All values in array into database?

Question

How would i put the $newd values into the database. The way I'm doing it now just puts one value in. I want all values.

<?php
$checked = $_POST['options'];
for($i=0; $i < count($checked); $i++){
    $newd = "" . $checked[$i] . ",";
}   
    if(isset($_POST['loginbtn'])){
        if(!empty($order)){

if($money){
    //making the sql command
    $sql = "INSERT INTO `orders`(`order`,`date`,`time`,`timepass`,`money`,`corder`,`cancel`,`category`) VALUES ('$order','$date','$timenow','$time','$money','$corder','$cancel','$newd')";

    //querying the sql
    $query = mysqli_query($db,$sql);

    $lastid = mysqli_insert_id($db);

    $twosql = "INSERT INTO `comments`(`order_id`, `comment`,`user`,`time`,`timepass`) VALUES ('$lastid','$comment','$username','$timenow','$time')";
    $twoquery = mysqli_query($db,$twosql);

    header("Location: moneyorder.php"); 

}
?>

Your code is vulnerable to SQL injection attacks. You should use mysqli or PDO prepared statements with bound parameters as described in this post. — Alex Howansky
– Alex Howansky, Commented Apr 28, 2017 at 18:17
Possible duplicate of How to insert array of data into mysql using php — vv01f
– vv01f, Commented Apr 28, 2017 at 18:25

mayersdesign · Accepted Answer · 2017-04-28 18:25:29Z

1

You need a dot here, you are not concatenating the new values :)

$newd .=

answered Apr 28, 2017 at 18:25

mayersdesign

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Amit Gaud · Accepted Answer · 2017-04-29 05:37:35Z

-1

You have to define first $newd variable on top else it will give error but please do the things as suggested by Alex Howansky in comment

$checked = $_POST['options'];
    $newd = '';
    for($i=0; $i < count($checked); $i++){
        $newd .= "" . $checked[$i] . ",";
    } 
$newd = rtrim($newd, ',');

   // OR you can use

    $newd = implode(",", $checked );

edited Apr 29, 2017 at 5:37

answered Apr 28, 2017 at 18:31

Amit Gaud

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