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houjunxiang
290
nx/helper/calcAnalysisValue.js
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290
nx/helper/calcAnalysisValue.js
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import { create, all, number, pow } from 'mathjs'
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const math = create(all)
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math.config({
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number: 'BigNumber',
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precision: 64
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})
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/*
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* 计算分析值*/
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export function calcAnalysisValue(group) {
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try {
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for (const g of group) {
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for (const ele of g.fields) {
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if (!ele.formula || ele.formula == '') continue
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if (ele.valueTypeManual && ele.valueTypeManual == '1') continue
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let formula = ele.formula
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let formulas = formula.split('|')
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let formulaVal = ''
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let hasNullVal = false
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formulas.forEach(f => {
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let value = ''
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if (f.charAt(0) === 'p') {
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// let o = elementParamValueData.find((i) => 'p' + i.paramNo === f);
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let o = findFieldInGroup(f, group, 'p')
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value = o.value
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} else if (f.charAt(0) === 'e') {
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// let o = elementParamValueData.find((i) => 'e' + i.paramNo === f);
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let o = findFieldInGroup(f, group, 'e')
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value = o.value
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} else if (f.charAt(0) === '<' || f.charAt(0) === '>') {
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value = "'" + f + "'"
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} else {
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value = f
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}
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if (typeof value == 'undefined' || value == null) {
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hasNullVal = true
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return true
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}
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formulaVal += value
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})
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if (hasNullVal) {
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ele.value = null
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continue
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}
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let v
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if (formulaVal.startsWith('Get')) {
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//计算公式为Get开头的，都是执行方法
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v = eval(formulaVal)
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} else {
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v = math.evaluate(formulaVal).toString()
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v = isFinite(v) ? v.toString() : ''
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}
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ele.value = handleRoundFiveNumber(v, ele.dataType)
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}
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}
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} catch (error) {
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console.log(error)
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}
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}
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const findFieldInGroup = function (paramNo, group, p) {
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for (const g of group) {
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for (const f of g.fields) {
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if (p + f.paramNo === paramNo) {
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return f
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}
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}
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}
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return { value: null }
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}
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const GetRecoveryRate = function (elementFormulaCode, value) {
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// //读取conRecoveryRateList
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const conRecoveryRateList = uni.getStorageSync('ConRecoveryRateList')
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if (!conRecoveryRateList) return
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let o = conRecoveryRateList.find(i => {
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//console.log("取得的" + i + "回收率：" + i);
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if (value === 0) {
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return i.elementFormulaCode === elementFormulaCode && i.minValue === value
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}
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return i.elementFormulaCode === elementFormulaCode && i.minValue < value && value <= i.maxValue
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})
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// console.log(o);
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const rate = o ? o.rate : 100
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// console.log("取得的" + elementFormulaCode + "回收率：" + rate);
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return rate
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// console.log("执行方法：GetRecoveryRate");
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// return 100;
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}
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/*
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* 根据硫值计算：支持碳、硝酸钾（滇中）
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* 当S%<22%时，需加入淀粉的量=(75-S%*22*M)/12
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* 当 S%>22%时，需加入硝酸钾的量=(S%*22*M-75)/4
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* 当 S%==22%时，硝酸钾为0，淀粉为空
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* sValueKey：硫值对应的dicKey
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* weightKey：重量对应的dicKey
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* percent：计算参数
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*
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* 举例：Get_C_KNO3_bySValue|(|p14|, |p2|, |>|)
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* */
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const Get_C_KNO3_bySValue = function (sValue, weight, operator) {
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//判断sValue是数字
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if (isNaN(sValue) || isNaN(weight)) {
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return ''
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}
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let v = number(sValue)
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const w = number(weight)
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if (w === 0) return ''
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v = v * 0.01
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//当S%<22%时，需加入淀粉的量=(75-S%*22*M)/12
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if (operator === '<' && v < 0.22) {
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// return ( 75 - v * 22 * w )/12;
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return accDiv(accSub(75, accMul(accMul(v, 22), w)), 12)
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}
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//S%>22%时，需加入硝酸钾的量=(S%*22*M-75)/4
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if (operator === '>=' && v >= 0.22) {
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//S%==22%时，硝酸钾为0，淀粉为空
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if (v === 22) return 0
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// return ( v * 22 * w - 75 ) + 44;
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return accDiv(accSub(accMul(v, accMul(22, w)), 75), 4)
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}
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return ''
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}
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/** 处理数值数据：四舍六入奇进偶不进
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* 1，如果取小数的最后一位为5，5前为奇数进位，为偶不进,五后非零就进一,五后皆零看奇偶,五前为偶应舍去,五前为奇要进一
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* 2.5后不为0时就入，为0时看5前，奇进偶不进
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* 3.四舍，六入
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* 传入数值和保留位数
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*/
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export function handleRoundFiveNumber(number, fixed = 0) {
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if (number == null || number === '' || isNaN(number)) return number
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if (fixed === -1) return number
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number = String(number)
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//可以考虑清掉末尾的0，暂时限制了不会有,因为传入数值，末尾0是去掉的
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const index = number.indexOf('.')
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if (index == -1) {
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if (fixed > 0) {
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number += '.'
