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通过java中的循环解决

[英]solve through a loop in java

I have a text box(having name b) and one submit button in index.jsp . 我在index.jsp有一个text box(having name b)和一个提交按钮。 I will enter something in text box then after clicking submit i get that value by in server side find.jsp . 我将在文本框中输入内容,然后单击“ submit ”,然后在服务器端find.jsp获得该值。 In find.jsp i get the value by request.getParameter("b"); 在find.jsp中,我通过request.getParameter("b");获取值request.getParameter("b"); In find.jsp calaculation is done as shown below: find.jsp计算如下所示:

double c=0;
double d=0;

7800.0/12 and 2640.0/12 are fixed.     

if request.getParameter("b")=1 then 

c=7800.0/12*5;// 5 is multiplied with above number(in first one)

d=2640.0/12*5;// 5 is multiplied with above number(in second one) 

if request.getParameter("b")=2 then 

c=7800.0/12*8;// here 5 gets incremented by 3 and became 8(in first one)

d=2640.0/12*8;// same also here( in second one) 

Similarly if i will enter 20 then accordingly c and d will be calculated. 同样,如果我输入20,则将相应地计算c和d。 I cannot use if-else as any value can be entered in client side and accordingly c and d will be calculated in server side. 我不能使用if-else,因为可以在客户端输入任何值,因此c和d将在服务器端计算。 How can i implement it in a loop? 我怎样才能在循环中实现它? Many thanks 非常感谢

try 尝试

    double devide=2.0
    devide = Integer.parseInt(request.getParameter("b"))*3+devide

    c=7800.0/(12*devide);
    d=2640.0/(12*devide);

You can use for any value to count below one logic. 您可以使用任何值来计算低于一个逻辑。

int value = Integer.parseInt(request.getParameter("b"));
double x=2.0;
double valueForDevide = (value*3)+x;

c=7800.0/(12*valueForDevide );
d=2640.0/(12*valueForDevide );

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