如何在将双精度值转换为字符串后删除小数点

Question

I want to remove a decimal point from a double value. I tried converting it to string

double intify(double a)
{
    String s=Double.toString(a);
    String s2=s.replaceAll(".","");
     a=Double.parseDouble(s2);
    
    return a;
    
}

If I pass 123.4567 it should return 1234567.

Answer 1

replaceAll replaces a regex. You should use replace instead:

String s2 = s.replace(".", "");

Answer 2

Instead of replaceAll use replace method of String:

String s2 = s.replace(".", "");

As "." has a special meaning in regex and replaceAll treats source string ie first parameter as regex.

Answer 3

The below solves the issue of an integer being passed in ( 123 or 123.0 ), or a double in scientific notation ( 1.23E7 ):

Double.parseDouble(Double.toString(a).replaceAll("[.](?0$)?",""));

In your code, just change the replaceAll string to [.](?0$)? , which states "Remove the decimal as well as the number 0 if it is right after the decimal and there's nothing after it".

Answer 4

    Double i=Double.parseDouble("String with double value");


    Log.i(tag,"display double "+i);


    try{
        NumberFormat nf = NumberFormat.getInstance();
        nf.setMaximumFractionDigits(0);// set as you need
        String myStringmax = nf.format(i);


        String result = myStringmax.replaceAll("[-+.^:,]","");
        Double i=Double.parseDouble(result);

       int max= Integer.parseInt(result);


    }catch(Exception e){
        System.out.println("ex="+e);
    }

Answer 5

您的问题对于一般答案的定义不够好，但是如果您想摆脱小数部分，只需将其解析为 double ，然后将其转换为int或long 。