User input in to variable


Posted on 16th Feb 2014 07:03 pm by admin

Hi all,

I'm sure this is very easy but I'm having another brain freeze!

Did you know?Explore Trending and Topic pages for more stories like this.
At the end of the code I have echoed 3 bits of info to screen $ip, $_POST['ssid'] and $ssid_id".

I would like to put $_POST['ssid'] into a variable, is this possible?

Jamie

Code: <?php
//run site query
require("confwifi.php") ;
{
$ext_site = $_POST['site'];

$extract = mysql_query ("SELECT * FROM site WHERE site='$ext_site'");
$numrows = mysql_num_rows ($extract);

while ($row = mysql_fetch_assoc($extract))
{
$id = $row['id'];
$site = $row['site'];
$ip = $row['ip'];
}

}

?>

<?php
//run site query
require("confwifi.php") ;
{
$ext_ssid = $_POST['ssid_name'];

$extract = mysql_query ("SELECT * FROM ssid WHERE ssid_name='$ext_ssid'");
$numrows = mysql_num_rows ($extract);

while ($row = mysql_fetch_assoc($extract))
{
$ssid_id = $row['ssid_id'];
$ssid_name = $row['ssid_name'];

}

}

?>


<form name="form1" method="POST" action="wifichange.php"><div>

<select name="site">
<option value="0" selected>(please select:)</option>
<option name="site">Llandaf</option>
<option name="site">Cyncoed</option>
<option name="site">Howard Gardens</option>
<option name="site">Colchester Avenue</option>
</select>


<select name="ssid_name">
<option value="0" selected>(please select:)</option>
<option name="ssid_name">Guest</option>
<option name="ssid_name">Conferences</option>
</select>
</form>
<p><input type="text" name="ssid"></div></P>

<div><input type="submit" value="submit"></div>
</form>




<?php echo "$ip"; ?><p></p>

<?php echo $_POST['ssid'];; ?><p></p>

<?php echo "$ssid_id"; ?><p></p>

<?php
No comments posted yet

Your Answer:

Login to answer
258 Like 41 Dislike
Previous forums Next forums
Other forums

Creating a unique 'control panel' for each user
Hi there,

I'm thinking of designing a site that will allow users to sign up and have their ow

Syntax Help
Code:


im having trouble with that code snipped
Parse error: syntax error, unexpec

Echoing a Variable from a Object
How do I get a variable from the new User Class to echo out in this clasS?

class MyApp
{

Simpler method of getting variables from mysql
Hi Guys,

I'm trying to streamline my CMS's code and as I was writing a new page it occured to

Two warning messages
Quote<b>Warning</b>: mysql_real_escape_string() expects parameter 1 to

iterating through an array and escape each value independently.
I have a set up where the variable being escaped is an array and it needs to be iterated and escaped

problem with query error
First Thanks to those who helped me on my previous posts, and the following code i'm using is not mi

Security Exception on pages using AJAX
I am getting the exception: attempted to perform an operation not allowed by the security policy on

Getting a variable to work in function params
I have this fuction which is inside a class:

Code: public static function generateEmbedCode($

php not working written for consumption of slots
1. Here is the function where i will be allowed to consume the slot booked by me.

2. Here i c

Sign up to write
Sign up now if you have flare of writing..
Login   |   Register
Follow Us
Indyaspeak @ Facebook Indyaspeak @ Twitter Indyaspeak @ Pinterest RSS



Play Free Quiz and Win Cash