这是我的php脚本,它从invoiceNo中选择invoiceNo不同的所有内容.
<?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);
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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.我是新编程的任何建议或帮助表示赞赏.谢谢:)
小智 2
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;
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这个查询解决了我的问题
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