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}
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for (let i = 0; i < fixed; i++) {
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number += '0'
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}
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//补0返回
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return number
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}
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//取到保留小数位的下一位，5.5555取保留俩位，那就是小数点后第三位
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const indexFixed = index + fixed + 1
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if (indexFixed >= number.length) {
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//如果小数位数不够, 补0直接返回
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const zerolen = indexFixed - number.length
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for (let i = 0; i < zerolen; i++) {
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number += '0'
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}
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return number
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}
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//取保留位数的后一位做判断，以下是有5的判断
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const endNumber = number.substr(indexFixed, 1)
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if (endNumber != '5') {
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//如果做判断的数不是五，就按正常的四舍五入，即忽略了5，后面补0的那些由于是数字传进来，0已经去掉，此处不做处理
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return Number(number).toFixed(fixed)
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}
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if (indexFixed != number.length - 1) {
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//由于当前判断位不是最后一位，而又去除了0，那么后面后的位数应该直接入位，五后不为0时入,由于会有小于五的，四舍五入肯定不行，那么只能截取到当前保留位数，然后转成数字加上10的负fixed的次方即可
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number = number.substring(0, indexFixed)
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if (number.indexOf('-') != -1) {
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//需要考虑到负数的情况
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return (Number(number) - Number(Math.pow(10, -fixed))).toFixed(fixed)
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}
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return (Number(number) + Number(Math.pow(10, -fixed))).toFixed(fixed)
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//return this.accAdd(number,Math.pow(10,-fixed))
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}
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//接下来就时五后没有值也就是0的需要看前面的奇入偶不入了，取当前位的上一位
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let twoNumber = number.substr(indexFixed - 1, 1)
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if (twoNumber == '.') {
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//取到小数点，再取一次，要取到小数点左边第一位
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twoNumber = number.substr(indexFixed - 2, 1)
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}
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if ('13579'.indexOf(twoNumber) != -1) {
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//奇进
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number = number.substring(0, indexFixed)
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if (number.indexOf('-') != -1) {
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//需要考虑到负数的情况
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return (Number(number) - Math.pow(10, -fixed)).toFixed(fixed)
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}
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return (Number(number) + Math.pow(10, -fixed)).toFixed(fixed)
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}
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//偶不进,将取值的当前位数,直接截取字符即可
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return number.substr(0, indexFixed)
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}
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/**
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** 除法函数，用来得到精确的除法结果
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** 说明：javascript的除法结果会有误差，在两个浮点数相除的时候会比较明显。这个函数返回较为精确的除法结果。
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** 调用：accDiv(arg1,arg2)
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** 返回值：arg1除以arg2的精确结果
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**/
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function accDiv(arg1, arg2) {
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var t1 = 0,
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t2 = 0,
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r1,
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r2
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try {
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t1 = arg1.toString().split('.')[1].length
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} catch (e) {}
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try {
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t2 = arg2.toString().split('.')[1].length
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} catch (e) {}
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r1 = Number(arg1.toString().replace('.', ''))
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r2 = Number(arg2.toString().replace('.', ''))
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return (r1 / r2) * pow(10, t2 - t1)
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}
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/**
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** 乘法函数，用来得到精确的乘法结果
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** 说明：javascript的乘法结果会有误差，在两个浮点数相乘的时候会比较明显。这个函数返回较为精确的乘法结果。
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** 调用：accMul(arg1,arg2)
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** 返回值：arg1乘以 arg2的精确结果
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**/
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function accMul(arg1, arg2) {
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var m = 0,
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s1 = arg1.toString(),
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s2 = arg2.toString()
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try {
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m += s1.split('.')[1].length
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} catch (e) {}
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try {
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m += s2.split('.')[1].length
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} catch (e) {}
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return (Number(s1.replace('.', '')) * Number(s2.replace('.', ''))) / Math.pow(10, m)
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}
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/**
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** 减法函数，用来得到精确的减法结果
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** 说明：javascript的减法结果会有误差，在两个浮点数相减的时候会比较明显。这个函数返回较为精确的减法结果。
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** 调用：accSub(arg1,arg2)
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** 返回值：arg1加上arg2的精确结果
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**/
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function accSub(arg1, arg2) {
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var r1, r2, m, n
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try {
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r1 = arg1.toString().split('.')[1].length
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} catch (e) {
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r1 = 0
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}
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try {
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r2 = arg2.toString().split('.')[1].length
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} catch (e) {
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r2 = 0
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}
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m = Math.pow(10, Math.max(r1, r2)) //last modify by deeka //动态控制精度长度
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n = r1 >= r2 ? r1 : r2
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return ((arg1 * m - arg2 * m) / m).toFixed(n)
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}
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/**
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** 加法函数，用来得到精确的加法结果
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** 说明：javascript的加法结果会有误差，在两个浮点数相加的时候会比较明显。这个函数返回较为精确的加法结果。
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** 调用：accAdd(arg1,arg2)
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** 返回值：arg1加上arg2的精确结果
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**/
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function accAdd(arg1, arg2) {
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var r1, r2, m, c
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try {
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r1 = arg1.toString().split('.')[1].length
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} catch (e) {
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r1 = 0
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}
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try {
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r2 = arg2.toString().split('.')[1].length
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} catch (e) {
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r2 = 0
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}
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c = Math.abs(r1 - r2)
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m = Math.pow(10, Math.max(r1, r2))
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if (c > 0) {
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var cm = Math.pow(10, c)
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if (r1 > r2) {
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arg1 = Number(arg1.toString().replace('.', ''))
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arg2 = Number(arg2.toString().replace('.', '')) * cm
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} else {
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arg1 = Number(arg1.toString().replace('.', '')) * cm
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arg2 = Number(arg2.toString().replace('.', ''))
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}
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} else {
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arg1 = Number(arg1.toString().replace('.', ''))
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arg2 = Number(arg2.toString().replace('.', ''))
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}
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return (arg1 + arg2) / m
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}
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