add data in database to a specific user

[amira90]

i'm trying to add data into specific user in database. the one who login in the form is admin and the admin will add information to the user's database..

 

here's the php code

 

 

<?php

session_start();

 

if ($_SESSION['UserId'] == "") {

 

echo "Please Login first";

 

} else {

 

echo "<h1> Appliances's adding list</h1>";

 

echo "<table>";

 

echo "<form action ='addappliancesprocess.php' method = 'POST'>";

 

echo "<tr>

 

<td>Appliances Name:

<input type='text' name='appliancesname'></td>

</tr>";

 

 

echo "<tr>

 

<td>Port Number: <select name = 'port' value='port'>

    <option value 'RB0'>RB0</option>

    <option value 'RB1'>RB1</option>

    <option value 'RB2'>RB2</option>

    <option value 'RB3'>RB3</option>

    <option value 'RB4'>RB4</option>

    <option value 'RB5'>RB5</option>

    <option value 'RB6'>RB6</option>

    <option value 'RB7'>RB7</option>

</select>

</tr>";

 

echo "<tr>

<td>Insert characters on:

<input type='text' name='charon'></td>

</tr>";

 

echo "<tr>

<td>Insert characters off:

<input type='text' name='charoff'></td>

</tr>";

 

echo "<tr>

<td colspan=2 align= 'right'><br>

<INPUT Type='button' VALUE='Back' onClick='history.go(-

 

1);return true;'>

<input type='submit' name='update' value='Submit'>

 

</td>

</tr>";

 

echo "</table>";

 

}

 

?>

 

 

and here is the process

 

 

<?php

session_start();

 

$appliancesname=$_POST['appliancesname'];

$port=$_POST['port'];

$username=$_REQUEST['username'];

$charon=$_POST['charon'];

$charoff=$_POST['charoff'];

 

$con = mysql_connect("localhost","root","");

 

if (!$con)

  {

 

  die('Could not connect: ' . mysql_error());

 

  }

 

mysql_select_db("myfyp", $con);

?>

 

<?php

 

$query="SELECT from user where UserId = '".$_SESSION ['UserId']."'";

 

//if ($username != UserId) {

 

//echo "no user";

 

//}else{

 

 

$sql="INSERT INTO appliances

 

VALUES

 

('','$appliancesname','$port','$username','','$charon','$charoff')";

 

echo $query;

 

if (!mysql_query($sql,$con))

  {

  die('Error: ' . mysql_error());

  }

 

 

echo "You have choose the following:<br>";

 

echo "<br>Appliances: $appliancesname</br>";

 

echo "<br>Port Number: $port</br>";

 

echo "<br>Character on: $charon</br>";

 

echo "<br>Character off: $charoff</br>";

 

echo "<br><INPUT Type='button' VALUE='Back' onClick='history.go(-

 

1);return true;'>";

//}

 

?>

 

 

when i try to add it will add to the database but the port is not updated and the userid is empty.. 

 

i'm sorry.. i am new to php.. there are lots that i don't know..

 

[lemmin]

$_REQUEST['username'] does not appear to be set and your HTML for the options' values is missing the equal signs:

<option value 'RB0'>RB0</option>

Needs to be:

<option value='RB0'>RB0</option>

[amira90]

Thank you.. Thank you for your help.. 

 

the port is already updated but the userID field is still empty.. my database look like this.. btw the AppliancesID should be increment but it also empty.. 

 

Edited May 1, 2013 by amira90

[lemmin]

I'm assuming that the UserId field is right after the username field, in which case, change your VALUES list to this:

('','$appliancesname','$port','$username','".$_SESSION['UserId']."','$charon','$charoff')";

[amira90]

its says like this..

 

SELECT from user where UserId = 'admin'Error: Duplicate entry '' for key 'PRIMARY'

 

i'm sorry.. my english is not good.. but is it possible for admin(who is the one who login) to add data to the user?  i want the query to be like 

SELECT from user where UserId = 'user' 

 

and i think because i use session so the query will always turned to be 

SELECT from user where UserId = 'admin'

[lemmin]

You are never using your SELECT query. You are just echoing it out and never running it. It would fail anyway, so you can just remove it. The query in question is the INSERT query. Duplicate entry means your are trying to insert an ID (I'm assuming AppliancesID is your primary key) that is already in the database.

 

In your case, that duplicate is likely to be the empty id. Change the AppliancesID field to use the AUTO_INCREMENT flag.