Shopping Cart [Sessions]

[Krux20]

Hi,

 

I kind of need help in how I can create a cart for a user, so that the user can add or delete the quantity of items in the cart. What do I need to do first?

 

Thanks

[MDCode]

Step 1. Create the system.

Step 2. Ask a question if something goes wrong.

 

If it's for a user, store it in mysql or cookies because they can log out and lose all their cart

Edited November 27, 2012 by SocialCloud

[Krux20]

Hi,

 

thanks for the reply and for some reason I'm getting this error Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in: I don't know what the problem is.

function products () {

   $con = mysqli_connect("localhost","root","");


   if (!$con) {
     die('Could not connect: ' . mysql_error());
    }


    mysql_select_db("bicycle");




 $result = mysqli_query($con,"SELECT BikeCode, Model  FROM bike");




 while($row = mysqli_fetch_array($result))
  {
   echo $row['BikeCode'] . " " . $row['Model'];
   echo "
";
 }

[codefossa]

Your query failed and it's returning false. You can't pass a boolean through the mysqli_fetch_array() function.

 

$sql = "SELECT BikeCode, Model FROM bike";

$result = mysqli_query($con, $sql) or die("Error: Failed to query database. ({$sql})");

Edited November 27, 2012 by Xaotique

[Krux20]

 if (!$con) {
     die('Could not connect: ' . mysql_error());
    }
   else {
        echo 'connected';
   } 

Thanks for the reply, but nothing seems to be wrong with my database and when I try the following code it says I'm connected to my database.

[PFMaBiSmAd]

You'll find that the error in your query is because you haven't selected a database.

 

You cannot mix mysql (no i) and mysqli (with an i) statements.

 

You either need to specify the database in the mysqli_connect statement or use a mysqli_select_db statement (or even specify the database name in your query statement.)

[Krux20]

I added a new mysqli_select_db statement, but it still gives me that error and I thought it was because my MySQL service wasn't running, but it seems to working fine.

 

 function products () {

   $con = mysqli_connect("localhost","root","");


   if (!$con) {
     die('Could not connect: ' . mysqli_error());
    }
   else {
        echo 'connected';
   } 


   $db= mysqli_select_db("bicycle");




 $sql = "SELECT BikeCode, Model FROM bike";
$result = mysqli_query($db, $sql) or die("Error: Failed to query database. ({$sql})"); 




 while($row = mysqli_fetch_array($result))
  {
   echo $row['BikeCode'] . " " . $row['Model'];
   echo "
";
 }










   }

[codefossa]

If it's the same as mysql functions, you have to specify the connection as the first parameter in the mysql_select_db (atleast I think it's a must, never tried without).

 

You could also double check your query in phpmyadmin to make sure it's valid.

Edited November 27, 2012 by Xaotique

[PFMaBiSmAd]

The mysqli_select_db statement requires the mysqli link resource ($con in your case) as the first parameter. Do you have php's error_reporting set to E_ALL so that all the php detected errors will be reported?

 

Also, the mysqli_error() statement requires the mysqli link resource as a parameter.

[Krux20]

Yes, I have it on E_ALL. http://tinypic.com/r/e5jybk/6

 

 

 

function products () {


$con = mysqli_connect("localhost","root","");


if (!$con) {
die('Could not connect: ' . mysqli_error());
}
else {
echo 'connected';
}


$db= mysqli_select_db($con,"bicycle");




$sql = "SELECT BikeCode, Model FROM bike ";
$result = mysqli_query($db, $sql) or die("Error: Failed to query database. ({$sql})");




while($row = mysqli_fetch_array($result))
{
echo $row['BikeCode'] . " " . $row['Model'];
echo "
";
}

}

Edited November 27, 2012 by Krux20

[PFMaBiSmAd]

If you are still getting an error at your mysqli_fetch_assoc statement, with the code you have been posting, it's because some of your code inside of your while(){} loop, that you have been cutting out of the post, is reusing and overwriting the $result variable.

[Krux20]

This is all my code

<?php
error_reporting(E_ALL);
session_start();



$page = 'index.php';



// $_SESSION['cart'. $_GET['add']] = 1;
// $_SESSION['cart'] += 1;






function products () {


$con = mysqli_connect("localhost","root","");


if (!$con) {
die('Could not connect: ' . mysqli_error());
}
else {
echo 'connected';
}


$db= mysqli_select_db($con,"bicycle");

$sql = "SELECT BikeCode,Model FROM bike ";
$result = mysqli_query($db, $sql) or die("Error: Failed to query database. ({$sql})");

while($row = mysqli_fetch_array($result))
{
echo $row['BikeCode'] . " " . $row['Model'];
echo "<br />";
}





}

?>

Edited November 27, 2012 by Krux20

[PFMaBiSmAd]

The error message you are getting now cannot possibly be the same one at the start of this thread, because you are using the WRONG variable name as the first parameter to the mysqli_query() statement. $db ISN'T the connection link. You should be getting a php error message at the mysql_query() statement.

[Krux20]

Thanks! you were right