科学技术法与数值之前的转换

一、JAVA

1、科学技术法转数值

方法1

BigDecimal za = zaVal1.divide(val2);// 9.05E-8

String za2 = za.toPlainString();//0.0000000905

方法2

String val2 = "9.05E-8";
BigDecimal val0 = new BigDecimal(a);//0.0000000905
String val2 = BigDecimal.valueOf(new Double(String.valueOf(val0))).toString();

其他问题:精度缺失

如果a和b为double类型，很可能出现精度丢失的情况：0.45699999999999999999
将a和b转为string类型，就不会出现精度丢失。

BigDecimal aa = new BigDecimal(Double.toString(a));
BigDecimal bb = new BigDecimal(Double.toString(b));
BigDecimal cc = aa.divide(bb);

2、数值转科学技术法

String za2 = "0.0000000905";

BigDecimal za4 = new BigDecimal(za2);// 9.05E-8     

String za5 = za.toString();//9.05E-8

上面的要是数据短一点E-6不会自动转过来，上面的不要用，用需要下面

String za2 = "0.000009056666666";

BigDecimal za4 = new BigDecimal(za2);// 0.0000090566666

String val = String.format("%E", val0);//9.05666666E-6     

//还需要保留keep位小数

String aa = val.substring(0,val.indexOf(".")+4);
BigDecimal bg = new BigDecimal(aa);
double cc = bg.setScale(keep, BigDecimal.ROUND_HALF_UP).doubleValue();
String bb = val.substring(val.indexOf("E"), val.length());
val = cc + bb;

二、JS

1、科学技术法转数值

function  convertNUM(num_str) {
   //科学计数法字符 转换 为数字字符， 突破正数21位和负数7位的Number自动转换
   // 兼容 小数点左边有多位数的情况，即 a×10^b（aEb），a非标准范围（1≤|a|<10）下的情况。如 3453.54E-6 or 3453.54E6
   var resValue = '',
      power = '',
      result = null,
      dotIndex = 0,
      resArr = [],
      sym = '';
   var numStr = num_str.toString();
   if (numStr[0] == '-') {  // 如果为负数，转成正数处理，先去掉‘-’号，并保存‘-’.
      numStr = numStr.substr(1);
      sym = '-';
   }
   myConsoleLog(numStr);
   if ((numStr.indexOf('E') != -1) || (numStr.indexOf('e') != -1)) {
      var regExp = new RegExp('^(((\\d+.?\\d+)|(\\d+))[Ee]{1}(([+-](\\d+))|(\\d+)))$', 'ig');
      result = regExp.exec(numStr);
      myConsoleLog(result);
      if (result != null) {
         resValue = result[2];
         power = result[5];
         result = null;
      }
      if (!resValue && !power) {
         return false
      }
      dotIndex = resValue.indexOf('.');
      resValue = resValue.replace('.', '');
      resArr = resValue.split('');
      if (Number(power) >= 0) {
         var subres = resValue.substr(dotIndex);
         power = Number(power);
         //幂数大于小数点后面的数字位数时，后面加0
         for (var i = 0; i < power - subres.length; i++) {
            resArr.push('0');
         }
         if (power - subres.length < 0) {
            resArr.splice(dotIndex + power, 0, '.');
         }
      } else {
         power = power.replace('-', '');
         power = Number(power);
         //幂数大于等于 小数点的index位置, 前面加0
         for (var i = 0; i <= power - dotIndex; i++) {
            resArr.unshift('0');
         }
         var n = power - dotIndex >= 0 ? 1 : -(power - dotIndex);
         resArr.splice(n, 0, '.');
      }
   }
   resValue = resArr.join('');
   myConsoleLog(sym + resValue);
   return sym + resValue;
}

2、数值转科学技术法

var dbValue = 9.05E-8;

var val = math2science(dbValue);//0.0000000905

math2science:function(math){
    var  result = toExponential(math,2);
    result = result.replace(/e/,"E");
    if (result.length == 7)
    {
        result = result.substr(0,6) + "0" + result.substr(6);
    }
    return result;
},

toExponential:function(_num,d){
    if(!this.IsNum(_num)) return _num;
    var _tmpNum = Math.abs(Number(_num));
    var _tmpS = _tmpNum.toExponential();
    var _tmpSNum = this.toFixed(Number(_tmpS.substr(0,_tmpS.indexOf('e'))),d);
    var _tmpSEStr =_tmpS.substr(_tmpS.indexOf('e'));
    var _resultValueABS = _tmpSNum + _tmpSEStr;
    return (_num > 0 )?_resultValueABS:"-" + _resultValueABS;
}