一旦选择并将其值传递给查询，如何在特定列中获取值？

I have a table and show its record in my page, then i put a link in one of the data in column like this. I made the drawing_part_number column as link:

我有一个表,并在我的页面中显示其记录,然后我在列中的一个数据中放置一个链接,如下所示。我将drawing_part_number列作为链接:

my problem is, how can I get the value of the drawing_part_number so I will be able to pass it in a query.

我的问题是,如何获取drawing_part_number的值,以便我能够在查询中传递它。

Thanks for your help. here is my code.

谢谢你的帮助。这是我的代码。

echo "<tr ".$row_color.">";
            echo '<td>' . $row['drawing_type'] . '</td>';
            echo '<td>' . $row['die_type'] . '</td>';
            echo '<td>' . $row['relay_type'] . '</td>';
            echo '<td>' . $row['die_name'] . '</td>';
            echo '<td>' . $row['part_name'] . '</td>';
            echo '<td>' . $row['drawing_number'] . '</td>';
            echo '<td><a href="show_versions.php">' . $row['drawing_part_number'] . '</td>';
            echo '<td>' . $row['sub_letter'] . '</td>';
            echo '<td>' . $row['specs'] . '</td>';
            echo '<td>' . $row['revision'] . '</td>';

2 个解决方案

#1

echo '<td><a href="show_versions.php?part='.urlencode($row['drawing_part_number']).'">' . $row['drawing_part_number'] . '</td>';

#2

I believe your trying to pass $row['drawing_part_number'] as a parameter to your show_versions.php

我相信你试图将$ row ['drawing_part_number']作为参数传递给你的show_versions.php

Do as follows.

做如下。

echo '<td><a href="show_versions.php?drawing_part_number=' . urlencode($row['drawing_part_number']) . '">' . $row['drawing_part_number'] . '</td>';

#1

echo '<td><a href="show_versions.php?part='.urlencode($row['drawing_part_number']).'">' . $row['drawing_part_number'] . '</td>';

#2