The best approach for the specific case here, in my opinion, is the one commonpike suggested. A little improvement I'd suggest that works in modern browsers is:
// aao is the "associative array" you need to "sort"
Object.keys(aao).sort(function(a,b){return aao[b]-aao[a]});
This could apply easily and work great in the specific case here so you can do:
let aoo={};
aao["sub2"]=1;
aao["sub0"]=-1;
aao["sub1"]=0;
aao["sub3"]=1;
aao["sub4"]=0;
let sk=Object.keys(aao).sort(function(a,b){return aao[b]-aao[a]});
// now you can loop using the sorted keys in `sk` to do stuffs
for (let i=sk.length-1;i>=0;--i){
// do something with sk[i] or aoo[sk[i]]
}
Besides of this, I provide here a more "generic" function you can use to sort even in wider range of situations and that mixes the improvement I just suggested with the approaches of the answers by Ben Blank (sorting also string values) and PopeJohnPaulII (sorting by specific object field/property) and lets you decide if you want an ascendant or descendant order, here it is:
// aao := is the "associative array" you need to "sort"
// comp := is the "field" you want to compare or "" if you have no "fields" and simply need to compare values
// intVal := must be false if you need comparing non-integer values
// desc := set to true will sort keys in descendant order (default sort order is ascendant)
function sortedKeys(aao,comp="",intVal=false,desc=false){
let keys=Object.keys(aao);
if (comp!="") {
if (intVal) {
if (desc) return keys.sort(function(a,b){return aao[b][comp]-aao[a][comp]});
else return keys.sort(function(a,b){return aao[a][comp]-aao[a][comp]});
} else {
if (desc) return keys.sort(function(a,b){return aao[b][comp]<aao[a][comp]?1:aao[b][comp]>aao[a][comp]?-1:0});
else return keys.sort(function(a,b){return aao[a][comp]<aao[b][comp]?1:aao[a][comp]>aao[b][comp]?-1:0});
}
} else {
if (intVal) {
if (desc) return keys.sort(function(a,b){return aao[b]-aao[a]});
else return keys.sort(function(a,b){return aao[a]-aao[b]});
} else {
if (desc) return keys.sort(function(a,b){return aao[b]<aao[a]?1:aao[b]>aao[a]?-1:0});
else return keys.sort(function(a,b){return aao[a]<aao[b]?1:aao[a]>aao[b]?-1:0});
}
}
}
You can test the functionalities trying something like the following code:
let items={};
items['Edward']=21;
items['Sharpe']=37;
items['And']=45;
items['The']=-12;
items['Magnetic']=13;
items['Zeros']=37;
//equivalent to:
//let items={"Edward": 21, "Sharpe": 37, "And": 45, "The": -12, ...};
console.log("1: "+sortedKeys(items));
console.log("2: "+sortedKeys(items,"",false,true));
console.log("3: "+sortedKeys(items,"",true,false));
console.log("4: "+sortedKeys(items,"",true,true));
/* OUTPUT
1: And,Sharpe,Zeros,Edward,Magnetic,The
2: The,Magnetic,Edward,Sharpe,Zeros,And
3: The,Magnetic,Edward,Sharpe,Zeros,And
4: And,Sharpe,Zeros,Edward,Magnetic,The
*/
items={};
items['k1']={name:'Edward',value:21};
items['k2']={name:'Sharpe',value:37};
items['k3']={name:'And',value:45};
items['k4']={name:'The',value:-12};
items['k5']={name:'Magnetic',value:13};
items['k6']={name:'Zeros',value:37};
console.log("1: "+sortedKeys(items,"name"));
console.log("2: "+sortedKeys(items,"name",false,true));
/* OUTPUT
1: k6,k4,k2,k5,k1,k3
2: k3,k1,k5,k2,k4,k6
*/
As I already said, you can loop over sorted keys if you need doing stuffs
let sk=sortedKeys(aoo);
// now you can loop using the sorted keys in `sk` to do stuffs
for (let i=sk.length-1;i>=0;--i){
// do something with sk[i] or aoo[sk[i]]
}
Last, but not least, some useful references to Object.keys and Array.sort