单击链接将值提交到SQL SELECT

Question

I am trying to get this to allow the user to click on a link, which will then change the value of what is submitted in the SQL SELECT Statement, if that makes sense.
My Code so far:

<div class="Tabs" >
<?php
require 'database/connect.php';
$sql = "SELECT DISTINCT Week FROM PMWUpdates";

$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {

echo '<a class="weeks"> Week' . $row{'Week'} . '</a>';
}
?>
</div>

<div class="Pages">
<?php
require 'database/connect.php';
$sql = "SELECT * FROM PMWUpdates WHERE Week='1'";
?>
</div>

I want the value of SELECT * FROM PMWUpdates WHERE Week="'1'"; to be what link the user has clicked on above, not just always 1. So if the user clicks on the link 2, the SQL changes to SELECT * FROM PMWUpdates WHERE Week="'2'";

I am basically trying to accomplish this , but instead, I want to display the number of tabs, based on the number of weeks in a database, and then, each week will display data from the database about that particular week.

Any help is appreciated.

Thank You

Answer 1

Your statement should be like this...

$sql = "SELECT DISTINCT Week FROM PMWUpdates";

$result = mysql_query($sql);

while ($row = mysql_fetch_array($result)) {
    echo '<a class="weeks" href="$row['Week_link']"> Week' . $row['Week'] . '</a>';
}

Answer 2

If you mean dynamically changing data on a page which requires PHP to be processed, you have to take a look at AJAX. You can't do it this way, once your script is loaded, there is no more php, only HTML. If you don't want to use AJAX, you will have to do it in 2 pages. First display links, then display your data related to whatever was selected.

Answer 3

The MySQL extension you are using is deprecated (>= 5.5). Use MySQLi or PDO instead.

What you are trying to accomplish is that where your "2" is listed, should change with the use of a variable. As you want to do it via Javascript, you will need a click event on those links which then submits the variable via AJAX.