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

[indhuma]

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.

<?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]

Hi.

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

$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]

$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:
[PHP]$queryresult = mysql_query (

SELECT day, category, title, article
FROM news
WHERE category = '$category'

)
or die(mysql_error());[/PHP]
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.