Using Forms in PHP-select option (Drop down list)

Hi,
I'm using the following code to show the category = thoughts. I have three categories, when i select thoughts only it has to show the thoughts category. If i select other category it has to show that particular category details. I don't know how i have to mention the particular category. Please help me.

Code: ( text )

<?php
if(isset($_POST['submit']) && ($_POST['submit']=="GO"))
{
$sqlquery2="select * from news where category like '%thoughts%'";
$queryresult2 = mysql_query($sqlquery2) or die(mysql_error());
while ($arr2=mysql_fetch_array($queryresult2))
{
echo $arr2["day"]."<br>";
echo $arr2["category"]."<br>";
echo $arr2["title"]."<br>";
echo $arr2["article"]."<br>";
}
}
?>
<form action="test2.php" method="post">
<hr>
<h4>"Category"</h4>
Select your Category:
<select name="category">
<option value="thoughts">thoughts</option>
<option value="travel">Travel</option>
<option value="important_events" selected="selected">Important_Events</option>
</select>
<input type="submit" name="submit" value="GO" />
</form>
</body>
</html>

Atli

Moderator

1,917 Posts

May 15th, 2008
07:25 AM
#2

Re: Using Forms in PHP-select option (Drop down list)

Hi.

Would it not work to simply put the option you selected into the query?

Code: ( text )

$category = $_POST['category'];
$sqlquery2="select * from news where category like '%{$category}%'";

Also...
Please use [code] tags when posting your code examples. (See How to ask a question)

[code=php] ...PHP code goes here... [/code]

Thanks.

TheServant

Needs Regular Fix

462 Posts

May 15th, 2008
10:17 PM
#3

Re: Using Forms in PHP-select option (Drop down list)

Quote:

Originally Posted by indhuma

Code: ( text )

$sqlquery2="select * from news where category like '%thoughts%'";
$queryresult2 = mysql_query($sqlquery2) or die(mysql_error());

Also, do you need all (*) of your information? You could replace the above by:

Code: ( text )

$queryresult = mysql_query (
SELECT day, category, title, article
FROM news
WHERE category = '$category')
or die(mysql_error());

This is probably more efficient I believe. Also, you can use some MySQL coding like I have done. I have capitals for SQL commands and then space it out appropriately for ease of reading.