Not sure if this got through

kazario — Sun, 27 Mar 2011 16:20:27 +0000

Not sure if this got through the first time I posted it. In general, here's how you post data from one form into two tables (action script):

<?php
$dbhost="server_name";
$dbuser="database_user_name";
$dbpass="database_password";
$dbname="database_name";

$con=mysql_connect($dbhost, $dbuser, $dbpass) or die('Error connecting to the database:' . mysql_error());

$mysql_select_db($dbname, $con);

$sql="INSERT INTO table1 (table1id, columnA, columnB) 
         VALUES (' ', '$_POST[columnA value]','$_POST[columnB value]')";

mysql_query($sql);

$lastid=mysql_insert_id();

$sql2=INSERT INTO table2 (table1id, table2id, columnA, columnB)
              VALUES ($lastid, ' ', '$_POST[columnA value]','$_POST[columnB value]')";

//tableid1 & tableid2 are auto-incrementing primary keys 

mysql_query($sql2);

mysql_close($con);
?>

//this example shows how to insert data from a form into multiples tables, I have not shown any security measures

In general, here's how you

kazario — Sun, 27 Mar 2011 16:18:22 +0000

In general, here's how you post data from one form into two tables:

<?php
$dbhost="server_name";
$dbuser="database_user_name";
$dbpass="database_password";
$dbname="database_name";

$con=mysql_connect($dbhost, $dbuser, $dbpass) or die('Error connecting to the database:' . mysql_error());

$mysql_select_db($dbname, $con);

$sql="INSERT INTO table1 (table1id, columnA, columnB) 
         VALUES (' ', '$_POST[columnA value]','$_POST[columnB value]')";

mysql_query($sql);

$lastid=mysql_insert_id();

$sql2=INSERT INTO table2 (table1id, table2id, columnA, columnB)
              VALUES ($lastid, ' ', '$_POST[columnA value]','$_POST[columnB value]')";

//tableid1 & tableid2 are auto-incrementing primary keys 

mysql_query($sql2);

mysql_close($con);
?>

//this example shows how to insert data from a form into multiples tables, I have not shown any security measures

Hi, Still I have no Idea

sandeep Kumar — Thu, 08 Jul 2010 07:04:34 +0000

Hi, Still I have no Idea about that, But I am Sure that, I'll Solve your Problem as soon as possible.

hello gentlemen, I am

learjet60xr — Sat, 03 Jul 2010 06:14:30 +0000

hello gentlemen,
I am experiencing the same problem, I am trying to create an application with multiple tables for my business that unfortunately is not doing that well and unfortunately I do not have the funds to support a programmer. I have some and less than basic php knowledge but quite a good understanding of mysql. My problem is that I am trying to insert data into different tables for keeping better record of my customers, employees and so on. What I did is that I created a table called users in mysql, also I created another table called user_addresses for giving the users the ability to add multiple addresses. What I did in order to identify what user has what address is that I pretty much created a second Id that is connected with the table of users which means, if the "users" table has a primary id called "ut_id" and every user gets assigned with an automated id then on the table user_addresses I added the same ID after the primary key id of the user addresses table. Then every time I add a record to the table user addresses I can assign a user. The question is how can I make the php script to do that automatically. ie. if a user fills up the form, obviously there will be fields that will add the data on the table "2" which is for the addresses. How can I make the script to get that ID from the table one and add it to the table 2 as well so the data will be assigned to that user automatically? I am really sorry if I am bothering with that question but I really need your help here Thank you so much in advance though.

I have tried to have this

busman — Mon, 22 Dec 2008 17:07:02 +0000

I have tried to have this data posted on the database and it just wont go. I dont understand this at all?

The first table data gets posted but the rest dont. How could I improve this code.

<?php
   
        $self = $_SERVER['PHP_SELF'];
        $hotel_name = $_POST['hotel_name'];
        $street = $_POST['street'];
        $city = $_POST['city'];
        $zip = $_POST['zip'];
        $state = $_POST['state'];
        $country = $_POST['country'];
        $telephone = $_POST['tel'];
        $fax = $_POST['fax'];
        $website = $_POST['www'];
        $rest_name = $_POST['rest'];
        $roulette = $_POST['roulette'];
        $bj = $_POST['bj'];
        $punto_banco = $_POST['punto'];
        $poker = $_POST['poker'];
        $three_card_poker = $_POST['tcp'];
        $spanish = $_POST['spanish'];
        $craps = $_POST['craps'];
        $card_room = $_POST['card'];
        $slots = $_POST['slots'];
?>

Casino Name:
Street address:
City:
Zip:
State:
Country:
Telephone:
Fax:
Website:

Restuarants:

Casino Games (enter number of tables)
Slots:
Roulette:
BJ:
Punto Banco:
Poker:
Three Card Poker:
Spanish 21:
Craps:
Card Room:

