我们可以在String中编写java代码来得到它的值。
String expressionStr = "x+y";
Map<String, Object> vars = new HashMap<String, Object>();
vars.put("x", 10);
vars.put("y", 20);
ExecutableStatement statement = (ExecutableStatement) MVEL.compileExpression(expressionStr);
Object result = MVEL.executeExpression(statement, vars);
我已经使用迭代解析和分流码算法,我真的很喜欢开发表达式求值器,你可以在这里找到所有的代码
https://github.com/nagaraj200788/JavaExpressionEvaluator
有73个测试用例,甚至工作于大整数,大小数
支持所有关系,算术表达式和两者的组合。 甚至支持三元运算符。
增加了增强,以支持有符号的数字,如-100+89,这是有趣的,详细信息请检查TokenReader.isUnaryOperator()方法,我已经更新了上面链接中的代码