Question

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我正在尝试基于查询与用户会话相关联的mysql db列数据来编写php mysqli sum查询。因此，脚本应该在A列中获取与会话用户名关联的所有数据库数据，然后从列B中提取变量，然后对与列B变量匹配的列C值求和。

继承我拥有的东西，结果产生0，而不是销售列C中的数据总和。

DB表看起来像：

Column A / Column B / Column C

Bill     / Sales    / 15.00

Helen    / Sales    / 32.00

James    / Accounts / 20.00

..

目前的代码如下：

<?php
    session_start();
    $db = new mysqli('DBHOST.com', 'DBUsername', 'DBPassword', 'DBNAME');
    if ($db->connect_errno > 0) {
        die('Unable to connect to database [' . $db->connect_error . ']');
    }
    $sql = mysqli_query($db, "SELECT * FROM Table_Name WHERE ColumnA = '" . $_SESSION["username"] . "'");
    $count = mysqli_num_rows($sql);
    if ($count == 1) {
        $output = mysqli_fetch_assoc($sql);
        $columnBresult = trim($output['ColumnB']);
    }
    $sumquery = mysqli_query($db, "SELECT SUM(ColumnC) as total FROM Table_Name WHERE ColumnB = '$columnBresult'");
    $row = mysqli_fetch_assoc($sumquery);
    $sum = $row['total'];
    echo $sum > 0 || $sum !== null ? $sum : "0";
?>

欢迎任何想法..

Answer 1

除了检查之外，$_SESSION['username']变量中还有任何用户名值。此外，您可以通过连接同一个表来实现单个查询的计算：

数据库表：

mysql> select * from table_name;
+---------+----------+---------+
| ColumnA | ColumnB  | ColumnC |
+---------+----------+---------+
| Bill    | Sales    |      15 |
| Helen   | Sales    |      32 |
| James   | Accounts |      20 |
+---------+----------+---------+
3 rows in set

查询：

SELECT
    SUM(t1.ColumnC) AS total
FROM
    table_name t1
    INNER JOIN table_name t2 ON t1.ColumnB = t2.ColumnB
WHERE
    t2.ColumnA = 'Bill'

结果：

+-------+
| total |
+-------+
| 47.00 |
+-------+
1 row in set

Answer 2

修正了它：

 <?php 

 session_start();

 $db = new mysqli('DBHost.com', 'DBUsername', 'DBPassword', 'DBName');
 if($db->connect_errno > 0){
 die('Unable to connect to database [' . $db->connect_error . ']');
 }

 $sql = mysqli_query($db, "SELECT ColumnB FROM Table_Name WHERE ColumnA = '".$_SESSION["username"]."'");
 $fetch = mysqli_fetch_assoc($sql);
 $dept = $fetch['ColumnB'];


 $result = mysqli_query($db, "SELECT SUM(ColumnC) AS total FROM Table_Name WHERE ColumnB = '$dept'");
 $row = mysqli_fetch_assoc($result);
 $sum = $row['total'];

 echo $sum > 0 || $sum!==null ? $sum : "0"; 
 ?>

这对我有用。

php mysqli sum基于与用户会话关联的列数据

2 个答案: