Question

这是我的php脚本，它从invoiceNo中选择所有内容，其中invoiceNo是不同的。

<?php

require 'init.php';
$query = 'SELECT * FROM  `selected_items` WHERE invoiceNo IN ( SELECT DISTINCT ( invoiceNo) AS invoiceNo FROM selected_items ) GROUP BY invoiceNo;';
$res = mysqli_query($con, $query);
$result = [];
while ($row = mysqli_fetch_array($res)) {
    array_push($result, [
        'custInfo'    => $row[0],
        'invoiceNo'   => $row[1],
        'barcode'     => $row[2],
        'description' => $row[3],
        'weight'      => $row[4],
        'rate'        => $row[5],
        'makingAmt'   => $row[6],
        'net_rate'    => $row[7],
        'itemTotal'   => $row[8],
        'vat'         => $row[9],
        'sum_total'   => $row[10],
        'bill_type'   => $row[11],
        'date'        => $row[12],
        'advance'     => $row[13],
        'balance'     => $row[14],
    ]);
}
echo json_encode(['result' => $result]);
mysqli_close($con);

现在这个脚本给了我sum_total的第一个值，即它给了我数据库的第一行，我怎么能得到最后一行。我是新手编程的任何建议或帮助表示赞赏。谢谢：）

Answer 1

Select * From (
SELECT t.*, 
       @rownum := @rownum + 1 AS rank
  FROM selected_items t, 
       (SELECT @rownum := 0) r order by rank DESC
) si GROUP BY si.invoiceNo;

此查询解决了我的问题

Answer 2

尝试这样：

$query ="SELECT * FROM  `selected_items` WHERE invoiceNo IN ( SELECT DISTINCT ( invoiceNo) AS invoiceNo FROM selected_items ) ORDER BY `sum_total` DESC";
$query ="SELECT max( `sum_total` ) FROM selected_items";

column_name可以变化的地方。

Answer 3

尝试这个，我认为这是你想要的，可能有所帮助

$query ="SELECT max( `sum_total` ) FROM `selected_items` GROUP BY invoiceNo;";

Answer 4

如果您只需要获取最后一次使用记录限制。

$query ="SELECT * FROM  `selected_items` GROUP BY invoiceNo ORDER BY `sum_total` DESC limit 1;

如果您需要获得最高至最低的sum_total记录，请尝试以下代码，

$query ="SELECT * FROM `selected_items` where `sum_total` = (SELECT max( `sum_total` ) FROM `selected_items` GROUP BY invoiceNo) GROUP BY invoiceNo ORDER BY `sum_total` DESC;

如何使用php从mysql数据库访问特定值？

4 个答案: