获取所有员工的最新交易类型详细信息。

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英文:

Get the details of latest transaction details of all employees based on selected transaction type

问题

$transaction_types = ['Add', 'Update', 'Read'];

$output = '';

$query = "SELECT * FROM transactions 
          INNER JOIN employees ON employees.employee_id = transactions.employee_id 
          WHERE transactions.transaction_type IN ('" . implode("','", $transaction_types) . "') 
          GROUP BY transactions.employee_id 
          ORDER BY DATE(transactions.transaction_date) DESC
          ";
$stmt = $connect->prepare($query);
if ($stmt->execute()) {
   if ($stmt->rowCount() > 0) {
      $output .= '<table>
                    <thead>
                       <tr>
                          <th>Transaction ID</th>
                          <th>Employee Name</th>
                          <th>Transaction Date</th>
                          <th>Transaction Type</th>
                          <th>Remarks</th>
                       </tr>
                    </thead>
                    <tbody>';
      $result = $stmt->fetchAll();
      foreach ($result as $row) {
         $output .= '<tr>
                        <td>' . $row->transaction_id . '</td>
                        <td>' . $row->employee_name . '</td>
                        <td>' . $row->transaction_date . '</td>
                        <td>' . $row->transaction_type . '</td>
                        <td>' . $row->remarks . '</td>
                     </tr>';
      }
      $output .= '</tbody></table>';
   }
}

echo $output;
英文:

everyone. Some one who can help me to get the latest transaction details of employees based on selected transaction type.
Below is the sample tables.

employees Table:

employee_id employee_name
1 John Doe
2 Jane Doe
3 Teri Dactyl
4 Allie Grater

transactions Table:

transaction_id employee_id transaction_date transaction_type remarks
1 1 2021-03-28 Add test 1
2 3 2022-09-07 Add test 2
3 2 2019-08-01 Add test 3
4 4 2023-06-05 Read test 4
5 4 2023-05-12 Add test 5
6 2 2020-02-01 Read test 6
7 3 2022-11-15 Update test 7
8 1 2020-06-14 Add test 8
9 1 2020-01-14 Update test 9
10 2 2023-12-31 Delete test 10

Then now, I want to get list of employees and their latest transaction details where transaction_type is either [Add, Update, Read]

Below is my desired output:

Transaction ID Employee Name Transaction Date Transaction Type Remarks
4 Allie Grater 2023-06-05 Read test 4
7 Teri Dactyl 2022-11-15 Update test 7
1 John Doe 2021-03-28 Add test 1

As you can see, Jane Doe is not in the list, because her last transaction type is Delete which is not in the selected transaction type [Add, Update, Read]

Below is my Code:

&lt;?php

$transaction_types = [&#39;Add&#39;, &#39;Update&#39;, &#39;Read&#39;];

$output = &#39;&#39;;

$query = &quot;SELECT * FROM transactions 
          INNER JOIN employees ON employees.employee_id = transactions.employee_id 
          WHERE transactions.transaction_type IN (&#39;&quot;.implode(&quot;&#39;, &#39;&quot;, $transaction_types).&quot;&#39;) 
          GROUP BY transactions.employee_id 
          ORDER BY DATE(transactions.transaction_date) DESC
          &quot;;
$stmt = $connect-&gt;prepare($query);
if ($stmt-&gt;execute()) {
   if ($stmt-&gt;rowCount() &gt; 0) {
      $output .= &#39;&lt;table&gt;
                    &lt;thead&gt;
                       &lt;tr&gt;
                          &lt;th&gt;Transaction ID&lt;/th&gt;
                          &lt;th&gt;Employee Name&lt;/th&gt;
                          &lt;th&gt;Transaction Date&lt;/th&gt;
                          &lt;th&gt;Transaction Type&lt;/th&gt;
                          &lt;th&gt;Remarks&lt;/th&gt;
                       &lt;/tr&gt;
                    &lt;/thead&gt;
                    &lt;tbody&gt;&#39;;
      $result = $stmt-&gt;fetchAll();
      foreach ($result as $row) {
         $output .= &#39;&lt;tr&gt;
                        &lt;td&gt;&#39;.$row-&gt;transaction_id.&#39;&lt;/td&gt;
                        &lt;td&gt;&#39;.$row-&gt;employee_name.&#39;&lt;/td&gt;
                        &lt;td&gt;&#39;.$row-&gt;transaction_date.&#39;&lt;/td&gt;
                        &lt;td&gt;&#39;.$row-&gt;transaction_type.&#39;&lt;/td&gt;
                        &lt;td&gt;&#39;.$row-&gt;remarks.&#39;&lt;/td&gt;
                     &lt;/tr&gt;&#39;;
      }
      $output .= &#39;&lt;/tbody&gt;&lt;/table&gt;&#39;;
   }
}

echo $output;

?&gt;

答案1

得分: 1

你首先需要从结果集中删除具有 Delete 的用户,然后使用 row_number() 窗口函数与 CTE 选择最大日期。

WITH cte
AS (
    SELECT t.transaction_id AS transaction_id
        ,t.transaction_date AS transaction_date
        ,row_number() OVER (
            PARTITION BY e.employee_name ORDER BY t.transaction_date DESC
            ) AS rn
        ,e.employee_name AS employee_name
        ,t.transaction_type AS transaction_type
        ,t.remarks AS remarks
    FROM employees e
    JOIN transactions t ON t.employee_id = e.employee_id
    WHERE e.employee_name NOT IN (
            SELECT e.employee_name
            FROM employees e
            JOIN transactions t ON t.employee_id = e.employee_id
            WHERE transaction_type = 'Delete'
            )
    )
SELECT cte.transaction_id
    ,cte.transaction_date
    ,cte.employee_name
    ,cte.transaction_type
    ,cte.remarks
FROM cte
WHERE rn = 1;

demo 这里

英文:

you first have to remove the users with Delete from your results set and then use the row_number() window function with CTE to select the max dates

WITH cte
AS (
	SELECT t.transaction_id AS transaction_id
		,t.transaction_date AS transaction_date
		,row_number() OVER (
			PARTITION BY e.employee_name ORDER BY t.transaction_date DESC
			) AS rn
		,e.employee_name AS employee_name
		,t.transaction_type AS transaction_type
		,t.remarks AS remarks
	FROM employees e
	JOIN transactions t ON t.employee_id = e.employee_id
	WHERE e.employee_name NOT IN (
			SELECT e.employee_name
			FROM employees e
			JOIN transactions t ON t.employee_id = e.employee_id
			WHERE transaction_type = &#39;Delete&#39;
			)
	)
SELECT cte.transaction_id
	,cte.transaction_date
	,cte.employee_name
	,cte.transaction_type
	,cte.remarks
FROM cte
WHERE rn = 1;

demo here

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  • 本文由 发表于 2023年6月5日 22:39:56
  • 转载请务必保留本文链接:https://go.coder-hub.com/76407560.html
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