<?php
     // connect to mysql
    $conn = @mysql_connect("localhost", "username", "password")
            or die("ERR: Conn");
            
     // select specified database
     $rs = @mysql_select_db("casino_list", $conn)
             or die ("ERR: DB");
            
    
     // create query
     $sql = "insert into casino_tbl (id, hotel_name, street, city, zip, state, country, telephone, fax, website)
              values ('', '".$hotel_name."', '".$street."', '".$city."', '".$zip."', '".$state."', '".$country."','".$telephone."', '".$fax."', '".$website."')";
    
    // execute query
     $rs = mysql_query($sql, $conn);
         
     $sql = "insert into restaurant_tbl (id, hotel_name, rest_name)
             values('', '".$hotel_name."', '".$rest_name."')";    
    // execute query
     $rs = mysql_query($sql, $conn);
     
     $sql = "insert into games_tbl (id, hotel_name, roulette, bj, punto-banco, poker, three-card-poker, spanish, craps, card-room)
             values('', '".$hotel_name."', '".$slots."', '".$roulette."', '".$bj."', '".$punto_banco."','".$poker."', '".$three_card_poker."', '".$spanish."', '".$craps."', '".$card_room."',)";
    
     // execute query
     $rs = mysql_query($sql, $conn);
     
     // confirm the added record details
     if($rs){echo("Record added: $hotel_name");}
     
?>

Thanks

Just another couple of

greg — Sat, 20 Dec 2008 23:56:00 +0000

Just another couple of notes.. As I cannot see your entire code and you may have only posted rough examples of what you have, this may be obsolete advice...but...

the problem with using or die("ERR: Connection"); in a mysql connect or table select code is it outputs a message to the screen and the script halts.
You shouldn't be using this in a public environment. Instead, a friendlier situation, you already use an @ before the code to stop any error outputting, so the next step is an error message you create from there being no data returned
EG - (taken from your code)

<?php
// execute query
$rs = mysql_query($sql, $conn);
    
// confirm the added record details
if($rs){
echo("Record added: $name $street $city $zip $state $country $telephone $fax $website");
}else{
echo "data could not be inserted, sorry, please try again blah etc";
}
?>

Or of course on mysql insert you can use mysql_affected_rows($rs), which returns a numerical value of the number of rows "affected" i.e. inserted, into a mysql table
Just make sure you check for 0 or less than 0 ('<=0') as sometimes mysql returns -1 (minus one) for no rows found/inserted.

And just adding to pr0gr4mm3rs good advice .. you need a mysql connection before using mysql_real_escape_string, so take that into account within your code, as currently your vars are set before your connection to the DB is done

So you need this

<?php
// connect to mysql
$conn = @mysql_connect("localhost", "user", "password")
or die("ERR: Connection");
?>

before this

<?php
 $variable = mysql_real_escape_string($variable);
?>

Also you don't have an else for this

<?php
if($name and $street and $city and $zip and $state and $country and $telephone and $fax and $website ) //ensure value exist
?>

so if those vars are not set, nothing will happen unless there is code underneath the } of the if
EG

<?php
}else{
echo "sorry, not all vars where set, please go back and input ALL fields";
}
?>

Or, send them back to the form page and tell them they didn't insert all fields

The following code only

pr0gr4mm3r — Sat, 20 Dec 2008 22:41:26 +0000

The following code only inputs data into one table. When I click Submit and all the fields in the form have been filled out the data is saved in the database, but if 1 field is left out the whole form is not recorded in the database, why?

It's because of this statement that checks the values before continuing:

if($name and $street and $city and $zip and $state and $country and $telephone and $fax and $website ) //ensure value exist

I would like to insert data into multiple tables. I have hit a road block and I cannot move forward.

If you want to enter the data in multiple tables, then you need to duplicate this line:

     // create query
     $sql = "insert into casino_tbl (id, name, street, city, zip, state, country, telephone, fax, website)
         values ('', '".$name."', '".$street."', '".$city."', '".$zip."', '".$state."', '".$country."','".$telephone."', '".$fax."', '".$website."')";

Replace casino_tbl with the other table you want to insert the data into.

Also, I should note that you should be escaping your data before inserting any data into your database. That first block of code you posted should be something like:

<?php
  
        $self = $_SERVER['PHP_SELF'];
        $name = mysql_real_escape_string($_POST['name']);
        $street = mysql_real_escape_string($_POST['street']);
        $city = mysql_real_escape_string($_POST['city']);
        $zip = mysql_real_escape_string($_POST['zip']);
        $state = mysql_real_escape_string($_POST['state']);
        $country = mysql_real_escape_string($_POST['country']);
        $telephone = mysql_real_escape_string($_POST['tel']);
        $fax = mysql_real_escape_string($_POST['fax']);
        $website = mysql_real_escape_string($_POST['www']);
?>

Welcome to the forums.