Java 字符串转换成算术表达式

今天遇到了一个需要将数据库中保存的表达式，替换其中的字符，并计算出值，java是不能直接计算的例如：

double d = (3+5-(2-4)*2)/24;没有问题

但是：遇到   "(3+5-(2-4)*2)/24" 字符串的格式，就不能使用了

java是无法解决，但是javaScript中有一个eval函数是可以执行的，所以，可以通过其他途径执行javaScript就可以做到，而ScriptEngine是java的一个javaScript实现类，所以就找到了方法，详细方法见下

public class MapJ {
	private String key;//替换的编号
	private String value;//值
	public String getKey() {
		return key;
	}
	public void setKey(String key) {
		this.key = key;
	}
	public String getValue() {
		return value;
	}
	public void setValue(String value) {
		this.value = value;
	}
	public MapJ(String key, String value) {
		super();
		this.key = key;
		this.value = value;
	}
	public MapJ() {
		super();
	}
	
}


//实现类
import java.util.List;

import javax.script.ScriptEngine;
import javax.script.ScriptEngineManager;
import javax.script.ScriptException;

public class JavaScript {
	ScriptEngineManager factory = new ScriptEngineManager();
    ScriptEngine engine = factory.getEngineByName("JavaScript");
    
    public Double getMathValue(List<MapJ> map,String option){
    	double d = 0;
		try {
			for(int i=0; i<map.size();i++){
				MapJ mapj = map.get(i);
				option = option.replaceAll(mapj.getKey(), mapj.getValue());
			}
			Object o = engine.eval(option);
			d = Double.parseDouble(o.toString());
		} catch (ScriptException e) {
			System.out.println("无法识别表达式");
			return null;
		}
		return d;
    }
}


//测试
import java.util.ArrayList;
import java.util.List;

public class JavaScriptTest {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		

		String sbt = "(B+D-(A-C)*A)/F";
		List<MapJ> all = new ArrayList<MapJ>();
		all.add(new MapJ("A","2"));
		all.add(new MapJ("B","3"));
		all.add(new MapJ("C","4"));
		all.add(new MapJ("D","5"));
		all.add(new MapJ("F","24"));
		JavaScript js = new JavaScript();
		Double d = js.getMathValue(all, sbt);
		if(d==null){
			System.out.println("                 无法计算这个表达式");
		}else{
			System.out.println(d*100+"%");
		}
	}

}

通过以上的方法即可实现此方法。
以上转载自：http://ruyuntao.iteye.com/blog/406935
----------------------------------------------------------------------------
下面是算法实现的：

package cn.anycall.test;

import java.math.BigDecimal;

public class TestSHU {

	/**  
     * 两个字符类型的小数进行相加为a+b;  
     *   
     * @param a  
     * @param b  
     * @return  
     */  
    public static String addBigDecimal(String a, String b) {   
        double a1 = Double.parseDouble(a);   
        double b1 = Double.parseDouble(b);   
        BigDecimal a2 = BigDecimal.valueOf(a1);   
        BigDecimal b2 = BigDecimal.valueOf(b1);   
        BigDecimal c2 = a2.add(b2);   
        String c1 = c2 + "";   
        return c1;   
    }  
    /**  
     * 两个字符类型的小数进行相减为a-b;  
     *   
     * @param a  
     * @param b  
     * @return  
     */  
    public static String reduceBigDecimal(String a, String b) {   
        double a1 = Double.parseDouble(a);   
        double b1 = Double.parseDouble(b);   
        BigDecimal a2 = BigDecimal.valueOf(a1);   
        BigDecimal b2 = BigDecimal.valueOf(b1);   
        BigDecimal c2 = a2.subtract(b2);   
        String c1 = c2 + "";   
        return c1;   
    }  
    /**  
     * 两个字符类型的数相乘 a*b=c；  
     *   
     * @param a  
     * @param b  
     * @return  
     */  
    public static String multipliedString(String a, String b) {   
        double a1 = Double.parseDouble(a);   
        double b1 = Double.parseDouble(b);   
        BigDecimal a2 = BigDecimal.valueOf(a1);   
        BigDecimal b2 = BigDecimal.valueOf(b1);   
        BigDecimal c2 = a2.multiply(b2);   
        String c1 = c2 + "";   
        return c1;   
    }  
    /**  
     * 两个字符类型的数相除 a/b=c；  
     *   
     * @param a  
     * @param b  
     * @return  
     */  
    public static String divideString(String a, String b) {   
        double a1 = Double.parseDouble(a);   
        double b1 = Double.parseDouble(b);   
        BigDecimal a2 = BigDecimal.valueOf(a1);   
        BigDecimal b2 = BigDecimal.valueOf(b1);   
        BigDecimal c2 = a2.divide(b2,a2.scale());   
        String c1 = c2 + "";   
        return c1;   
    }  

