First off thanks in advance, everyone here has been a great help in the past and know you will be this time..

What i'm trying to achomplish is to submit data from a form into multipule tables in the Mysql DB. However I want the user to select which table the data gets inserted too by using check boxes..

I got the insert working but using the checkboxes are a little past my knowledge base.

Here is the insert code i have so far

<?php
require('FC_DB_connection.php');

//check if form has been submited
if(isset($_POST['submit'])){

//new code sent from form
 $code=$_POST['code']; 
 
 //to protect from mysql injections
 $code = stripslashes($code);

   
 //inserting data order   
 $order = mysql_query("SELECT * FROM imob WHERE code='$code'");
 		if(mysql_num_rows($order) > 0 ){
		
		echo "code already in DB";
		}
		else if(mysql_query("INSERT INTO table1 (code, time)   
             VALUES ('$code', NOW())"));
		   
 //declare in the order variable   
$result = mysql_query($order);  //order executes   
if($result){   
     echo("<br>New Code added");   
 } 
} 

 ?>

Form code:

<form name="addcode" action="<?php $_SERVER['PHP_SELF'];?>" method="POST">
 <br />
 <input name="code" type="text" value="Enter Code">
 <br />
 1:<input type="checkbox" name="1">
 2:<input type="checkbox"  name="2">
 3:<input type="checkbox" name="3">
 4:<input type="checkbox" name="4">
 <br />
 <input type="submit" name="submit" value="Add Code">
 </form>

so basically the data from the text box needs to be inserted into all the tables where a checkbox is checked.

Instead of having different names for your checkboxes, use a checkbox array and on submit, loop through all the selected checkbox and insert a record to the table.
For eg.,

<form name="addcode" action="<?php $_SERVER['PHP_SELF'];?>" method="POST">
 <br />
 <input name="code" type="text" value="Enter Code">
 <br />
 1:<input type="checkbox" name="checkbox" value="table1">
 2:<input type="checkbox"  name="checkbox" value="table2">
 3:<input type="checkbox" name="checkbox" value="table3">
 4:<input type="checkbox" name="checkbox" value="table4">
 <br />
 <input type="submit" name="submit" value="Add Code">
 </form>

Then, On submit,

<?php
require('FC_DB_connection.php');

//check if form has been submited
if(isset($_POST['submit'])){
//new code sent from form
$textbox_value = mysql_real_escape_string($_POST['code']);
foreach($_POST['checkbox'] as $tablename_value) {
$query = "insert into $tablename_value (col1) values ('$textbox_value')";
mysql_query($query);
}
..... etc....
 ?>

I tried the following code, it will echo each variable from the loop just fine but it won't insert the data into the database. Everything looks fine but maybe I missed something.

// sent from form
$textbox_value = ($_POST['code']);
$checkbox = ($_POST['checkbox']);
foreach($checkbox as $tablename) {
$query = "INSERT INTO $tablename (code, time) VALUES ('$textbox_value', NOW())";

$result = mysql_query($query);
if($result) {
     echo("code added to $tablename database <br />");
}
else
     echo "could not add code to $tablename database <br />";
}
?>

Print out the query, execute it in phpmyadmin, Or, simply add, or die(mysql_error()); ie.,

// sent from form
$textbox_value = ($_POST['code']);
$checkbox = ($_POST['checkbox']);
foreach($checkbox as $tablename) {
$query = "INSERT INTO $tablename (code, time) VALUES ('$textbox_value', NOW())";
$result = mysql_query($query) or die(mysql_error());
if($result) {
     echo("code added to $tablename database <br />");
}
else
     echo "could not add code to $tablename database <br />";
}
?>

This will make the code to print the error message on the screen if there is any.

My database connection was selecting the wrong DB. Its all working now Thanks for all your help.

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