从查询中检索特定键值并在查询中获取它们对的计数

Question

我有一个表lead_submission，其中包含特定格式的用户值，例如

agent_name     qa_details
xxx      1001:|1083:|504:Yes|1009:|
ccc      504:Yes|1083:No|1008:|1009:|

现在我想从两行中获取只说 504:Yes 的计数

这些值来自另一个表paid_response

qno    paid_response
504     Yes
1083    No
1083    Possibly

<?php
//db connection goes here

$sql=mysql_query("select qno,paid_response from paid_response where qno='504' ");
while($rows=mysql_fetch_array($sql)) {    
$exqnos= $rows['qno'].'|'.$rows['paid_response'];
}

list($key,$val)=explode('|',$exqnos);
$exqno[$key]=$val;

foreach($exqno as $qno=>$value) {
$string .="qa_details LIKE '%|$qno:$value|%' ";  
}

$sql=mysql_query("SELECT count(agent_name) as agent_cnt,count($string) as ppicount FROM `lead_submission` WHERE $string "); ?>

               <table border="1">
                <thead>
                  <tr>
                    <th>CountAgent</th>
                    <th>504-COUNT</th>                      
                  </tr>

<?php
while($row=mysql_fetch_array($sql)) { ?>

     <tr style="color:red" >            
        <td><?php echo $row['agent_cnt']; ?></td>
        <td><?php echo $row['ppicount']; ?></td>            
        </tr>

<?php
}
?> 

现在通过这样做，我将 504:Yes 计为 2

CountAgent  504-COUNT
 2            2        //as u can see that `504:Yes` has occured two times in lead_submission table.

我的意思是，我如何计算另一个组合，例如 1083:No 并在同一张表中显示计数

NB:- cant we just fetch the combination like `504:Yes` or `1083:No` or `1083:Yes` from paid_response table to maintain stability so that i dont have to change the query everytime. 

CountAgent  504-COUNT   1083-Count
  2           2           1        //how to get this count `1083:No` . as u can see it only appeared 1 times in `lead_submission` table

Answer 1

好的，这是不完整的答案。不完整，因为我懒得为你写具体的查询。但我正在写如何做到这一点。

注意：阅读规范化。你会看到你的错误。

如果您可以通过在数据开头添加| 来修改您的数据，即将1001:|1083:|504:Yes|1009:| 设置为|1001:|1083:|504:Yes|1009:| 对您有很大帮助。

现在，您可以轻松地在 %|<ID>:% 上搜索文本搜索，这可以确保您找到 504 或 1009 或任何数字，方法是用该数字替换 <ID>。

你读过规范化吗？无论如何。

一旦你这样做了，你现在就可以做 Row to Column 概念（我认为它被称为 PIVOT Queries，当你 google 规范化时用 google 搜索它。）。这里给出了一些例子

MySQL Row to Column 要么 Mysql query to dynamically convert rows to columns

抱歉，答案不完整，但文本搜索是您最好的选择，最好先创建一个视图，然后再将其用于复杂查询。但实际上获取数据会非常慢，并且不推荐。

希望你找到规范化...如果不试试这个：https://www.google.co.in/#q=normalization+in+database

哦，别忘了为该列建立索引...

Answer 2

试试这个：

SELECT COUNT(DISTINCT ls.agent_name), 
       SUM(CASE WHEN pr.qno = 504 AND pr.paid_response = 'Yes' THEN 1 ELSE 0 END) AS '504-Count', 
       SUM(CASE WHEN pr.qno = 1083 AND pr.paid_response = 'No' THEN 1 ELSE 0 END) AS '1083-Count'
FROM lead_submission ls 
INNER JOIN paid_response pr
ON CONCAT('|', ls.qa_details, '|') LIKE CONCAT('%|', pr.qno, ':', pr.paid_response , '|%');

查看SQL FIDDLE DEMO

输出

| COUNTAGENT | 504-COUNT | 1083-COUNT |
|------------|-----------|------------|
|          2 |         2 |          1 |

::EDIT::

首先执行下面的查询

SELECT GROUP_CONCAT('SUM(CASE WHEN pr.qno = ', qno, ' AND pr.paid_response = ''', paid_response,''' THEN 1 ELSE 0 END) AS ''', qno, '-Count''') 
FROM paid_response;

使用此查询的输出并构建您的最终查询，如下所示：

query = 'SELECT COUNT(DISTINCT ls.agent_name), ' + outputOfAboveQuery + ' FROM lead_submission ls NNER JOIN paid_response pr ON CONCAT('''|''', ls.qa_details, '''|''') LIKE CONCAT('''%|''', pr.qno, ''':''', pr.paid_response , '''|%''');';

