尝试用jQuery按多个字段对JSON进行排序

Question

对于this posting，我可以按两个字段对JSON对象数组进行排序。这篇文章还说：

“若要添加要排序的其他列，可以在数组比较中添加其他项。”

//THIS WORKS BUT NEEDS TO SORT BY ALL 5 FIELDS..
function sortRecords(a, b){
//note the minus before -cmp, for descending order
// field1 (Y or N), field2 (numeric), field3 (numeric), 
// field4 (Y or N), field5 (Y or N)
return cmp( 
    [-cmp(a.field1, b.field1), cmp(a.field2, b.field2)], 
    [-cmp(b.field1, a.field1), cmp(b.field2, a.field2)]
 );}

//THIS ONLY SORTS BY field1 and field2
function sortRecords(a, b){
//note the minus before -cmp, for descending order
// field1 (Y or N), field2 (numeric), field3 (numeric), 
// field4 (Y or N), field5 (Y or N)
return cmp( 
    [-cmp(a.field1, b.field1), cmp(a.field2, b.field2), cmp(a.field3, b.field3)], 
    [-cmp(b.field1, a.field1), cmp(b.field2, a.field2), cmp(b.field3, a.field3)]
 );}

function cmp(x,y){
return x > y ? 1 : x < y ? -1 : 0;
}


///implementation
data = $(data).sort(sortRecords);

我在这里做错什么了？

Answer 1

我一直在使用StackOverflow，我结合了几个如何排序/排序的例子；这个函数似乎给了我我想要的.

将这个JSON对象数组作为数据。

在{“名称”：“JOE”、"field1":"Y“、"field2":3、"field3":79、"field4":"Y”、"field5":"Y"} {"name":"BEN“、"field1":"Y”、"field2":15、"field3":113、"field4":"Y“、"field5":"N"} {”名称“：”SUE“、"field1":"Y”、“Y”"field2":35，"field3":395，"field4":"Y"，"field5":"Y"}

我可以调用sortRecords()来对我想要的进行排序。

在 {"name":"JOE“、"field1":"Y”、"field2":3、"field3":79、"field4":"Y“、"field5":"Y"} {"name":"SUE”、"field1":"Y“、"field2":35、"field3":395、"field4":"Y”、"field5":"Y"} {名称：“BEN”、"field1":"Y“、”Y“"field2":15，"field3":113，"field4":"Y"，"field5":"N"}

function sortRecords(a, b){
    if (a.field1 < b.field1){
        return 1;
    } else if (a.field1 > b.field1){
        return -1;
    } else if (a.field2 < b.field2){
        return 1;
    } else if (a.field2 > b.field2){
        return -1;
    } else if (a.field3 < b.field3){
        return 1;
    } else if (a.field3 > b.field3){
        return -1;
    } else if (a.field4 < b.field4){
        return -1;
    } else if (a.field4 > b.field4){
        return 1;
    } else if (a.field5 < b.field5){
        return -1;
    } else if (a.field5 > b.field5){
        return 1;
    } else {
        return 0;
    }
}

data = $(data).sort(sortRecords);