help me in sql queries

elle_girl

if ($result) { 

		$data = "SELECT itemId, qty, price, cartId FROM cart ";
		$result1 = @mysql_query($data) or die (mysql_error());

		 if ($row = mysql_fetch_array($result1)) {
			$itemId = $row['itemId'];
			$qty = $row['qty'];
			$price = $row['price'];
			$cartId = $row['cartId'];
         				$query1 = "INSERT INTO view (itemId, price, qty, cartId) VALUES ('$itemId','$price','$qty','$cartId' )"; 
        				$result2 = mysql_query($query1) or die(mysql_error());// Run the query. 

			if ($result2) {
				$insert = "UPDATE view SET account_no = '$aa', token_id = '$t', name = '$n', address = '$ad', state = '$s', country = '$c' WHERE cartId = '$cartId'";
				$result_insert = mysql_query ($insert) or die(mysql_error());
			} else {
				echo '<p><font color="red" size="+1">Cannot update to the database</font></p>';

			}

			// Successful add the new customer. 
     				echo '<h3>Successful the transaction</h3>'; 
    				include ('includes/header_bank.html'); 
     				exit(); 
		} else {
			echo '<p><font color="red" size="+1">Unsucessful transaction</font></p>';
		}




        		} else { // The account number does not match with token ID. 
     			echo '<p><font color="red" size="+1">The account number does not match with the reference ID in the database</font></p>'; 
  		}

My syntax is correct. No error encountered. The problem is that ..
in my cart table it has 2 data but when I want to insert the cart table to view table it just appear 1 data... it does not insert all the data. Can you please help me in the sql queries to solve my problem? I really appreciated your help.

rgermain

Ok, if I am getting this you are tellin me that you are pulling 2 sets of records from the cart table. Then you are trying to insert those to sets of records into the view table.

If this is the case then where is the while statement. You are using if statements, which only run once. For it to run through all the records it found from cart table you would need something like the following:

<?
//ALL QUERY INFORMATION...
while($row = mysql_fetch_array($result){
  If($comparison == "2"){
//DO SOMETHING
}elseif($comparison == "1"){
//DO SOMETHING ELSE
}else{//DO SOMETHING DIFFERENT}

}//END OF WHILE STATEMENT
?>

What this would do if $result contained, lets say 4 records, it would run through the if statements 4 times doing whatever it was supposed to depending on what was in the comparison variable.

Hopes this helps and doesn't confuse you more!

elle_girl

I make a little modification.


// Make sure the token available. 
      		$query = "SELECT * FROM customer_bank WHERE token_id = '$t' AND account_no = '$aa' "; 
      		$result = @mysql_query ($query); 


  		if ($result) { 

		$data = "SELECT itemId, price, qty FROM cart ";
		$result1 = @mysql_query($data) or die (mysql_error());


		 while($row = mysql_fetch_row($result1)) {

			$comp = "SELECT COUNT (*) FROM cart ";
			$comparison = @mysql_query($data) or die (mysql_error());
			$itemId = $row['itemId']; // line 93
			$price = $row['price']; // line 94
			$qty = $row['qty']; // line 95
			$cartId = $row['cartId']; // line 96


			if ($comparison == "1") {

				$query1 = "INSERT INTO view (itemId, price, qty, cartId) VALUES ('$itemId','$price','$qty','$cartId' )"; 
        					$result2 = mysql_query($query1) or die(mysql_error());// Run the query. 
				if ($result2) {
					$insert = "UPDATE view SET account_no = '$aa', token_id = '$t', name = '$n', address = '$ad', state = '$s', country = '$c' WHERE cartId = '$cartId'";
					$result_insert = mysql_query ($insert) or die(mysql_error());
					// Successful add the new customer. 
     						echo '<h3>Successful the transaction</h3>'; 
    						include ('includes/header_bank.html'); 
     						exit(); 
				} else {
					echo '<p><font color="red" size="+1">Cannot update to the database</font></p>';

				}
			} else if ($comparison == "2"){

				$query2 = "INSERT INTO view (itemId, price, qty, cartId) VALUES ('$itemId','$price','$qty','$cartId' )"; 
        					$result3 = mysql_query($query2) or die(mysql_error());// Run the query. 
				if ($result3) {
					$insert1 = "UPDATE view SET account_no = '$aa', token_id = '$t', name = '$n', address = '$ad', state = '$s', country = '$c' WHERE cartId = '$cartId'";
					$result_insert1 = mysql_query ($insert1) or die(mysql_error());
				} else {
					echo '<p><font color="red" size="+1">Cannot update to the database</font></p>';

				}
				$query3 = "INSERT INTO view (itemId, price, qty, cartId) VALUES ('$itemId','$price','$qty','$cartId' )"; 
        					$result4 = mysql_query($query3) or die(mysql_error());// Run the query. 
				if ($result4) {
					$insert2 = "UPDATE view SET account_no = '$aa', token_id = '$t', name = '$n', address = '$ad', state = '$s', country = '$c' WHERE cartId = '$cartId'";
					$result_insert2 = mysql_query ($insert2) or die(mysql_error());
					// Successful add the new customer. 
     						echo '<h3>Successful the transaction</h3>'; 
    						include ('includes/header_bank.html'); 
     						exit(); 
				} else {
					echo '<p><font color="red" size="+1">Cannot update to the database</font></p>';

It make this error

Notice: Undefined index: itemId in C:\Program Files\Apache Group\Apache2\htdocs\change_checkout.php on line 93

Notice: Undefined index: price in C:\Program Files\Apache Group\Apache2\htdocs\change_checkout.php on line 94

Notice: Undefined index: qty in C:\Program Files\Apache Group\Apache2\htdocs\change_checkout.php on line 95

Notice: Undefined index: cartId in C:\Program Files\Apache Group\Apache2\htdocs\change_checkout.php on line 96

can you please help me to solve this problem? I really appreciated your help?