why does this code run twice ????

sander_ESP

hello,

i wrote some code to enter some date into a datebase.

<?
ob_start();
$host = "******";
$user = "******";
$pass = "******";
$db = "*******_cooking";

$conn = mysql_connect($host, $user, $pass)
OR DIE (mysql_error());
@mysql_select_db ($db, $conn) OR DIE (mysql_error());

echo $date_ID;



$sqldate ="select date from calender where cal_ID =".$date_ID;
$result=mysql_query($sqldate);

$sqlID  ="select cal_ID from calender where cal_ID =".$date_ID;
$resultID=mysql_query($sqlID);
list ($listID) = mysql_fetch_array ($resultID);
list ($resulttest) = mysql_fetch_array ($result);




$sqlIDinsert  = " INSERT INTO details "; 
$sqlIDinsert .= " (cal_ID, timestamp) VALUES "; 
$sqlIDinsert .= " ('$listID', NOW()) "; 		


    #execute SQL statement 
    $resultIDinsert = mysql_query($sqlIDinsert); 




if ($_SERVER['REQUEST_METHOD'] == "POST")  

    { 
        # escape data and set variables 
        $Datecooking = addslashes($_POST["date"]); 
        $people = addslashes($_POST["people"]); 
        $datebooking = addslashes($_POST["booking"]); 
   		$who = addslashes($_POST["who"]); 
		$price = addslashes($_POST["price"]);
		$cal_ID = addslashes($_POST["cal_ID"]);


$sqlselect="Select * from details WHERE timestamp = max(timestamp)";
mysql_query($sqlselect);

$sqlupdate ="UPDATE details SET price='$price', date='$datebooking', number='$people', who='$who'"; 
mysql_query($sqlupdate);


$sqlsoldadapt = "Select sold from calender where cal_ID =".$date_ID;
$soldadaptresult= mysql_query($sqlsoldadapt);
$soldadapt=mysql_result($result,$i,"sold");
$newsold = $soldadapt + $people;

$sqlnewsold =" UPDATE calender SET sold='newsold' WHERE cal_ID=".$date_ID;
mysql_query($sqlnewsold);


$URL="http://www.***********.com/bookings/";
header ("Location: $URL");
ob_end_flush();





    # check for error 
    if (mysql_error()) { print "Database ERROR: " . mysql_error(); } 

}
?>

<form name="fa" action="add.php" method="POST">      

    <table>                                                                                                         

    <tr><td><b>DATE of cooking class</b> </td><td><? echo $resulttest ?></td></tr>
    <tr><td><b>number of people </b> </td><td><input type="text" name="people" size=2></td></tr>
	  <tr><td><b>Price</b> </td><td><input type="text" name="price" size=2></td></tr>
    <tr><td><b>date of booking</b> </td><td><input type="text" name="booking" size=10></td></tr> 
    <tr><td valign=top><b>booked by </td><td> Travelbar <INPUT NAME="who" TYPE=radio VALUE="travelbar">
The Shop<INPUT NAME="who" TYPE=radio VALUE="The Shop">
</td></tr> 
    <tr><th colspan=2><p><input type="submit" value="Add Record"></p></th></tr>  

    </table> 
    </form>

however, this code runs twice and therefor i get two entries into the database instead of one..

the page gets loaded when people click on a link on another page

<A HREF='add.php?date_ID=$date_ID&action=add'>Add

i have no idea why its runs twice though..

Piranha

I can't see how it could be inserted twice. However I see a few other things that should be fixed, the first one could cause the problem you describe if you have only 2 rows in the database.

You miss a WHERE-part in the first UPDATE query. It will cause all rows in the table to be updated. If it is as I guess that you want to update the record you just made then it is a better idea to insert everything in the record at once, no need to update it at all. And use [man]mysql_last_id[/man] to get the row, not max(timestamp) since there may have been something inserted after your insert.

Don't use short tags. Use the long version instead. In other words, use <?php instead of <?. One reason to do it is that short tags will not be supported from PHP 6.

