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Getting the error Undefined index but the column exists

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I added a column to my database with SQL as follows:

ALTER TABLE comment_table ADD date_col Datetime NOT NULL;

I try to insert into the database with the following PHP but nothing gets inserted. 


$date_col = "test";//this will be DateTime later
$name = $_POST["name"];

mysqli_query($connection, "INSERT INTO comment_table (name, date_col) VALUES ('$name', '$date_col')"); 
$comsql = "SELECT * FROM comment_table";
$comres = mysqli_query($connection, $comsql);
while($comr = mysqli_fetch_assoc($comres)){
<div class="row">
<p>Name: <strong><?php echo $comr['name']; ?></strong>This is the code that is being pointed to as undefined index. <?php echo $comr['date_col']; ?> </p>
<?php } ?>

The data I'm using for the date_col column is varchar and it should be Datetime but I don't know if that's the reason for the error. I would like to set $date_col equal to a Datetime expression for testing purposes but the formatting I chose wasn't working either.

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I changed the code so that it is using prepared statements but I just get new errors now. I altered the table so that the date_col column has the data type of varchar. I get the following warning:

mysqli_stmt_bind_param(): Number of variables doesn't match number of parameters in prepared statement

if(isset($_REQUEST["submit"]) ) {
$date = new DateTime('', new DateTimeZone('America/New_York'));
$name = mysqli_real_escape_string($connection, $_REQUEST["name"]);
$date_col = $date->format('M d, Y H:i');
$website = $_REQUEST["website"];
$comment = mysqli_real_escape_string($connection, $_REQUEST["comment"]);
$sql = "INSERT INTO comment_table (name,date_col,website,comment) VALUES ('$name', '$date_col','$website','$comment')";  
if($stmt = mysqli_prepare($connection, $sql)){
    // Bind variables to the prepared statement as parameters
    // Attempt to execute the prepared statement
        echo "Records inserted successfully.";
    } else{
        echo "ERROR: Could not execute query: $sql. " . mysqli_error($connection);
} else{
    echo "ERROR: Could not prepare query: $sql. " . mysqli_error($connection);
// Close statement
// Close connection


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The solution posted to a different forum was:

$sql = "INSERT INTO comment_table (name,date_col,website,comment) VALUES (?, ?, ?, ?)";

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