PHP - Getting Mysql_insert_id() From Multiple Inserts

Hi all.

Im trying to insert from one form 3 items of data into to different tables.  I will post the code as most of my notes are in there but first.......

What is happening is the form is filled in then the first part of the form where it updates my clubs table with category ID's works fine. Nothing in the
if ($result) {   }
seems to run.
I have create some fake groups so the drop down is populated with groups already created and I have tried to select one of these groups so it can be entered into the clubs table and nothing.  No errors just no answers.  Does anybody no what I am doing wrong?


first Ill post the form and the querys that make fields in the form... 

Querys

//gets
$validation = $_GET['new_club'];
//Querys
    $qGetClub = "SELECT * FROM clubs WHERE validationID = '$validation'"; 
$rGetClub = mysql_query($qGetClub);
$Club = mysql_fetch_array($rGetClub);
//Query for category by name
$qGetCat = "SELECT * FROM club_category WHERE catID = ".$Club['cat']."";
$rGetCat = mysql_query($qGetCat);
$CatName = mysql_fetch_array($rGetCat);
//query for related sub categorys.
$qGetSub = "SELECT * FROM sub_categorys WHERE catID =".$Club['cat']."";
$rSubCat = mysql_query($qGetSub);
// query for groups created
$Groupq = mysql_query("SELECT * FROM groups WHERE memberID = '".$User['memberID']."'");



  then we have the form


<form action="" method="post"  id="insert_clubfrm">
  
                        <tr valign="baseline">
                                <td colspan="2"  valign="top"><br/><p>Your Chosen category was <strong><?php echo $CatName['categorys'];?></strong> Please select a sub category.</p><br></td>
                              </tr>
                              
                              <tr valign="baseline">
                                <td width="132"  valign="top" >Club Sub Category:</td>
                                <td width="256"  valign="top"  ><select name='subcat'>
                            <option value="None">Please Select One</option>
<?php while ($New_SubCat = mysql_fetch_assoc($rSubCat)) { ?>

                                <option value="<?php echo $New_SubCat['subID']; ?>"><?php echo $New_SubCat['sub_categorys']; ?></option>"<?php  }  ?>

</select>
  </td>
                              </tr>
                              <tr valign="baseline">
                                 <td colspan="2"  valign="top" ><br/><p>If your sub category does not exist in the list please select "none" above and enter your new sub categroy below.</p><br></td>
                              </tr>
                              <tr valign="baseline">
                                <td  valign="top" class="left_padding">New Sub Category</td>
                                <td valign="top" class="left_padding"><input type="text" name="newSubCat" id="new_cat" value="" /></td>
                              </tr>
                              <tr valign="baseline">
                                <td colspan="2"  valign="top" class="left_padding"><br/>
                                 If you had/have more than one site and you had to give your group of companies a name what would that name be?
                                <br/></td>
                              </tr>
                              <tr valign="baseline">
                                <td  valign="top" class="left_padding">Group Name:</td>
                                <td  valign="top" class="left_padding"><label for="other_groups"></label>
                                  <select name="other_groups" id="other_groups">
                                  <option>None</option>
                                  <?php while ($group = mysql_fetch_assoc($Groupq)) { ?>
                                    <option value="<?php echo $group['groupID']; ?>"><?php echo $group['group']; ?></option>"<?php  }  ?>  
                                </select></td>
                              </tr>
                              <tr valign="baseline">
                                <td  valign="top" class="left_padding">Group Name:</td>
                                <td  valign="top" class="left_padding"><label for="new_groups"></label>
                                 <input type="text" name="new_group" value="">

  
</td>
                              </tr>
                                                        <tr valign="baseline">
                                <td valign="top" nowrap="nowrap">&nbsp;</td>
                                <td valign="top"><div align="left">
                                  <input type="submit" id="insert_clubbtn1" name="insert_clubbtn1" value="Insert record" />
                                </div></td>
                              </tr>
                            </table>
                            </form>






now the code to make the form work...



if(isset($_POST['insert_clubbtn1'])){ 
//Process data for validation     
$subcat     = trim($_POST['subcat']);  
$NewSubCat = trim($_POST['NewSubCat']); 