Don't use addslashes to try and make the variables safe for database injection. It is to general, meaning that it will cause troubles and are not safe. Use the right way according to your database, with mysql it is [man]mysql_real_escape_string[/man].

sander_ESP

well the idea of the first insert is to enter a new record in the database with the right cal_ID. and then update this with the information submitted through the form.

i tried doing both things in one query, but for some reason the cal_id variable gets emptied when users click on SUBMIT. therefor i get a cal_ID of 0 in the database and not the value that is set on the other page via

<A HREF='add.php?date_ID=$date_ID&action=add'>Add

so i came up with this workaround to first insert the cal_id and then update the other values in this entry..

it works except for the fact that i get two entries in the database, one with all the values and the right cal_id and one with all the values and a cal_id of 0.

sander_ESP

ok i have adapted the code

its like this now

<?PHP
ob_start();
$host = "localhost";
$user = "*********";
$pass = "**********";
$db = "***********";

$conn = mysql_connect($host, $user, $pass)
OR DIE (mysql_error());
@mysql_select_db ($db, $conn) OR DIE (mysql_error());

echo $date_ID;



$sqldate ="select date from calender where cal_ID =".$date_ID;
$result=mysql_query($sqldate);

$sqlID  ="select cal_ID from calender where cal_ID =".$date_ID;
$resultID=mysql_query($sqlID);
list ($listID) = mysql_fetch_array ($resultID);
list ($resultID) = mysql_fetch_array ($result);

echo "listID = $listID";

echo "resultid = $resultID";







if ($_SERVER['REQUEST_METHOD'] == "POST")  

    { 
        # escape data and set variables 
        $Datecooking = ($_POST["date"]); 
        $people = ($_POST["people"]); 
        $datebooking = ($_POST["booking"]); 
   		$who = ($_POST["who"]); 
		$price = ($_POST["price"]);
		$cal_ID = ($_POST["cal_ID"]);


$query = "INSERT INTO details"
. "(cal_ID, price, date, number, who) ";
$query.= "VALUES (";
$query.= "'$listID', '$price', '{$datebooking}', '$people', '$who')";
$answer = mysql_query($query);


$sqlsoldadapt = "Select sold from calender where cal_ID =".$date_ID;
$soldadaptresult= mysql_query($sqlsoldadapt);
$soldadapt=mysql_result($result,$i,"sold");
$newsold = $soldadapt + $people;

$sqlnewsold =" UPDATE calender SET sold='newsold' WHERE cal_ID=".$date_ID;
mysql_query($sqlnewsold);


$URL="http://www.***********.com/bookings/";
header ("Location: $URL");
ob_end_flush();





    # check for error 
    if (mysql_error()) { print "Database ERROR: " . mysql_error(); } 

}

?>

<form name="fa" action="add.php" method="POST">      

    <table>                                                                                                         

    <tr><td><b>DATE of cooking class</b> </td><td><? echo $resulttest ?></td></tr>
    <tr><td><b>number of people </b> </td><td><input type="text" name="people" size=2></td></tr>
	  <tr><td><b>Price</b> </td><td><input type="text" name="price" size=2></td></tr>
    <tr><td><b>date of booking</b> </td><td><input type="text" name="booking" size=10></td></tr> 
    <tr><td valign=top><b>booked by </td><td> Travelbar <INPUT NAME="who" TYPE=radio VALUE="travelbar">
The Shop<INPUT NAME="who" TYPE=radio VALUE="The Shop">
</td></tr> 
    <tr><th colspan=2><p><input type="submit" value="Add Record"></p></th></tr>  

    </table> 
    </form>

it doesnt work! there is 2 problems.. the first is dat the cal_id is not entered into the database.. instead the default value of 0 seems to be inserted..

the variable $listID seems to get reset when submit is pressed.

the second problem is that all the values in the database get overwritten with the last form values...

help!!! its driving me insane..