PHP problem here, I’ve made a login/logout kind code, with a insert and delete function that stand by login and logout.
So the problem is that after I insert the the text I simply cannot delete it, cause the delete button is like a simple
turn back botton, and doesn’t make his work, nothing
in the if(isset($_POST['delete'])) condition seems to work.
May the problem be that I’m using two void action
that refer to the same page? cause the first button
work and the second not.
Anyone can understand why?
<html>
<header></header>
<body>
<!-- START PHP -->
<?php
//If not submit i put the submit form
if(!isset($_POST['send'])){
echo "<form name='send' action='' method='POST'>
<input type='text' name='text' value=''/>
<input type='submit' name='send' value='send' />
</form>";
}<!-- IF END -->
//If submit was set I insert $text into the db and I render
//the delete button
else {
$conn= mysql_connect('localhost','root','');
mysql_select_db('db_try',$conn ) or die(mysql_error());
$dato=$_POST['dato'];
mysql_query(" INSERT INTO test (value) VALUES ('$text') ") or die(mysql_error());
echo "Operation complete";
//Now i render the delete submit button...
echo "<form name='delete' action='' method='POST'>
<input type='submit' name='delete' value='delete' />
</form>";
//...and if i push it NOTHING, like it's only
//a return to the first form button
if(isset($_POST['delete'])){
mysql_query(" DELETE FROM test WHERE value='$text' ") or die(mysql_error());
echo "<br>Text'".$text."' deleted";
}
}<!-- ELSE END-->
?><!-- END PHP -->
</body>
</html>
Here is the right way to do this, it is a quick tip, you need to work a little more on mysql insert security etc.