javascript更改函数参数变量名称并返回其值

Question

Hi i have a function and i want to return the function parameter with an added value to the parameter. So instead of writing this:

function(response) {
    return response.links_1;
}

function(response) {
    return response.links_2;
}

function(response) {
    return response.links_3;
}

I want to make a for loop that iterates through and adds the number, something like this:

function(response) {
    var counter = 3;
    for(var i = 0; i < counter; i++) {
        return response.links_ +i;
    }
}

The important part is that response.link_ must not be a string! Then it looses the function parameter value.

I tried with doing this:

function(response) {
    var i = 1,  
    resp = 'response.links_',
    endResp = resp + i;
    return endResp ;
    }
}

And console.log(endResp); returns the correct string, but thats just it, its a string.. i want the value of the variable response.links_1 not the string value response.links_1.

I have just tried the following without any luck: (the parse: is just a backbone method)

parse: function(response) {
    var counter = 3;
 for(var i = 0; i < counter; i++) {
   return response[links_ + i];
}
}

Any help is welcome..

Answer 1

您应该尝试response [“ links_” + 3]

Answer 2

To access the keys of a JSON hash, there are two ways

// For the following object
var obj = {
  links_1: 'value1',
  links_2: 'value1',
  links_3: 'value1',
};

// key1 can be accessed like
console.log(obj.links_1);

// Or
console.log(obj['links_1']);

So, in your case, you can use the second method

var counter = 3;
for(var i = 0; i < counter; i++) {
  return response['links_' + i];
}

Also, you can get the total count by

var counter = Object.keys(response).length;

And, if you don't need to do anything else with the counter variable, and the response object only contains the links, you can use something like

for (link in response) {
    return response[link];
}

Answer 3

The object dereference operator . doesn't work with dynamically generated property names, so you have to use the array dereference operator [] instead:

return response['links_' + i];