为另一列中的每个值获取一列中的相似值的总和

问题描述：

ok, I have a mysql table named "sales_records" like so:

  User    saleType
+-------+-------+------+
| Jeff  | Sale_Closed  |
+-------+-------+------+
| Jeff  | Sale_null    |
+-------+-------+------+
| Sean  | sale_closed  |
+-------+-------+------+
| Jeff  | sale_closed  |
+-------+-------+------+
| Sean  | sale_closed  |
+-------+-------+------+
| Mark  | Sale_null    |
+-------+-------+------+
| Mark  | sale_closed  |
+-------+-------+------+
| Josh  | sale_closed  |
+-------+-------+------+
| Josh  | sale_closed  |
+-------+-------+------+

I'm trying to count up each saleType for each User.. In other words.. Jeff has 2 "sale_closed" and 1 "sale_null. Mark has 1 "sale_closed" and 1 "sale_null", and so on..

and store it into an array.. (Json) like so.:

{"jeff":{"sale_closed":2,"sale_null":1},"sean":{"sale_closed":2},"Mark":{"sale_closed":1, "sale_null":1},"Josh":{"sale_closed":2}

Ok, so here's my attempt:

<?php
// assuming database connection has already been established..

$result = mysql_query("SELECT DISTINCT `USER' FROM `sales_records`");
while($row = mysql_fetch_assoc($result)) {
    $user_array[] = $row['User']; 
}

foreach ($user_array as $user) {
    $json_array[$user] = mysql_query("SELECT saleType FROM sales_records WHERE user='$user'");

    while ($row = mysql_fetch_array($json_array[$user])) { // AND HERE IS WHERE i GET STUCK!!!
        $json_array[$user][$username][] = $row['saleType']; // I DON'T KNOW WHERE TO GO FROM HERE...
    }
} // AM I APPROACHING THIS CORRECTLY?..

//...
// Once I have all the 'saleType's for each user, maybe I could use:
array_count_values()... or something.

?>

好吧，我有一个名为“sales_records”的mysql表，如下所示： p>

 用户saleType 
 + ------- + ------- + ------ + 
 | 杰夫|  Sale_Closed | 
 + ------- + ------- + ------ + 
 | 杰夫|  Sale_null | 
 + ------- + ------- + ------ + 
 | 肖恩|  sale_closed | 
 + ------- + ------- + ------ + 
 | 杰夫|  sale_closed | 
 + ------- + ------- + ------ + 
 | 肖恩|  sale_closed | 
 + ------- + ------- + ------ + 
 | 马克|  Sale_null | 
 + ------- + ------- + ------ + 
 | 马克|  sale_closed | 
 + ------- + ------- + ------ + 
 | 乔希|  sale_closed | 
 + ------- + ------- + ------ + 
 | 乔希|  sale_closed | 
 + ------- + ------- + ------ + 
  code>  pre> 
 
 我想算一算 为每个用户提供每个saleType ..换句话说..杰夫有2个“sale_closed”和1个“sale_null .Mark有1个”sale_closed“和1个”sale_null“，等等..  p> 
 
 并将其存储到数组中..（Json）就像这样。： p> 
nn 
  {“jeff"：{"sale_closed"：2 ,"sale_null”：1}，  “sean”：{“sale_closed”：2}，“Mark”：{“sale_closed”：1，“sale_null”：1}，“Josh”：{“sale_closed”：2} 
  code>  pre  > 
 
 好的，所以这是我的尝试： p> 
 
 
 ＆lt;？php 
 //假设数据库连接已经建立.. 
 
  $ result = mysql_query（“SELECT DISTINCT`USER'from`sales_records`”）; 
while（$ row = mysql_fetch_assoc（$ result））{
 $ user_array [] = $ row ['User']; 
} \  n 
foreach（$ user_array as $ user）{
 $ json_array [$ user] = mysql_query（“SELECT saleType FROM sales_records WHERE user ='$ user'”）; 
 
 while（$ row = mysql_fetch_array（$ json_array）  [$ user]））{//和这里我去哪里!!! 
 $ json_array [$ user] [$ username] [  ] = $ row ['saleType'];  //我不知道从哪里开始... 
} 
} //我正确地接近这个问题？.. 
 
 // ... ... 
 //一旦我拥有所有的' 每个用户的saleType，也许我可以使用：
array_count_values（）......或其他东西。
 
？＆gt; 
  code>  pre> 
  div>

答

You can get all the data you need in a single query:

SELECT user, saleType, count(*) totalSales FROM sales_record
GROUP BY user, saleType

That will return all the results in 3 columns that should be easy to turn into JSon.

为另一列中的每个值获取一列中的相似值的总和

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