$otherg = trim($_POST['other_groups']);
$newg = trim($_POST['new_groups']);     
//Prepare data for db insertion      
$newg = mysql_real_escape_string($newg);  
$subcat = mysql_real_escape_string($subcat); 
//find the new category 
//insert
$result = mysql_query("UPDATE clubs  SET 
`sub_category` = '$subcat'
WHERE validationID ='$validation'") or die(mysql_error());
if ($result) {
//if an item other than none from the list is selected then update the club with an ID relating to the group it belongs to
if ($otherg !=='None') { $groupsq = mysql_query("UPDATE `clubs` SET groupID ='$otherg' WHERE validationID ='$validation')") or die (mysql_error()); }
// If none is selected then $newg must have a value so create a new group in the groups table and then on the next page I will add the group in the club table 
if ($otherg == 'None') {
$groupsq = mysql_query("INSERT INTO `groups` (memberID, group, clubID) VALUES ('".$User['memberID']."', '$newg', ''".$Club['clubID']."')");
} }


if ($NewSubCat !="") { mail("email","New Sub Category Request","Dear Owner,
\n\nThe club in the name of $name with a validation code of $validationID would like a new sub category called $new_cat\n\n  
\nTeam Me\n\n\n\n"); }

$url = "/create/create_clubp3.php?new_club=$validation";  
header("Location: $url");     

Hi Everyone,

I have a code that stores an ID to a cookie named "cookieuser" and inserts it to a table named "cookieuser" wheneven someone new to my site visited. When someone revisits, it should only access the cookie in the computer and retrieve data in the table based on the cookie. The problem is, the code inserts multiple rows with different IDs (which is autoincremented). The codes works for returning visitors which only reads the cookie ID value. I noticed that the error reacts differently with different browsers: When in localhost, mozilla works as expected, ie inserts 4 rows, and chrome inserts 2 rows. In live server, mozilla creates 4 rows as well. The code I used is shown below. Hope someone knows the problem. Thank you in advance.

Code: [Select]
if (isset($_COOKIE['cookieuser'])) {
   $cookieuserx = $_COOKIE['cookieuser'];
   $sql = "SELECT userid FROM cookieuser WHERE userid = $cookieuserx";
   $res = mysql_query($sql) or die(mysql_error());
   if(mysql_num_rows($res) == 0){
      $date = date("Y-m-d H:i:s",time());
      $sqlcookie = "INSERT INTO cookieuser (userid,lastgood,lastbad,lastneutral,lastvisit,hitcount,date_registered) VALUES('','$date','$date','$date','$date','1','$date')";
      $rescookie = mysql_query($sqlcookie) or die(mysql_error());
      $newcookieuser = mysql_insert_id();
      $expire=time()+60*60*24*360;
      setcookie("cookieuser", $newcookieuser, $expire,"/");   
      $cookieuser = $newcookieuser;
   }else{
      $cookieuser = $_COOKIE['cookieuser'];
   }
   
}else{
   $date = date("Y-m-d H:i:s",time());
   $sqlcookie = "INSERT INTO cookieuser (userid,lastgood,lastbad,lastneutral,lastvisit,hitcount,date_registered) VALUES('','$date','$date','$date','$date','1','$date')";
   $rescookie = mysql_query($sqlcookie) or die(mysql_error());
   $newcookieuser = mysql_insert_id();
   $expire=time()+60*60*24*360;
   setcookie("cookieuser", $newcookieuser, $expire,"/");
   $cookieuser = $newcookieuser;
}

how do i  stop multiple duplications in database on my PHP script?



ok i have attached a screenshot of what the database looks like after a few runs of my script. the script is designed to pull api information, input into 1 database and update another user table.
i have made it run as a cron job every 60 minutes.

here is my code:
<?php

/* You need multiple instances of this script. Each instance runs once every hour so 6 instances means one 
runs every 10 mins. Remember to change the API URL to reflect the different accounts or characters.*/
include "connect.php";

$columns = "`date` , `refID`, `refType`, `ownerName1`, `ownerName2`, `argName1`, `amount`, `balance`, `reason`";

//Live URL is 

//Assumeing that they are only donating at this time and no one is being paid to reduce the balance. Balance reduction can be done in the prize claim script so its not API delayed.
if ( ($data[2] == "Player Donation") && ($data[4] == "Ship Lotto")){
$reUsed = mysql_query("SELECT * FROM bank WHERE refID='$data[1]';");
if(!empty($reUsed)){
$import="INSERT into bank($columns) values('$data[0]','$data[1]','$data[2]','$data[3]','$data[4]','$data[5]','$data[6]','$data[7]','$data[8]')";
mysql_query($import) or die(mysql_error());
}

