Insert multiple checkbox values in MysQl

Question

I have the following checkbox items and I want to insert them in the database.

<form action="test.php" method="post">
<input type="checkbox" name="check_list[]" value="rice:10">
<input type="checkbox" name="check_list[]" value="milk:8">
<input type="checkbox" name="check_list[]" value="orange:4">
<input type="submit" />
</form>

The following is the php code:

if(!empty($_POST['check_list'])) {
    foreach($_POST['check_list'] as $check) { 
        $mealA = explode(":",$check);
        $meal = $mealA[0];
        $sql1 = "INSERT INTO Order (User, Meal, Total) VALUES ('$name' , '$meal', '$total')";
        if (mysqli_query($mysqli,$sql1)) {
            echo "You order has been recieved";
        } else {
            echo "Error: " . $sql1;
        }
    }
}

Note that the value of each checkbox means item name:price.

The problem is that I got error "Error: " . $sql1;

Also, try adding mysqli_error to your code: php.net/manual/en/mysqli.error.php — Maximus2012
– Maximus2012, Commented Apr 28, 2015 at 18:31
Change "Error: " . $sql1 to "Error:" . mysqli_error($mysqli) so you can have a better understanding of why it fails. — kidA
– kidA, Commented Apr 28, 2015 at 18:33
echo the three values $name, $meal, and $total to see what exactly is contained in them. If they look fine try running the failing sql command manually in mysql to see if it still gives an error. — kojow7
– kojow7, Commented Apr 28, 2015 at 18:38
I changed what @kidA suggested and got this error Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Order (User, Meal, Total) VALUES ('bero' , 'margherita', )' at line 1Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Order (User, Meal, Total) VALUES ('bero' , 'sicilian', )' at line 1 — Nasser
– Nasser, Commented Apr 28, 2015 at 19:21

kidA · Accepted Answer · 2015-04-28 20:04:23Z

2

Your query fails because you have not set a value for $total. So either you have to remove the Total field from the query resulting in

$sql1 = "INSERT INTO Order (User, Meal) VALUES ('$name' , '$meal')";

which will give a NULL value to the Total column for each successful new insert

OR

based on the code you provided I 'm guessing you wanted to do something similar to this?

$mealA = explode(":",$check);
$meal = $mealA[0];
$total = $mealA[1]; // setting the value of $total otherwise query is going to fail
$sql1 = "INSERT INTO Order (User, Meal, Total) VALUES ('$name' , '$meal', '$total')";

edited Apr 28, 2015 at 20:04

answered Apr 28, 2015 at 19:59

kidA

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Nasser · Accepted Answer · 2015-04-29 06:59:22Z

0

The problem was the the insert statement:

INSERT INTO Order (User, Meal, Total) VALUES ('$name' , '$meal', '$total')

I changed the table name from Order to Orders and I removed Total since it is null and there is no value stored in it.

Thanks for your suggestions.

answered Apr 29, 2015 at 6:59

Nasser

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