Node.js node_redis使用内核获取zrange

Node.js node_redis使用内核获取zrange,node.js,redis,node-redis,Node.js,Redis,Node Redis,有人知道如何通过节点redis获得具有分数的成员吗? 我试过这样的方法: client.ZRANGE(key, 0, -1, withscores, function(err, replies) { }); 谢谢。这段代码看起来不错。查看以下链接以检索所需内容: 在此处添加了该链接示例中的代码,以防被删除。 var rc=require('redis').createClient(); var _=require('underscore'); rc.zincrby('myset', 1,

有人知道如何通过节点redis获得具有分数的成员吗? 我试过这样的方法:

client.ZRANGE(key, 0, -1, withscores, function(err, replies) {

});

谢谢。

这段代码看起来不错。查看以下链接以检索所需内容:

在此处添加了该链接示例中的代码,以防被删除。

var rc=require('redis').createClient();
var _=require('underscore');

rc.zincrby('myset', 1, 'usera');
rc.zincrby('myset', 5, 'userb');
rc.zincrby('myset', 3, 'userc');
rc.zrevrange('myset', 0, -1, 'withscores', function(err, members) {
        // the resulting members would be something like
        // ['userb', '5', 'userc', '3', 'usera', '1']
        // use the following trick to convert to
        // [ [ 'userb', '5' ], [ 'userc', '3' ], [ 'usera', '1' ] ]
        // learned the trick from
        // http://stackoverflow.com/questions/8566667/split-javascript-array-in-chunks-using-underscore-js
    var lists=_.groupBy(members, function(a,b) {
        return Math.floor(b/2);
    });
    console.log( _.toArray(lists) );
});
rc.quit();

看来你的代码是对的。以下是获取zrange的语法

不得分:

redisClient.zrange(keyName,start,stop,function(err,result){
   //result is array
   // every index will give you member name
})
例:

分数:

redisClient.zrange(keyName,start,stop,'withscores',function(err,result){
   //result is array
   // here even index will hold member
   // odd index will hold its score
})
例:


^2.0版本的正确方法

var args=[key,to,from];

redisClient.zrevrangebyscore(args,function(err,data){
我尝试了之前接受的答案,但没有得到想要的结果,后来我尝试了以下代码并得到了适当的结果

原始输出:

[ 'player:522',
    '19685',
    'player:164',
    '19594',
    'player:807',
    '19171',
    'player:694',
    '19165',
    'player:905',
    '19108',
    'player:859',
    '19087',
    'player:432',
    '18973',
    'player:515',
    '18831',
    'player:163',
    '18750',
    'player:4',
    '18552' ]
预期产出:

{
    "player:522": "19685",
    "player:164": "19594",
    "player:807": "19171",
    "player:694": "19165",
    "player:905": "19108",
    "player:859": "19087",
    "player:432": "18973",
    "player:515": "18831",
    "player:163": "18750",
    "player:4": "18552"
}
解决方案:

redisClient.ZREVRANGE('daily', 1, 10, 'WITHSCORES', function(err, result) {
     result = _.fromPairs(_.chunk(result, 2));
     return res.status(200).json(result);
});
香草JS溶液 Redis呼叫:

redisClient.zrange(keyName, start, stop, 'withscores', function(err, result) {
  // result.reduce ... (See below)
}
这是一个我很快想到的普通JS解决方案。 就我个人而言,导入下划线或任何其他库来执行如此简单的任务是没有意义的:

result.reduce(function (a, c, i) {
  var idx = i / 2 | 0;
  if (i % 2) {
    a[idx].score = c;
  } else {
    a[idx] = { id: c };
  }

  return a;
}, []);
假设这一输入:

['player1', 13, 'player2', 11, 'player4', 7, 'player3', 3, 'player5', 0]
此函数产生:

[
  { id: 'player1', score: 13 },
  { id: 'player2', score: 11 },
  { id: 'player4', score: 7 },
  { id: 'player3', score: 3 },
  { id: 'player5', score: 0 }
]
下面是另一个将结果转换为二维数组的方法: 将生成以下数组:

[
  [ 'player1', 13 ],
  [ 'player2', 11 ],
  [ 'player4', 7 ],
  [ 'player3', 3 ],
  [ 'player5', 0 ]
]
单线lambda版本:
香草JS reduce在这里很管用

const result = [
    'player:522',
    '19685',
    'player:164',
    '19594',
    'player:807',
    '19171',
    'player:694',
    '19165',
    'player:905',
    '19108',
    'player:859',
    '19087',
    'player:432',
    '18973',
    'player:515',
    '18831',
    'player:163',
    '18750',
    'player:4',
    '18552'
]

const map = result.reduce((map, k, i, res) => {
    if (i % 2 !== 0) {
        map[res[i - 1]] = Number(k);
    }
    return map;
}, {})
地图现在是:

{
  'player:522': 19685,
  'player:164': 19594,
  'player:807': 19171,
  'player:694': 19165,
  'player:905': 19108,
  'player:859': 19087,
  'player:432': 18973,
  'player:515': 18831,
  'player:163': 18750.65468,
  'player:4': 18552
}
[
  [ 'player1', 13 ],
  [ 'player2', 11 ],
  [ 'player4', 7 ],
  [ 'player3', 3 ],
  [ 'player5', 0 ]
]
result.reduce((a, c, i) => i % 2 ? (a[i / 2 | 0].data = c, a) : (a[i / 2 | 0] = { id: c }, a), []);
const result = [
    'player:522',
    '19685',
    'player:164',
    '19594',
    'player:807',
    '19171',
    'player:694',
    '19165',
    'player:905',
    '19108',
    'player:859',
    '19087',
    'player:432',
    '18973',
    'player:515',
    '18831',
    'player:163',
    '18750',
    'player:4',
    '18552'
]

const map = result.reduce((map, k, i, res) => {
    if (i % 2 !== 0) {
        map[res[i - 1]] = Number(k);
    }
    return map;
}, {})
{
  'player:522': 19685,
  'player:164': 19594,
  'player:807': 19171,
  'player:694': 19165,
  'player:905': 19108,
  'player:859': 19087,
  'player:432': 18973,
  'player:515': 18831,
  'player:163': 18750.65468,
  'player:4': 18552
}
var data = []
results.map((result, index) => {
    if (index % 2 == 0) {
        data.push(results[index] = { player: results[index], score: results[index+1] })
    }
})

console.log(data)