使用Java的Float类。

float f = Float.parseFloat("25");
String s = Float.toString(25.0f);

要比较，最好是将字符串转换为浮点数，然后作为两个浮点数进行比较。这是因为对于一个浮点数有多个字符串表示形式，当将其作为字符串进行比较时是不同的(例如。"25" != "25.0" != "25.00"等等)

您可以尝试以下示例代码:

public class StringToFloat
{

  public static void main (String[] args)
  {

    // String s = "fred";    // do this if you want an exception

    String s = "100.00";

    try
    {
      float f = Float.valueOf(s.trim()).floatValue();
      System.out.println("float f = " + f);
    }
    catch (NumberFormatException nfe)
    {
      System.out.println("NumberFormatException: " + nfe.getMessage());
    }
  }
}

在这里找到

- string . valueof ()

float amount=100.00f;
String strAmount=String.valueOf(amount);
// or  Float.toString(float)

String to Float - Float. parsefloat ()

String strAmount="100.20";
float amount=Float.parseFloat(strAmount)
// or  Float.valueOf(string)

我相信下面的代码会有所帮助:

float f1 = 1.23f;
String f1Str = Float.toString(f1);      
float f2 = Float.parseFloat(f1Str);

这是一个可能的答案，这也会给出精确的数据，只需要改变小数点在所需的形式。

public class TestStandAlone {

    /**
     * 
This method is to  main
     * @param args  void
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        try {
            Float f1=152.32f;
            BigDecimal roundfinalPrice = new BigDecimal(f1.floatValue()).setScale(2,BigDecimal.ROUND_HALF_UP);

            System.out.println("f1 --> "+f1);
            String s1=roundfinalPrice.toPlainString();
            System.out.println("s1 "+s1);
        } catch (Exception e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
    }

}

输出将是

f1 --> 152.32
s1 152.32

这个方法不是很好，但是很简单，不建议使用。也许我应该说这是最无效的方法，也是最糟糕的编码实践，但是，使用起来很有趣，

float val=10.0;
String str=val+"";

空引号，将一个空字符串添加到变量str，将'val'上转换为字符串类型。

String str = "1234.56";
float num = 0.0f;

int digits = str.length()- str.indexOf('.') - 1;

float factor = 1f;

for(int i=0;i<digits;i++) factor /= 10;

for(int i=str.length()-1;i>=0;i--){

    if(str.charAt(i) == '.'){
        factor = 1;
        System.out.println("Reset, value="+num);
        continue;
    }

    num += (str.charAt(i) - '0') * factor;
    factor *= 10;
}

System.out.println(num);

如果你要找的是小数点后两位。 浮动f = (Float)12.34; 字符串s = new DecimalFormat("#.00")。格式(f);

有三种方法将float转换为String。

"" + f Float.toString (f) String.valueOf (f)

有两种方法将字符串转换为float

Float.valueOf（str） Float.parseFloat（str）;

例子:-

public class Test {

    public static void main(String[] args) {
        System.out.println("convert FloatToString " + convertFloatToString(34.0f));

        System.out.println("convert FloatToStr Using Float Method " + convertFloatToStrUsingFloatMethod(23.0f));

        System.out.println("convert FloatToStr Using String Method " + convertFloatToStrUsingFloatMethod(233.0f));

        float f = Float.valueOf("23.00");
    }

    public static String convertFloatToString(float f) {
        return "" + f;
    }

    public static String convertFloatToStrUsingFloatMethod(float f) {
        return Float.toString(f);
    }

    public static String convertFloatToStrUsingStringMethod(float f) {
        return String.valueOf(f);
    }

}

To go the full manual route: This method converts doubles to strings by shifting the number's decimal point around and using floor (to long) and modulus to extract the digits. Also, it uses counting by base division to figure out the place where the decimal point belongs. It can also "delete" higher parts of the number once it reaches the places after the decimal point, to avoid losing precision with ultra-large doubles. See commented code at the end. In my testing, it is never less precise than the Java float representations themselves, when they actually show these imprecise lower decimal places.

/**
 * Convert the given double to a full string representation, i.e. no scientific notation
 * and always twelve digits after the decimal point.
 * @param d The double to be converted
 * @return A full string representation
 */
public static String fullDoubleToString(final double d) {
    // treat 0 separately, it will cause problems on the below algorithm
    if (d == 0) {
        return "0.000000000000";
    }
    // find the number of digits above the decimal point
    double testD = Math.abs(d);
    int digitsBeforePoint = 0;
    while (testD >= 1) {
        // doesn't matter that this loses precision on the lower end
        testD /= 10d;
        ++digitsBeforePoint;
    }

    // create the decimal digits
    StringBuilder repr = new StringBuilder();
    // 10^ exponent to determine divisor and current decimal place
    int digitIndex = digitsBeforePoint;
    double dabs = Math.abs(d);
    while (digitIndex > 0) {
        // Recieves digit at current power of ten (= place in decimal number)
        long digit = (long)Math.floor(dabs / Math.pow(10, digitIndex-1)) % 10;
        repr.append(digit);
        --digitIndex;
    }

    // insert decimal point
    if (digitIndex == 0) {
        repr.append(".");
    }

    // remove any parts above the decimal point, they create accuracy problems
    long digit = 0;
    dabs -= (long)Math.floor(dabs);
    // Because of inaccuracy, move to entirely new system of computing digits after decimal place.
    while (digitIndex > -12) {
        // Shift decimal point one step to the right
        dabs *= 10d;
        final var oldDigit = digit;
        digit = (long)Math.floor(dabs) % 10;
        repr.append(digit);

        // This may avoid float inaccuracy at the very last decimal places.
        // However, in practice, inaccuracy is still as high as even Java itself reports.
        // dabs -= oldDigit * 10l;
        --digitIndex;
    }

    return repr.insert(0, d < 0 ? "-" : "").toString(); 
}

注意，虽然使用StringBuilder是为了提高速度，但是这个方法可以很容易地重写为使用数组，因此也适用于其他语言。