這樣：

var data = {

"rows": [{

"name": "張三",

"time": "2011/4/1 0:00:00",

},{

"name": "李四",

"time": "2015/5/6 12:30:00",

},{

"name": "王五",

"time": "2012/10/1 22:10:00",

},{

"name": "趙六",

"time": "2011/9/1 22:10:00",

}]

};

var rows = data.rows;

rows.sort(function(a,b){

return Date.parse(a.time) - Date.parse(b.time);//時間正序

});

for(var i =0,l=rows.length;i

console.log(rows[i].name + " | " + rows[i].time);

}

擴充套件資料：

注意事項

1、獲取兩個陣列時間的字串，轉成時間格式比較大小 [(NSDate*) compare:(NSDate*)]

2、根據時間大小排序此陣列 sortedArrayUsingComparator

(NSMutableArray *)[arrData sortedArrayUsingComparator:^NSComparisonResult(id obj1, id obj2) {

NSDateFormatter *formatter = [[NSDateFormatter alloc] init];

[formatter setDateFormat:@"yyyy-MM-dd HH:mm:ss"];

NSString *string1 = obj1[@"time"]; //獲取時間

NSString *string2 = obj2[@"time"];

NSDate *date1 = [formatter dateFromString:string1];

NSDate *date2 = [formatter dateFromString:string2];

NSComparisonResult result = [date1 compare:date2]; //比較時間大小

return result == NSOrderedAscending; //倒敘排列 返回陣列

}];