Question

我的阵列就是这样:

[["1", "apple 1"], ["1", "pear 2"], ["2", "lemon 1"], ["12", "lemon 12"]]

我愿分别以<代码>one”代替1、2和12个阵列指数0和one>、、2“和>“twelve”。



[["one", "apple 1"], ["one", "pear 2"], ["two", "lemon 1"], ["twelve", "lemon 12"]]


我如何这样做?

Answer 1


var arr = [
    [  1 ,  apple 1  ], [  2 ,  pear 2  ], [  3 ,  lemon 3  ]
]

var numbers = [  one ,  two ,  three  ]

arr.forEach(function (el) {
    el[0] = numbers[el[0]]
})

arr // [ [  one ,  apple 1  ], [  two ,  pear 2  ], [  three ,  lemon 3  ] ]


射线指数实际上是插图,为什么 编号[2]( 2_ a string)将检索第三名成员。

在阵列上,可使用for-loop,但
                
                    时间：2012-05-03 06:32:53
                
                
            
   
      

   
   	问题回答
   
      
   	
例如:

var obj = [["1", "apple 1"], ["1", "pear 2"], ["2", "lemon 1"], ["12", "lemon 12"]];

for ( i = 0; i < obj.length ; i++ ) {
    switch(obj[i][0]) {
        case "1" : obj[i][0] = "one"; break;
        case "2" : obj[i][0] = "two"; break;
        case "12" : obj[i][0] = "twelve"; break;
    }
}


http://jsfiddle.net
    
       
            
                
                    时间：2012-05-03 06:32:12


<><>Edit>>

我想提一下的是,这一解决办法要求你把阵列起来:。 Arr = JSON.stringification(theArr).replace(.)。页: 1

我可以最快的是:

var str =  [["1", "apple 1"], ["1", "pear 2"], ["2", "lemon 1"], ["12", "lemon 12"]] .replace(/([")12/g, $1twelve ).replace(/([")2/g, $1two ).replace(/([")1/g, $one );

But there are a couple of catches here: if write the replace(/([")12/g, $1twelve ) after you ve replaced the ones, it won t work. so to get around this, you could do:

str =  [["1", "apple 1"], ["1", "pear 2"], ["2", "lemon 1"], ["12", "lemon 12"]] .replace(/([")12"/g, $1twelve" ).replace(/([")2"/g, $1two" ).replace(/([")1"/g, $one" );

添加最后引文。然而,更可取的解决办法是:

var replace = { 1 : one , 2 : two , 12 : twelve };

并将相应的数字作为财产名称。

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Answer 2

例如:

var obj = [["1", "apple 1"], ["1", "pear 2"], ["2", "lemon 1"], ["12", "lemon 12"]];

for ( i = 0; i < obj.length ; i++ ) {
    switch(obj[i][0]) {
        case "1" : obj[i][0] = "one"; break;
        case "2" : obj[i][0] = "two"; break;
        case "12" : obj[i][0] = "twelve"; break;
    }
}

http://jsfiddle.net

Answer 3


<><>Edit>>

我想提一下的是,这一解决办法要求你把阵列起来:。 Arr = JSON.stringification(theArr).replace(.)。页: 1

我可以最快的是:

var str =  [["1", "apple 1"], ["1", "pear 2"], ["2", "lemon 1"], ["12", "lemon 12"]] .replace(/([")12/g, $1twelve ).replace(/([")2/g, $1two ).replace(/([")1/g, $one );

But there are a couple of catches here: if write the replace(/([")12/g, $1twelve ) after you ve replaced the ones, it won t work. so to get around this, you could do:

str =  [["1", "apple 1"], ["1", "pear 2"], ["2", "lemon 1"], ["12", "lemon 12"]] .replace(/([")12"/g, $1twelve" ).replace(/([")2"/g, $1two" ).replace(/([")1"/g, $one" );

添加最后引文。然而,更可取的解决办法是:

var replace = { 1 : one , 2 : two , 12 : twelve };

并将相应的数字作为财产名称。

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