Question

我在每个表格中有3个包含restaurant_id的表格，我试图在uhd_restaurant加入uhd_user_order和uhd_order_history

上获取数据

这是我的代码：

$this->db->DISTINCT();
$this->db->select("t1.restaurant_id,t1.restaurant_name,t1.restaurant_code, COUNT(t1.restaurant_id)as total");
$this->db->from("uhd_restaurant as t1");
$this->db->join("uhd_user_order as t2","t1.restaurant_id = t2.restaurant_id","left");
$this->db->join("uhd_order_history as t3","t1.restaurant_id = t3.restaurant_id");
$this->db->group_by("t1.restaurant_name");
$this->db->order_by("total","desc");
$res = $this->db->get()->result_array();
return $res;

该代码将返回：

array(6) {
  [0]=>
  array(4) {
    ["restaurant_id"]=>
    string(3) "365"
    ["restaurant_name"]=>
    string(8) "Yukihira"
    ["restaurant_code"]=>
    string(4) "MG99"
    ["total"]=>
    string(2) "88"
  }
  [1]=>
  array(4) {
    ["restaurant_id"]=>
    string(3) "367"
    ["restaurant_name"]=>
    string(5) "alice"
    ["restaurant_code"]=>
    string(4) "ZF42"
    ["total"]=>
    string(1) "3"
  }
  [2]=>
  array(4) {
    ["restaurant_id"]=>
    string(3) "363"
    ["restaurant_name"]=>
    string(6) "takumi"
    ["restaurant_code"]=>
    string(4) "ZO09"
    ["total"]=>
    string(1) "2"
  }
  [3]=>
  array(4) {
    ["restaurant_id"]=>
    string(3) "368"
    ["restaurant_name"]=>
    string(5) "test1"
    ["restaurant_code"]=>
    string(4) "WS94"
    ["total"]=>
    string(1) "2"
  }
  [4]=>
  array(4) {
    ["restaurant_id"]=>
    string(3) "366"
    ["restaurant_name"]=>
    string(6) "nakiri"
    ["restaurant_code"]=>
    string(4) "XN27"
    ["total"]=>
    string(1) "1"
  }
}

在我的表uhd_user_order中有3个restaurant_id，其中

364有1个，365有8个，366有1个

并在我的uhd_order_history中有4 restaurant_id

365有11个，363有2个，368有2个，367有3个

所以总数应该是

365 = 19，363 = 2，364 = 1，366 = 2 367 = 3且368 = 2 < / p>

但是我返回了错误的数据，请将365的返回总数与88一起查看，但19我的366只返回1而我无法退回restaurant_id 364

所以数据应该返回到：

array(5) {
  [0]=>
  array(4) {
    ["restaurant_id"]=>
    string(3) "365"
    ["restaurant_name"]=>
    string(8) "Yukihira"
    ["restaurant_code"]=>
    string(4) "MG99"
    ["total"]=>
    string(2) "19"
  }
  [1]=>
  array(4) {
    ["restaurant_id"]=>
    string(3) "367"
    ["restaurant_name"]=>
    string(5) "alice"
    ["restaurant_code"]=>
    string(4) "ZF42"
    ["total"]=>
    string(1) "3"
  }
  [2]=>
  array(4) {
    ["restaurant_id"]=>
    string(3) "363"
    ["restaurant_name"]=>
    string(6) "takumi"
    ["restaurant_code"]=>
    string(4) "ZO09"
    ["total"]=>
    string(1) "2"
  }
  [3]=>
  array(4) {
    ["restaurant_id"]=>
    string(3) "368"
    ["restaurant_name"]=>
    string(5) "test1"
    ["restaurant_code"]=>
    string(4) "WS94"
    ["total"]=>
    string(1) "2"
  }
  [4]=>
  array(4) {
    ["restaurant_id"]=>
    string(3) "366"
    ["restaurant_name"]=>
    string(6) "nakiri"
    ["restaurant_code"]=>
    string(4) "XN27"
    ["total"]=>
    string(1) "2"
  }
  [5]=>
  array(4) {
    ["restaurant_id"]=>
    string(3) "364"
    ["restaurant_name"]=>
    string(6) "test"
    ["restaurant_code"]=>
    string(4) "WY58"
    ["total"]=>
    string(1) "1"
  }
}

uhd_restaurant中的P.S restaurant_id uhd_user_order和uhd_order_history

中的所有

function executeQuery( stmt, data ) {
  var outhjk = "";
  pool.connect(function(err, client, done) {
    if(err) {
      console.error('error fetching client from pool', err);
    }
    client.query(stmt, data, function(err, result) {
      //call `done()` to release the client back to the pool
      done();

      if(err) {
        return console.error('error running query', err);
      }
      outhjk = "just work please";
    });
  });
  return outhjk;
}

你可以帮助我如何获得正确的数据总量吗？

Answer 1

$this->db->select("t1.restaurant_id,t1.restaurant_name,t1.restaurant_code, COUNT(t1.restaurant_id)as total");
$this->db->from("uhd_restaurant as t1");
$this->db->join("uhd_user_order as t2","t1.restaurant_id = t2.restaurant_id","left");
$this->db->join("uhd_order_history as t3","t1.restaurant_id = t3.restaurant_id");
$this->db->group_by("t1.restaurant_name");
$this->db->order_by("total","desc");
$res = $this->db->get()->result_array();
return $res;

Answer 2

请在CodeIgniter中尝试类似此查询的内容.. [虽然您的理想结果集中存在一些错误，但它会给您提示...]

SELECT t1.restaurant_id,t1.restaurant_name,t1.restaurant_code, SUM(total) FROM ( SELECT t1.restaurant_id,t1.restaurant_name,t1.restaurant_code, COUNT(t1.restaurant_id)as total FROM uhd_restaurant as t1 LEFT JOIN uhd_order_history as t3 ON t1.restaurant_id = t3.restaurant_id GROUP BY t1.restaurant_id UNION ALL SELECT t1.restaurant_id,t1.restaurant_name,t1.restaurant_code, COUNT(t1.restaurant_id)as total FROM uhd_restaurant as t1 LEFT JOIN uhd_user_order as t2 ON t1.restaurant_id = t2.restaurant_id GROUP BY t1.restaurant_id) as t1 GROUP BY t1.restaurant_id ORDER BY total DESC

如何在codeigniter活动记录中使用oder_by count返回数据

2 个答案: