uniapp 解决计算时精度丢失问题

用法 let num = jian(a,b)

加法函数，用来得到精确的加法结果

jia(arg1, arg2) {
  var r1, r2, m;
  try { r1 = arg1.toString().split(".")[1].length } catch (e) { r1 = 0 }
  try { r2 = arg2.toString().split(".")[1].length } catch (e) { r2 = 0 }
  m = Math.pow(10, Math.max(r1, r2));
  return (arg1 * m + arg2 * m) / m;

}
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乘法函数，用来得到精确的乘法结果

cheng(arg1, arg2) {
  var m = 0, s1 = arg1.toString(), s2 = arg2.toString();
  try { m += s1.split(".")[1].length } catch (e) { }
  try { m += s2.split(".")[1].length } catch (e) { }
  return Number(s1.replace(".", "")) * Number(s2.replace(".", "")) / Math.pow(10, m);
}
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除法函数，用来得到精确的加法结果

chu(arg1, arg2) {
  var t1 = 0, t2 = 0, r1, r2;
  try { t1 = arg1.toString().split(".")[1].length } catch (e) { }
  try { t2 = arg2.toString().split(".")[1].length } catch (e) { }
  r1 = Number(arg1.toString().replace(".", ""))
  r2 = Number(arg2.toString().replace(".", ""))
  return (r1 / r2) * Math.pow(10, t2 - t1);

}
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减法函数，用来得到精确的加法结果

jian(arg2, arg1) {
  var r1, r2, m, n;
  try { r1 = arg1.toString().split(".")[1].length } catch (e) { r1 = 0 }
  try { r2 = arg2.toString().split(".")[1].length } catch (e) { r2 = 0 }
  m = Math.pow(10, Math.max(r1, r2));
  //lastmodifybydeeka  
  //动态控制精度长度  
  n = (r1 >= r2) ? r1 : r2;
  return ((arg2 * m - arg1 * m) / m).toFixed(n);
}
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