	public static String yunsuanjibie(String s) {
		String r = "";
		int p = 0;
		for (int i = 0; i < s.length(); i++) {
			if (s.charAt(i) == '+' || s.charAt(i) == '-' || s.charAt(i) == '*'
					|| s.charAt(i) == '/') {
				p++;
			}
		}
		String k[] = new String[2 * p + 1];
		int k1 = 0;
		int first = 0;
		for (int i = 0; i < s.length(); i++) {
			if (s.charAt(i) == '+' || s.charAt(i) == '-' || s.charAt(i) == '*'
					|| s.charAt(i) == '/') {
				k[k1] = s.substring(first, i);
				k1++;
				k[k1] = "" + s.charAt(i);
				k1++;
				first = i + 1;
			}
		}
		k[k1] = s.substring(first, s.length());
		int kp = p;
		while (kp > 0) {
			for (int i = 0; i < k.length; i++) {
				if (k[i].equals("*") || k[i].equals("/")) {
					int l;
					for (l = i - 1; l > -1; l--) {
						if (!(k[l].equals("p")))
							break;
					}
					int q;
					for (q = i + 1; q < k.length; q++) {
						if (!(k[l].equals("p")))
							break;
					}
					if (k[i].equals("*")) {
						k[i] = ""+ multipliedString(k[l],k[q]);
						k[l] = "p";
						k[q] = "p";
						kp--;
					} else {
						k[i] = ""+divideString(k[l],k[q]);
						k[l] = "p";
						k[q] = "p";
						kp--;
					}
					break;
				}
			}
			for (int i = 0; i < 2 * p + 1; i++) {
				if (k[i].equals("+") || k[i].equals("-")) {
					int l;
					for (l = i - 1; l > -1; l--) {
						if (!(k[l].equals("p")))
							break;
					}
					int q;
					for (q = i + 1; q < k.length; q++) {
						if (!(k[q].equals("p")))
							break;
					}
					if (k[i].equals("+")) {
						k[i] = ""+addBigDecimal(k[l],k[q]);
						k[l] = "p";
						k[q] = "p";
						kp--;
					} else {
						k[i] = ""+reduceBigDecimal(k[l],k[q]);
						k[l] = "p";
						k[q] = "p";
						kp--;
					}
					break;
				}
			}
			for (int i = 0; i < k.length; i++) {
				if (!(k[i].equals("p"))) {
					r = k[i];
					break;
				}
			}
		}
		return r;
	}

	public static void sizeyunsuan(String s) {
		while (true) {
			int first = 0;
			int last = 0;
			for (int i = 0; i < s.length(); i++) {
				if (s.charAt(i) == '(')
					first = i;
				if (s.charAt(i) == ')') {
					last = i;
					break;
				}
			}
			if (last == 0) {
				System.out.println(yunsuanjibie(s));
				return;
			} else {
				String s1 = s.substring(0, first);
				String s2 = s.substring(first + 1, last);
				String s3 = s.substring(last + 1, s.length());
				s = s1 + yunsuanjibie(s2) + s3;
			}
		}
	}

	public static void main(String[] args) {
		String s = "1.4+2*32/(3-2.1)";
		sizeyunsuan(s);
	}

}

转载自：http://yafei.iteye.com/blog/725582

评论

1 楼 zhou85xin 2014-12-04  

option = option.replaceAll(mapj.getKey(), mapj.getValue());

这个就不对,map中key如果contains，很容易就替换错了