Sorting array after firstName and lastName, javascript - Stack Overflow

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Currently I am sorting an array on firstName.

currentUsers = currentUsers.sort(function(a, b) {
    if (a.firstName > b.firstName) {
        return 1;
    } else if (a.firstName < b.firstName) {
        return -1;
    }
};

But I would like to have the lastName sorted as well. So that Bob Anderson would be sorted above Bob Bobson. I looked at another question here on stackoverflow which suggested that I should add:

currentUsers = currentUsers.sort(function(a, b) {
    if (a.firstName > b.firstName) {
        return 1;
    } else if (a.firstName < b.firstName) {
        return -1;
    }

    if (a.lastName > b.lastName) {
        return 1;
    } else if (a.lastName < b.lastName) {
        return -1;
    } else {
        return 0;
    }
};

//Edit The problem I have is that the sorting keeps sorting on every letter in firstName. How would I only sort on the first letter of firstName and then move on to the lastName?

Currently I am sorting an array on firstName.

currentUsers = currentUsers.sort(function(a, b) {
    if (a.firstName > b.firstName) {
        return 1;
    } else if (a.firstName < b.firstName) {
        return -1;
    }
};

But I would like to have the lastName sorted as well. So that Bob Anderson would be sorted above Bob Bobson. I looked at another question here on stackoverflow which suggested that I should add:

currentUsers = currentUsers.sort(function(a, b) {
    if (a.firstName > b.firstName) {
        return 1;
    } else if (a.firstName < b.firstName) {
        return -1;
    }

    if (a.lastName > b.lastName) {
        return 1;
    } else if (a.lastName < b.lastName) {
        return -1;
    } else {
        return 0;
    }
};

//Edit The problem I have is that the sorting keeps sorting on every letter in firstName. How would I only sort on the first letter of firstName and then move on to the lastName?

• javascript
• arrays
• sorting

Share Improve this question Follow edited Mar 24, 2021 at 7:19 Reality-Torrent asked Oct 9, 2015 at 6:36 Reality-TorrentReality-Torrent 35833 silver badges1515 bronze badges 4

• 1 Your first condition always fires return 1 or return -1 if firstName isnt equal. – Joakim M Commented Oct 9, 2015 at 6:38
• 1 jsfiddle/arunpjohny/awao21ug/1 ? – Arun P Johny Commented Oct 9, 2015 at 6:40
• I don't see what the problem is... – Hayley Guillou Commented Oct 9, 2015 at 6:41
• 1 Just concatenate them if ( a.firstName+a.lastName > b.firstName+b.lastName ) – Shanimal Commented Oct 9, 2015 at 6:42

Add a ment  | 

5 Answers 5

Sorted by: Reset to default 7

You can try this ES6 version

const currentUsers = [{
  firstName: "Bob",
  lastName: "Adler"
}, {
  firstName: "Barney",
  lastName: "Jones"
}, {
  firstName: "Freddie",
  lastName: "Crougar"
}, {
  firstName: "Bob",
  lastName: "Adams"
}, {
  firstName: "Joe",
  lastName: "Lewis"
}, {
  firstName: "Joseph",
  lastName: "Lewis"
}];

const sortedUsers = currentUsers.sort((a, b) => {
          const result = a.firstName.localeCompare(b.firstName);

          return result !== 0 ? result : a.lastName.localeCompare(b.lastName);
        })

console.log(sortedUsers);

Just concatenate them with an underscore

var currentUsers = [{
  firstName: "Bob",
  lastName: "Adams"
}, {
  firstName: "Barney",
  lastName: "Jones"
}, {
  firstName: "Freddie",
  lastName: "Crougar"
}, {
  firstName: "Bobby",
  lastName: "Anderson"
}, {
  firstName: "Joe",
  lastName: "Lewis"
}, {
  firstName: "Joseph",
  lastName: "Lewis"
}];

currentUsers = currentUsers.sort(sortOnFirstAndLast)

function sortOnFirstAndLast(a,b) {
  var aa = a.firstName + ", " + a.lastName,
    bb = b.firstName + ", " + b.lastName;
  if (aa > bb)
    return 1;
  else if (aa < bb)
    return -1;
  return 0;
}

var $r = $("#result");
for (var i in currentUsers) {
  var div = $("<div/>").html(i + ":" + currentUsers[i].firstName + ":" + currentUsers[i].lastName);
  $r.append(div)
}

<script src="https://ajax.googleapis./ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="result"></div>

Try this;

var currentUsers = [ {
  firstName: "Bob",
  lastName: "Bobson"
},{
  firstName: "Bob",
  lastName: "Anderson"
}, {
  firstName: "Amy",
  lastName: "Jackson"
}];

var sorted = currentUsers.sort(function(a, b) {
  var aFirstChar = a.firstName.charAt(0);
  var bFirstChar = b.firstName.charAt(0);
  if (aFirstChar > bFirstChar) {
    return 1;
  } else if (aFirstChar < bFirstChar) {
    return -1;
  } else {
    var aLastChar = a.lastName.charAt(0);
    var bLastChar = b.lastName.charAt(0);
    if (aLastChar > bLastChar) {
      return 1;
    } else if (aLastChar < bLastChar) {
      return -1;
    } else {
      return 0;
    }    
  }
});

alert(JSON.stringify(sorted));

Just for the sake of alternatives. If you're only paring the first char, you could use charCodeAt instead of charAt to get a numeric value for another way of paring:

currentUsers  = currentUsers.sort(function(a, b) {
    return a.firstName.charCodeAt(0) - b.firstName.charCodeAt(0) 
        || a.lastName.charCodeAt(0) - b.lastName.charCodeAt(0);
});

The || executes only if the firstname first chars are equal (diff = 0)

The best, shortest and most readable code I could figure out was

currentUsers = currentUsers.sort((a, b) =>
    a.firstName.localeCompare(b.firstName) ||
    a.lastName.localeCompare(b.lastName)
);

'||' allows the other parison only execute when first localeCompare returns 0, aka. both names being equal value.

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