為什么我的程序提示未定義的值

Question

var cars = ["nano", "bmw"];
var bikes = ["pulsar", "splander"];
var con = [];
var foundCars = [];
var foundBikes = [];
var notFoundTransport =[];
for(var i = 0; i<cars.length; i++){
    con = prompt("enter your tranport name");
    if(con[i] == cars[i]){
    foundCars.push(con[i]);
    } 
}
for(var i = 0; i<2; i++){
alert(foundCars[i]);
}

為什么在警告foundcars數組后顯示未定義的值，例如nano bmw，它是警報，未定義，未定義的請讓我知道我錯了。

Answer 1

首先，您需要將提示結果添加到con數組中的適當位置。

然后，您應該考慮到foundCars的長度（而不是硬編碼的2）來遍歷數組。

var cars = ["nano", "bmw"];
var bikes = ["pulsar", "splander"];
var con = [];
var foundCars = [];
var foundBikes = [];
var notFoundTransport =[];
for(var i = 0; i<cars.length; i++){
    con[i] = prompt("enter your tranport name");
    if(con[i] == cars[i]){
        foundCars.push(con[i]);
    } 
}
for(var i = 0; i<foundCars.length; i++){
    alert(foundCars[i]);
}

Answer 2

jsfiddle

因為foundCars為空

您應該將這樣的車用於發現的汽車。

for(var i = 0; i<cars.length; i++){
    con = prompt("enter your tranport name");
    for(var p = 0; p<cars.length; p++){
       if(con == cars[p]){
         foundCars.push(con);
       } 
    }
}

並使用for(var i = 0; i<foundCars.length; i++) for(var i = 0; i<2; i++)

Answer 3

請檢查：

var cars = ["nano", "bmw"];
var bikes = ["pulsar","splander"];
var con = [];
var foundCars = [];
var foundBikes = [];
var notFoundTransport =[];
var found = false;
var checkExist = false;


for(var i = 0; i<2; i++){
    con[i] = prompt("enter your tranport name");
}

for(var i = 0; i<con.length; i++){
    for(var j = 0; j<con.length; j++){
        if(con[i] == cars[j]){
            foundCars.push(con[i]);
            found = true;
            checkExist = true;
        }else if(con[i] == bikes[j]){
            foundBikes.push(con[i]);
            found = true;
            checkExist = true;
        }else if(con[i]!=cars[j]&&con[i]!=bikes[j]&&found==false&&j!=0){
                notFoundTransport.push(con[i]);
                found = true;
        }
    }
        found = false;
}

console.log(foundCars);
console.log(foundBikes);
console.log(notFoundTransport);