将数组 2 和 3 访问到一个字符串数组中

Question

如何输出：

ID: 0001
Name: Mike
Birthday: London 21/05/1989
Hobby: Reading

我下面的代码是未定义的，我希望数组城市+日期在生日时在一起。

我的代码不是，请检查下面的代码：

var input = [
                ["0001", "Mike", "London", "21/05/1989", "Reading"],
                ["0002", "Sara", "Manchester", "10/10/1992", "Swimming"],
                ["0003", "John", "Kansas", "25/12/1965", "Cooking"],
                ["0004", "Dave", "Nevada", "6/4/1970", "going to gym"]
            ];

var data = ["ID: ", "Name: ", "Birthday: ", "Hobby: "];




for(var i = 0 ; i <= input.length ; i++){
  for(var j = 0  ; j <= input.length ; j++){
  for(var i = 0 ; i <= data.length; i++){
  console.log(data[i] + input[j][i])
    };
  };
};

有什么建议可以解决这个逻辑吗？为此，我只想使用循环。

Answer 1

您可以使用array.map

var input = [
                ["0001", "Mike", "London", "21/05/1989", "Reading"],
                ["0002", "Sara", "Manchester", "10/10/1992", "Swimming"],
                ["0003", "John", "Kansas", "25/12/1965", "Cooking"],
                ["0004", "Dave", "Nevada", "6/4/1970", "going to gym"]
            ];
            
var expectedOutput = input.map(a=>{
 return {ID:a[0],Name:a[1],Birthday:a[2] + ' ' + a[3],Hobby:a[4]}
})

console.log('string output',JSON.stringify(expectedOutput));
console.log(expectedOutput);

Answer 2

试试这个

//contoh input
var input = [
                ["0001", "Mike", "London", "21/05/1989", "Reading"],
                ["0002", "Sara", "Manchester", "10/10/1992", "Swimming"],
                ["0003", "John", "Kansas", "25/12/1965", "Cooking"],
                ["0004", "Dave", "Nevada", "6/4/1970", "going to gym"]
            ];

var data = ["ID: ", "Name: ", "Birthday: ", "Hobby: "];

// for(var i = 0 ; i < data.length ; i++){
//   console.log(data[i]);

var k = 0;
for(var i = 0 ; i < input.length ; i++){
   for(var j = 0; j <= data.length ; j++){
       if(j == 2 ){
          console.log(data[k] + input[i][j]+ " " + input[i][j+1]);
          j++;
       }
       else
          console.log(data[k] + input[i][j]);
       k++;
   }k=0;
}

Answer 3

let output = input.map( item => {
return ({ID:item[0],name:item[1],birthDay:item[2]+item[3],hobby:item[4]})
})

希望对你有帮助

Answer 4

由于数组中的索引是基于 0 并且 i=0 你必须改变

i <= input.length

到

i < input.length

//contoh input
var input = [
    ["0001", "Mike", "London", "21/05/1989", "Reading"],
    ["0002", "Sara", "Manchester", "10/10/1992", "Swimming"],
    ["0003", "John", "Kansas", "25/12/1965", "Cooking"],
    ["0004", "Dave", "Nevada", "6/4/1970", "going to gym"]
];

var data = ["ID: ", "Name: ", "Birthday: ", "Hobby: "];

for(var i = 0 ; i < input.length ; i++){
  for(var j = 0  ; j < input.length ; j++){
    for(var i = 0 ; i < data.length; i++){
      if(i == 2)
        console.log(data[i] + input[j][i] +' '+ input[j][i+1])
      else if(i == 3)
        console.log(data[i] + input[j][i+1])
      else
        console.log(data[i] + input[j][i])
    };
    console.log('=================')
  };
};

