为node.js中的变量赋值

Question

我是node.js的初学者，在评估evalMath（）函数后我遇到了问题。特别是，将 Math2的值指定为yScale 。

Math2是网页的输入框，引入了一个简单的等式，如“Ir = Vr / R＆＃39;”。此字符串值从网页传递到node.js上实现的Web服务器。在那里，执行函数evalMath（），它将Vr（带字符串）和R替换为相应的整数值。执行此功能后，获得类似这样的＆＃39; ANI0 [i] -ANI5 [i] / 10＆＃39;，它存储在变量Math2中。该功能完美无缺。但是，当Math2分配给yScale时，只保存＆＃39; Ir = Vr / R＆＃39;而不是执行evalMath（）后获得的字符串 - ＆gt; ＆＃39; ANI0 [I] -ANI5 [I] / 10＆＃39 ;.我不确切知道为什么。

 if (yScale == 'Math2') {
    console.log('Math2 selected');
    evalMath(textAD0,textAD1,textAD2,textAD3,textAD4,Math2,Rb,Rc,R,yScale);
    yScale = Math2;
    console.log ("yScale: "+ yScale);
} else {
    console.log("Nothing to change :) - No eqn");
}

function evalMath(labelAD0, labelAD1, labelAD2, labelAD3,
                labelAD4, entry, entry1, entry2, entry3, entry4) {
    iAD0="",iAD1="",iAD2="",iAD3="",iAD4="",iR="",iRb="", iRc="", iEq="";

    var iAD0 = entry.search(labelAD0);
    var iAD1 = entry.search(labelAD1);
    var iAD2 = entry.search(labelAD2);
    var iAD3 = entry.search(labelAD3);
    var iAD4 = entry.search(labelAD4);
    var iR = entry.search('R');
    var iRb = entry.search('Rb');
    var iRc = entry.search('Rc');
    var iEq = entry.search('=');    

    if (iAD0 > "0"){ entry = entry.replace(labelAD0,'(ANI0[i]-ANI5[i])'); }
    else{console.log("Nothing to change AD0")}
    if (iAD1 > "0"){ entry = entry.replace(labelAD1,'(ANI1[i]-ANI5[i])'); }
    else{console.log("Nothing to change AD1")}
    if (iAD2 > "0"){ entry = entry.replace(labelAD2,'(ANI2[i]-ANI5[i])'); }
    else{console.log("Nothing to change AD2")}
    if (iAD3 > "0"){ entry = entry.replace(labelAD3,'(ANI3[i]-ANI5[i])'); }
    else{console.log("Nothing to change AD3")}
    if (iAD4 > "0"){ entry = entry.replace(labelAD4,'(ANI4[i]-ANI5[i])'); }
    else{console.log("Nothing to change AD4")}
    if (iRb > "0"){ entry = entry.replace('Rb',entry1); }
    else{console.log("Nothing to change Rb")}
    if (iRc > "0"){ entry = entry.replace('Rc',entry2); }
    else{console.log("Nothing to change Rc")}
    if (iR > "0"){ entry = entry.replace('R',entry3); }
    else{console.log("Nothing to change R")}
    if (iEq > "0"){ entry = entry.slice(iEq+1,entry.length); }
    else{console.log("Nothing to change iEq")}

    console.log("entry: " + entry);
    return entry;
}

Answer 1

将参数传递给函数时，只传递值而不传递变量的引用（对于string，int，float等）。如果您希望能够修改参数的值，请将其传递给对象：

var a = "hello";
var b = "moto"; 

var obj = {a: "hello", b: "moto"};

function c(p1, p2){ p1 = "yala"; p2="yele";}

c(a, b); // a == hello, b == moto, values not changed

function c2(p1) { p1.a = "yala"; p1.b = "yele";}

c2(obj); // obj.a == "yala", obj.b == "yele"

因此，在您的情况下，请执行以下操作：

var entries = {
 Math2: Math2,
 Rb: Rb,
 Rc: Rc,
 R:  R
 yScale: yScale
};


evalMath(textAD0,textAD1,textAD2,textAD3,textAD4,entries);
entries.yScale = entries.Math2;
console.log ("yScale: "+ entries.yScale);

然后用

替换evalMath的原型

   function evalMath(labelAD0, labelAD1, labelAD2, labelAD3,
            labelAD4, entries) {

然后在evalMath中，用entry.Math2替换entry1的所有出现，用entry.Rb替换entry2的所有出现，具有entries.Rc的entry3的所有出现，具有entries.R的entry4的所有出现以及具有entry5的所有出现entries.yScale

警告如果影响条目的某些值。它将被修改为函数