Insert to database into two tables - php

Question

This code really made me confused.
The first and second time I ran it, it worked perfectly but after that it stopped working
Let me explain it:
I work with 2 tables.
The first table I insert to it the current date, current time and the id of the user the id I take it from the session.
Which I believe works fine.
My problem is in the second table the error I get is the error i typed in the " print " after the second insert.
this is my code :
session_start();
//Check whether the session variable SESS_MEMBER_ID is present or not
if(!isset($_SESSION['con_id'])) {
header("location: login.html");
exit();
}
$DB_USER ='root';
$DB_PASSWORD='';
$DB_DATABASE='';
$con= mysql_connect($DB_HOST ,$DB_USER , $DB_PASSWORD);
if (!$con) {
die('Failed to connect to server :'.mysql_error());
}
$db=mysql_select_db($DB_DATABASE);
if (!$db) {
die("unable to select database");
}
//first table
$qry="insert into shipment values('',NOW(),CURTIME(),'".$_SESSION['con_id']."');";
$resultop=mysql_query($qry);
//to take the id frome last insert because i need it in the second insert
$SNo=mysql_insert_id();
if ($resultop) {
$options=$_POST['op'];//this is the name of the check boxe's
if (empty($options)) {
header("location: manage_itemsE.php");}
// this is the second table .. my reaaal problem
$qun=$_POST['Quantit'];
$size =count($options);
for ($i =0; $i<$size; $i++) {
$qqry="insert into shipmentquantity values('".$options[$i]."','".$SNo."','".$qun[$i]."');"; // $options is array of the id's which i took from the checkbox's in the html ... $qun is array of the values i took form html ... i sure this is right ;)
$resultqun=mysql_query($qqry);
}
if ($resultqun) {
header("location: shipment_order.php");
}
else print "error in the Quantity";
}
else print "error in the shipmet";

Answer 1

Just add some debug statements to find out what is going wrong. Something like -
$resultqun = mysql_query($qqry) or print mysql_error();
You need to do some reading about SQL injection as this script is vulnerable. Checkout these pages on the use of prepared statements - PDO::prepare and mysqli::prepare
UPDATE - here is an example using PDO to interact with your db -
<?php
session_start();
//Check whether the session variable SESS_MEMBER_ID is present or not
if(!isset($_SESSION['con_id'])) {
header("location: login.html");
exit();
}
$DB_USER ='root';
$DB_PASSWORD='';
$DB_DATABASE='';
$db = new PDO("mysql:dbname=$DB_DATABASE;host=127.0.0.1", $DB_USER, $DB_PASSWORD);
//first table
$qry = "INSERT INTO shipment VALUES(NULL, CURRENT_DATE, CURRENT_TIME, ?)";
$stmt = $db->prepare($qry);
$resultop = $stmt->execute(array($_SESSION['con_id']));
if(!$resultop){
print $stmt->errorInfo();
} else {
$SNo = $db->lastInsertId();
$options = $_POST['op'];//this is the name of the check boxe's
if (empty($options)) {
header("location: manage_itemsE.php");
exit;
}
// this is the second table .. my reaaal problem
$qun = $_POST['Quantit'];
$size = count($options);
$stmt = $db->prepare("INSERT INTO shipmentquantity VALUES(?, ?, ?)");
for($i = 0; $i < $size; $i++) {
$resultqun = $stmt->execute(array($options[$i], $SNo, $qun[$i]));
}
if($resultqun) {
header("location: shipment_order.php");
} else {
print $stmt->errorInfo();
}
}

Answer 2

What is your primary key for the 'shipmentquantity' table? It looks like you are trying to enter two values of '3' for the primary key and that's where it's going awry.
DESCRIBE `shipmentquanitity`