Answer 5

var input = [
                ["0001", "Mike", "London", "21/05/1989", "Reading"],
                ["0002", "Sara", "Manchester", "10/10/1992", "Swimming"],
                ["0003", "John", "Kansas", "25/12/1965", "Cooking"],
                ["0004", "Dave", "Nevada", "6/4/1970", "going to gym"]
            ];
            
let output= input.map(([ID, Name, Country, DOB, Hobby]) =>{
    return({
      ID, 
      Name, 
      Birthday: `${Country} ${DOB}`, 
      Hobby
    })
})

console.log(output)

Answer 6

这应该可行。只需在打印前检查必要的条件。 并检查是否打印超出范围的索引所指向的数组值。这就是它显示未定义的原因。

var input = [
  ["0001", "Mike", "London", "21/05/1989", "Reading"],
  ["0002", "Sara", "Manchester", "10/10/1992", "Swimming"],
  ["0003", "John", "Kansas", "25/12/1965", "Cooking"],
  ["0004", "Dave", "Nevada", "6/4/1970", "going to gym"]
];

var data = ["ID: ", "Name: ", "Birthday: ", "Hobby: "];

for (var j = 0; j < input.length; j++) {
  for (var i = 0; i < input[j].length; i++) {
    if (i === 2)
      console.log(data[i] + input[j][i] + " " + input[j][i + 1]);
    else if (i === 3)
      console.log(data[i] + input[j][i+1]);
  };
};

Answer 7

你可以先合并input数组中的城市+日期，然后循环会更容易

var expectedOutput = input.map(a=>{
 return [a[0], a[1], a[2]+' '+a[3], a[4]]
})

Answer 8

使用原生 ForLoop 并在 one 循环中您可以这样做：

//contoh input
var input = [
                ["0001", "Mike", "London", "21/05/1989", "Reading"],
                ["0002", "Sara", "Manchester", "10/10/1992", "Swimming"],
                ["0003", "John", "Kansas", "25/12/1965", "Cooking"],
                ["0004", "Dave", "Nevada", "6/4/1970", "going to gym"]
            ];

var data = ["ID: ", "Name: ", "Birthday: ", "Hobby: "];

for(var i = 0 ; input[i] && input[i].length ? i < input[i].length : null ; i++) {
  console.log(data[0] + input[i][0]+ ',', data[1] + input[i][1]+ ',', data[2]+input[i][2] + ' '+ input[i][3]+ ',', data[3]+ input[i][4]);
};

PS： 使用 ES6 运算符如 map 和 forEach 更优雅..

Answer 9

尝试在第三个“for”语句中更改变量名。我没有对循环进行深入研究，但我相信您不想在第三个循环中使用变量“i”。尝试将其命名为“var k”

Answer 10

//contoh input
var input = [
                ["0001", "Mike", "London", "21/05/1989", "Reading"],
                ["0002", "Sara", "Manchester", "10/10/1992", "Swimming"],
                ["0003", "John", "Kansas", "25/12/1965", "Cooking"],
                ["0004", "Dave", "Nevada", "6/4/1970", "going to gym"]
            ];

var data = ["ID: ", "Name: ", "Birthday: ", "Hobby: "];

// for(var i = 0 ; i < data.length ; i++){
//   console.log(data[i]);


  var check = 0;
  for(var j = 0  ; j < input.length ; j++){
  var count = 0;
  outerloop:
  
  for(var i = 0 ; i < data.length; i++){
  if(count==2){
    console.log(data[i] + input[j][i] +' '+ input[j][i+1]);
  count = 0;
  check = 1;
    continue outerloop;
  } if(check==1){
    count++;
     console.log(data[i] + input[j][i+1]);
    check = 0; 

}
else{
    count++;
     console.log(data[i] + input[j][i]);
  }
    };
  };

简介： 我使用了一个 if condition 和 1 个 ifelse condition，并为这两个条件使用了 2 个变量，第一个变量仅在条件是计数变量时检查索引是否为生日，如果是，则连接当前值和它的相邻值，然后我使用了一个检查变量，该变量在 if 条件下为真，我们将继续循环而不深入该迭代。 check 变量仅用于识别您在 check 为 0 时实现的正常迭代，如果为 1，则将显示相邻索引。希望对你有帮助。