Sort an object array by custom order

问题:

I have an array of objects which have a property called 'CODE'.

[
  {
   ID: 168,
   NAME: "First name",
   CODE: "AD"
  },
  {
   ID: 167,
   NAME: "Second name",
   CODE: "CC"
  },
  {
   ID: 169,
   NAME: "Third name",
   CODE: "CCM"
  },
  {
   ID: 170,
   NAME: "Fourth name",
   CODE: "CR"
  },
]

How do I order the array by a customized order like:

var item_order = ["CCM","CR","AD","CC"];

Been trying various methods with no success. Please help.

回答1:

You can use the function sort along with the function indexOf.

var array = [  {   ID: 168,   NAME: "First name",   CODE: "AD"  },  {   ID: 167,   NAME: "Second name",   CODE: "CC"  },  {   ID: 169,   NAME: "Third name",   CODE: "CCM"  },  {   ID: 170,   NAME: "Fourth name",   CODE: "CR"  }],
    item_order = ["CCM","CR","AD","CC"];

array.sort((a, b) => item_order.indexOf(a.CODE) - item_order.indexOf(b.CODE));

console.log(array);

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回答2:

For huge arrays, I suggest to use an object for the indices.

var array = [{ ID: 168, NAME: "First name", CODE: "AD" }, { ID: 167, NAME: "Second name", CODE: "CC" }, { ID: 169, NAME: "Third name", CODE: "CCM" }, { ID: 170, NAME: "Fourth name", CODE: "CR" }],
    item_order = ["CCM", "CR", "AD", "CC"],
    order = item_order.reduce((r, k, v) => Object.assign(r, { [k]: v }), {});

array.sort((a, b) => order[a.CODE] - order[b.CODE]);

console.log(array);

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回答3:

You're going to use array.sort(customSort), where:

function customSort(a,b)
{
    a = item_order.indexOf(a.CODE);
    b = item_order.indexOf(b.CODE);

    return a - b;
}

回答4:

var array = [
  {
   ID: 168,
   NAME: "First name",
   CODE: "AD"
  },
  {
   ID: 167,
   NAME: "Second name",
   CODE: "CC"
  },
  {
   ID: 169,
   NAME: "Third name",
   CODE: "CCM"
  },
  {
   ID: 170,
   NAME: "Fourth name",
   CODE: "CR"
  },
];

var sortOrder =  ["CCM","CR","AD","CC"];

var sorted = array.sort((a, b) => sortOrder.indexOf(a.CODE) - sortOrder.indexOf(a.CODE));

console.log(sorted);

回答5:

If you have to do something like this often, you might write a small utility to help:

const array = [{ ID: 168, NAME: "First name", CODE: "AD" }, { ID: 167, NAME: "Second name", CODE: "CC" }, { ID: 169, NAME: "Third name", CODE: "CCM" }, { ID: 170, NAME: "Fourth name", CODE: "CR" },{ ID: 166, NAME: "Fifth name", CODE: "CCM" }, { ID: 171, NAME: "Sixth name", CODE: "XXX" }, { ID: 172, NAME: "Seventh name", CODE: "CR" }]

const sortOn = (prop, list) => {
  const order = list.reduce((obj, key, idx) => Object.assign(obj, { [key]: idx + 1}), {});
  const getVal = item => order[item[prop]] || Infinity
  
  return (a, b) => getVal(a) - getVal(b)
}

array.sort(sortOn('CODE', ["CCM", "CR", "AD", "CC"]))
console.log(array)

The order object is much like what Nina Scholz suggested. The reason for idx + 1 rather than just idx is to simplify the next line. That line uses Infinity as a way to sort to the end those whose key value is either undefined or not in the sort list. If you want them at the beginning, you can use 0 or -Infinity.