在您的代码中执行此字符串，以便获取动态查询以获取计数

Answer 3

假设数据的格式保持不变，你可以尝试这样的事情（未经测试）：

$query[] = "SELECT COUNT(agent_name) as agent_cnt"; // number of agents in total

// Counts of each question/answer.
foreach ($exqno as $qno => $value) {
    $query[] = "(SELECT COUNT(agent_name) FROM lead_submission WHERE qa_details LIKE '%|$qno:$value|%') AS count_{$qno}";  
}

$full_query = implode(', ', $query) . " FROM lead_submission";

$sql = mysql_query( $full_query );

Answer 4

SELECT productOwner_org_id, 
(SELECT COUNT(*) FROM tblProducts P2 WHERE P1.product_id=p2.product_id AND productDescription='Leisure at PER01 IVR Gas') AS '504-data',
(SELECT COUNT(*) FROM tblProducts P3 WHERE P1.product_id=p3.product_id AND productDescription='Leisure Plus at Centerpoint') AS '1083-data'
FROM tblProducts p1
WHERE productOwner_org_id = 'AEG01'
AND
(SELECT COUNT(*) FROM tblProducts P2 WHERE P1.product_id=p2.product_id AND productDescription='Leisure at PER01 IVR Gas') != 0
OR
(SELECT COUNT(*) FROM tblProducts P3 WHERE P1.product_id=p3.product_id AND productDescription='Leisure Plus at Centerpoint') != 0


;

如你所见，它有点丑。

A) 您更好的选择是重新组织您的数据 要么 B) 当使用更多 php 逻辑以不同方式呈现/格式化数据时

Answer 5

呼应别人的cmets。您可能应该标准化您的模型。

也就是说，提取您需要的结果并非不可能，只是该解决方案效率低下、不可扩展、新开发人员难以理解且扩展性不强。

此外，提取长格式的数据比提取宽格式的数据更容易，即

# wide data format
CountAgent  504-COUNT   1083-Count
  2           2           1  

对比

# Long data format
dimension   count
CountAgent    2
504-Count     2
1083-Count    1

在 php 中从长到宽的转换更容易（甚至可能不需要）。

SELECT 
CONCAT(pr.qno, ":", pr.paid_response) dimension,
COUNT(*) `count`
FROM lead_submission ls 
JOIN paid_response pr 
  ON ls.qa_details LIKE CONCAT("%", pr.qno, ":", pr.paid_response, "%")
-- insert where clause 
GROUP BY 1
UNION
SELECT 'count-agent' dimension,
COUNT(DISTINCT ls.agent_id) `count`
FROM lead_submission ls 
JOIN paid_response pr 
  ON ls.qa_details LIKE CONCAT("%", pr.qno, ":", pr.paid_response, "%")
-- insert where clause        
GROUP BY 1

在上面的查询中，where clause 对于两个联合选择应该是相同的，我认为对于您的情况，它应该采用以下形式：

WHERE CONCAT(pr.qno, ":", pr.paid_response) IN (<key-value pair 1>, <key-value pair 2>, ...)

这将返回以下结果：

DIMENSION       COUNT
1083:No         1
504:Yes         2
count-agent     2

这是sqlfiddle demo

Answer 6

计数代理在每个搜索案例中可能不同。如果在上面的示例中搜索“504”、“1083:No”，那么在这两种情况下都不是 2。我建议你像这样修改你的脚本：

<?php
$array = array('504','1083');
//db connection goes here
$th="";
$tr="";
foreach($array as $arr) {
$srch=$arr;
$sql=mysql_query("select qno,paid_response from paid_response where qno=$srch ");
while($rows=mysql_fetch_array($sql)) {    
$exqnos= $rows['qno'].'|'.$rows['paid_response'];
}

list($key,$val)=explode('|',$exqnos);
$exqno[$key]=$val;

foreach($exqno as $qno=>$value) {
$string .="qa_details LIKE '%|$qno:$value|%' ";  
}
$sql=mysql_query("SELECT count(agent_name) as agent_cnt,count($string) as ppicount FROM `lead_submission` WHERE $string ");
        $th.="<th>CountAgent(".$arr.")</th><th>(".$arr.")-COUNT</th>";                      
while($row=mysql_fetch_array($sql)) {         
        $tr.="<td>".$row['agent_cnt']."</td><td>".$row['ppicount']."</td>";           
}
}
?> 
   <table border="1">
     <thead>
     <tr>
     <?php echo $th; ?>                     
     </tr>
     </thead>
     <tbody>
     <tr style="color:red" >            
     <?php echo $tr; ?>             
     </tr>
     </tbody>
     </table>