/*check to see if the player has already been credited. It checks the last recorded reference #
and checks to see if the new ref # is greater, else skips the processing. You need to check since 
the API gives you the last 1000 journal entries or 1 week, what ever is shorter. Not just what is 
new since last check. Check is performed by seeing if the record in the database for the user is 
less then or equal to the new once. This works only because CCP's reference #s are auto increasing 
so they only go up if they are newer, never down.*/
$name = $data[3];
//echo "Updating account of ".$name."<br />";
$queryLastRef = mysql_query("SELECT lastRef FROM users WHERE username='$name';") or die(mysql_error());
//echo $queryLastRef;
$arraylastRef = mysql_fetch_assoc($queryLastRef);
$lastRef = $arraylastRef["lastRef"];
//echo "The last reference # was: ".$lastRef."<br />";
$currentRef = $data[1];
//echo "The current reference # is: ".$currentRef."<br />";
if($lastRef<$currentRef){
$amount = $data[6];
//echo "Player deposited ISK in the amount of: ".$amount."<br />";
$queryBal = mysql_query("SELECT user_iskbalance FROM users WHERE username='$name';") or die(mysql_error());
//echo "Executing the SQL command to query balance ID#: ".$queryBal."<br />";
$getBal = mysql_fetch_assoc($queryBal);
//echo "Executing the SQL command to get balance amount: ".$getBal["user_iskbalance"]."<br />";
$deposit = $amount+$getBal["user_iskbalance"];
//echo "Depositing ISK in the ammount of: ".$deposit."<br />";
$importBal= "UPDATE users SET user_iskbalance=$deposit WHERE username='$name';";
//echo "Executing the SQL command to desposit: ".$importBal."<br />";
mysql_query($importBal) or die(mysql_error());
$importRefID= "UPDATE users SET lastRef='$currentRef' WHERE username='$name';";
//echo "Executing the SQL command to set the new reference: ".$currentRef."<br />";
mysql_query($importRefID) or die(mysql_error());
//echo "Success!"."<br />";

//For the sake of stats tracking update the total isk on deposit. The payout script will subtract.

$queryiskDeposit = mysql_query("SELECT iskDeposit FROM stats;") or die(mysql_error());
//echo "Executing the SQL command to query the ISK deposited : ".$queryiskDeposit."<br />";
$arrayiskDeposit = mysql_fetch_assoc($queryiskDeposit);
$getiskDeposit = $arrayiskDeposit["iskDeposit"];
//echo "Got total isk on deposit of: ".$getiskDeposit."<br />";
$iskDeposit = $getiskDeposit+$deposit;
//echo "Inserting: ".$iskDeposit." ISK"."<br />";
$importiskDeposit= "UPDATE stats SET iskDeposit='$iskDeposit';";
//echo "Executing the SQL command to desposit: ".$importBal."<br />";
mysql_query($importiskDeposit) or die(mysql_error());
//echo "<br />";
//echo "<br />";
//echo "NEXT!<br />";
//echo "<br />";
}
else{
//echo "There is no update for ".$name." because ".$lastRef." is not less then or equal to ".$currentRef."<br />";
//echo "<br />";
//echo "<br />";
//echo "NEXT!<br />";
//echo "<br />";
}
//echo "DEBUG for ".$name." lastRef ".$lastRef." and currentRef ".$currentRef."<br />";

//update the time that last update ran
$today = date("Ymd G:i");
mysql_query("UPDATE stats SET iskLastUpdate='$today';") or die(mysql_error());
//echo "Updating Date to: ".$today;
//echo "<br />";
//echo "<br />";
//echo "NEXT!<br />";
//echo "<br />";
}
}
?>


can anyone help me stop it duplicating the entries in the database please?

Hello dear friends,

say i've database table with  (id,name) and want to add more informations
by insert new names within an file for example has

Code: [Select]
$q1 = "INSERT INTO `mytable` VALUES (???,name1);
mysql_query($q1) or die(mysql_error()." at row ".__LINE__);

$q2 = "INSERT INTO `mytable` VALUES (???,name2);
mysql_query($q2) or die(mysql_error()." at row ".__LINE__);

ect.....

how then it automatic detect the last id to go on after it ?

some says use mysql_insert_id but i didn't understand how to apply it in this way also the example at php.net is bad not explain much

please how can i use it
thanks

How can you debug to find out why it's not getting the mysql_insert_id number? I have it echoed all the queries and all with all the correct values from my form but the only problem is that its not getting the insert id number of the id.

$query1 = "INSERT INTO `efed_bio` (charactername,username,posername,style_id,gender,status_id,division_id,alignment_id,sortorder) VALUES ('".$charactername."','".$username."','".$posername."','".$style."','".$gender."','".$status."','".$division."','".$alignment."','".$sort."')";
            mysql_query($query1);
            $query1_id = mysql_insert_id(); 
            echo $query1;
            echo $query1_id;
            $query2 = "INSERT INTO `efed_bio_allies` (bio_id) VALUES (".$query1_id.")";
            mysql_query($query2); 
            echo $query2;
            $query3 = "INSERT INTO `efed_bio_rivals` (bio_id) VALUES (".$query1_id.")";
            mysql_query($query3);  
            echo $query3;
            $query5 = "INSERT INTO `efed_bio_singles` (bio_id) VALUES (".$query1_id.")";
            mysql_query($query5);  
            echo $query5;