// 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?
Share Improve this question edited Mar 24, 2021 at 7:19 Reality-Torrent asked Oct 9, 2015 at 6:36 Reality-TorrentReality-Torrent 3583 silver badges15 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
5 Answers
Reset to default 7You 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.