我有一个javascript函数返回一个值。但是,我无法让这个值返回到函数之外。我已经研究过并且无法弄清楚这个问题。它说最后一行代码中的grossResults是未定义的。回报是否会使其无法访问?Javascript函数返回值无法访问
"use script";
var hourlyRate = prompt("Enter hourly pay rate:", hourlyRate);
var totalHrsWorked = parseInt(prompt("Enter number of hours worked:", totalHrsWorked));
//function to calculate gross wages for an hourly employee
function grossWages (hourlyRate, totalHrsWorked){
var OTHours = totalHrsWorked - 40;
var regPay = (totalHrsWorked - OTHours) * hourlyRate;
var OTPay = (OTHours * (hourlyRate * 1.5)) + regPay;
grossResults = (OTHours > 0) ? OTPay : regPay;
return grossResults;
};
document.writeln("<br /><br />");
document.writeln("Pay rate entered: " + hourlyRate + "<br />");
document.writeln("Hours entered: " + totalHrsWorked + "<br />");
document.writeln("Gross Pay: $" + grossResults + "<br /><br />")
你在哪里调用'grossWages'您只需定义的函数,你没有把它的代码函数内部从未执行过,我推荐阅读[** JavaScript教程**](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide)b在继续进行之前,特别是[关于函数](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Functions)。 – 2014-10-02 20:19:43
我相信最后一行的意思是:'document.writeln(“Gross Pay:$”+ grossWages(hourlyRate,totalHrsWorked)+“
”)' – LcSalazar 2014-10-02 20:26:11