我经营一个小型服务台，我需要每月向我的直线经理提供一份工单报告。
我希望升级我们生成此报告的方式，因为目前，我使用 Navicat 每月运行我手动编写的以下查询:
SELECT
updates.incidentid AS `Incident ID`,
updates.bodytext AS `Call Status`,
updates.duration AS `Total Minutes`,
software.`name` AS 'Support Type',
incidents.title AS Description,
contacts.forenames AS `First Name`,
contacts.surname AS `Last Name`,
FROM_UNIXTIME(incidents.opened, '%d.%m.%Y') AS `Date Logged`,
sites.`name`,
users.realname
FROM
updates
INNER JOIN incidents ON updates.incidentid = incidents.id
JOIN contacts ON incidents.contact = contacts.id
INNER JOIN sites ON sites.id = contacts.siteid
INNER JOIN users ON incidents.`owner` = users.id
INNER JOIN software ON incidents.softwareid = software.id
WHERE
updates.bodytext = 'Incident Closed'
AND FROM_UNIXTIME(incidents.opened, '%m') = '07'
AND FROM_UNIXTIME(incidents.opened, '%Y') = '2016'
ORDER BY contacts.siteid ASC
我想将这一切放在一个非常简单的网页中，允许用户选择月份和年份，然后为他们提供 HTML 格式的下载。
我最好将其放在 PHP 中还是可以用 HTML 创建它？请友善，我是一个新手，我的思维过程有点偏离。
非常感谢您的时间和耐心!
最佳答案
我将向您展示基本的:首先，一个 HTML 文件，用户在其中输入月份和年份作为数字(注意对“report.php”的调用):
<html>
<body>
<form method="post" action="report.php">
Enter month (1..12) <input type="text" name="month"/>
<br/>
Enter year (four digits) <input type="text" name="year"/>
<br/>
<input type="submit" value="Display report"/>
</form>
</body>
</html>
现在 PHP 文件“report.php”(注意如何在开头捕获月份和年份并将其存储在变量 $month 和 $year 中):
<?php
$month = $_POST["month"]; // PARAMETERS FROM
$year = $_POST["year"]; // THE HTML FILE.
$cnx = mysqli_connect( "localhost","user","password","databasename");
$data = mysqli_query( $cnx,"YOUR BIG SELECT HERE" ) or die( mysqli_error( $cnx ) );
echo "<table>" .
" <tr>" .
" <td>Incident id</td>" .
" <td>Call status</td>" .
" <td>Description</td>" .
" <td>Date logged</td>" .
" <td>Site name</td>" .
" <td>User</td>" .
" </tr>";
while ( $row = mysqli_fetch_array( $data ) ) // LOOP TO DISPLAY DATA.
echo "<tr>" .
" <td>{$row["IncidentID"]}</td>" .
" <td>{$row["CallStatus"]}</td>" .
" <td>{$row["Description"]}</td>" .
" <td>{$row["DateLogged"]}</td>" .
" <td>{$row["name"]}</td>" .
" <td>{$row["realname"]}</td>" .
"</tr>";
echo "</table>"; // TABLE END.
?>
您必须将文本“YOUR BIG SELECT HERE”替换为您的大查询。这是在查询中插入变量 $month 和 $year 的方法:
SELECT
updates.incidentid AS `Incident ID`,
updates.bodytext AS `Call Status`,
updates.duration AS `Total Minutes`,
software.`name` AS 'Support Type',
incidents.title AS Description,
contacts.forenames AS `First Name`,
contacts.surname AS `Last Name`,
FROM_UNIXTIME(incidents.opened, '%d.%m.%Y') AS `Date Logged`,
sites.`name`,
users.realname
FROM
updates
INNER JOIN incidents ON updates.incidentid = incidents.id
JOIN contacts ON incidents.contact = contacts.id
INNER JOIN sites ON sites.id = contacts.siteid
INNER JOIN users ON incidents.`owner` = users.id
INNER JOIN software ON incidents.softwareid = software.id
WHERE
updates.bodytext = 'Incident Closed'
AND FROM_UNIXTIME(incidents.opened, '%m') = '$month' ◄■■■■
AND FROM_UNIXTIME(incidents.opened, '%Y') = '$year' ◄■■■■
ORDER BY contacts.siteid ASC
将之前的代码复制粘贴到名为“report.html”和“report.php”的两个文件中，然后在浏览器中打开“report.html”。
有很多事情需要改进，例如，使用 jQuery，您可以改进用户输入月份和年份的方式，但这是另一个问题。
关于php - php 中的 sql 查询，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/38899864/