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Connecting two mysql tables in PHP script

 

Connecting two mysql tables in PHP script

neeno

26 Aug, 2008 - 03:23 AM
Post #1

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Joined: 14 Aug, 2008
Posts: 2

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Hello people.. its my first post on this gr8 forum, i hope i placed it fine.
Anyway to the point, i know i aint much of an artist thats why i will try to explain the picture i posted.
So i have this table with events wich is shown to users, i made it so that certain event shows following:
[name of event] [date][time][details] , [name of event] is clickable and takes you to details site wich cointas a sign up textboxes. Now what do i need. I need a line of MySQl code wich will somehow bound the users who signed up to that specific event they clicked. I'm really a begginer, i wrote all this alone with a help of this book i bought and i kinda got stuck at this point.

I hope everything is somehow clear. Thanks for help.

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CTphpnwb

RE: Connecting Two Mysql Tables In PHP Script

26 Aug, 2008 - 05:48 AM
Post #2

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My Contributions
When the user clicks, they're taken to another page, or the page is refreshed. You can use $_POST or $_GET to transfer information about what was clicked (and other information) to the new/refreshed page. Show us your code and we can offer suggestions as to what you want to post.


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neeno

RE: Connecting Two Mysql Tables In PHP Script

26 Aug, 2008 - 06:19 AM
Post #3

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ok first i got "admin" site where events are generated:

CODE
<?php

/**
* @neeno
* @copyright 2008
*/
//database connecting
mysql_connect("xx, "xx", "xx") or die(mysql_error());
mysql_select_db("xx") or die(mysql_error());
//checking fields and running ze code
if (isset($_POST['submit'])) {

//This makes sure they did not leave any fields blank
if (!$_POST['type'] | !$_POST['date'] | !$_POST['time'] | !$_POST['link'] | !$_POST['details'] ) {
die('You did not complete all of the required fields');
}
//inserting the event
$insert = "INSERT INTO dogodki (type, date, time, link, details)
VALUES ('".$_POST['type']."', '".$_POST['date']."','".$_POST['time']."','".$_POST['link']."','".$_POST['details']."')";
$add_member = mysql_query($insert);
}
?>


<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
        <title>Untitled Document</title>
    </head>
    <body>
    <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
    <fieldset>
        <legend>Data</legend>
            <p>
                <label>Type of event</label>
                <select name= "type">
                    <option>GvG</option>
                    <option>TA</option>
                    <option>HA</option>
                    <option>AB</option>
                    <option>Guild meeting</option>
                    </select>
            </p>
            <p>
                <label>Date of event</label>
                <input type = "text" name="date"/>
            </p>
            <p>
                <label>Time of event</label>
                <input type = "text" name="time"/>
            </p>
            <p>
                <label>Link to the build</label>
                <input type = "text" name="link"/>
            </p>
            <p>
                <label>Details</label>
                <textarea name = "details" rows = "10" cols = "30"></textarea>
            </p>
            <p>
            <input type="submit" name="submit" value="Add"/>
            </p>
    </fieldset>
    </form>
        
    </body>
</html>


Then i have the site where these generated events are shown:

CODE

<html>
<head>
</head>
<body>
<?php

/**
* @author
* @copyright 2008
*/

$conn = mysql_connect("xxx", "xxx", "xxx") or die (mysql_error());
mysql_select_db("xxx");

$sql = "SELECT type FROM dogodki";
$rezultat = mysql_query($sql, $conn) or die (mysql_error());
$sql1 = "SELECT date, time  FROM dogodki";
$rezultat1 = mysql_query($sql1, $conn) or die (mysql_error());
while($row = mysql_fetch_assoc($rezultat)){
    $row2 = mysql_fetch_assoc($rezultat1);
    foreach ($row as $name => $value){
    print "<a href = 'details.php'>$name: $value</a> ";}
    foreach ($row2 as $name => $value){
    print "$name: $value ";
    
    }
    print "<br/>";
    }
    
?>
</body>
</html>


These shown events are clickable and takes you to the next site:

CODE

<html>
<head>
</head>
<body>
<?php
$conn = mysql_connect("xxx", "xxx", "xxx") or die (mysql_error());
mysql_select_db("xxx");

$sql = "SELECT details FROM dogodki";
$rezultat = mysql_query($sql, $conn) or die (mysql_error());
while($row = mysql_fetch_assoc($rezultat)){
    foreach ($row as $name => $value){
    print "$name: $value ";
    
    }
    print "<br/>";
    }
    
?>

<form action="<?php echo $_SERVER['signup.php']; ?>" method="post">
    <fieldset>
        <legend>Sign up</legend>
        <p>
                
                <input type = "text" name="signup"/>
                <label>Sign up</label>
                
            </p>
        <p>
            
            <input type = "text" name= "late"/>
            <label>Minutes Late</label>
        </p>
        <p>
            <input type="submit" name="submit" value="Sign Up"/>
            </p>
    </fieldset>
    </form>
</body>


Here the users are suppose to sign up to the event they clicked. And this is where i got stuck.. i need this form in details.php to be submited and saved into database and the data has to be somehow connected to the events.
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CTphpnwb

RE: Connecting Two Mysql Tables In PHP Script

26 Aug, 2008 - 02:49 PM
Post #4

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My Contributions
First, NEVER insert a $_POST directly into the database. That's a huge security hole.

Second, your code is all over the place. You need to organize it much more efficiently. There is nothing I see that would prevent this all from being on one page, and that would have the side benefit of causing you to see some of the obvious problems sooner.


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