在php中更新mysql表，并为每个受影响的行存储ID [重复]

Question

This question already has an answer here:

• Create array in PHP from mysql 2 answers
• php update to mysql with array not updating 3 answers

I'm trying to use the results of a MySQL select in a php script and make some updates in the same table.

For the following query, I need to look at each record and say "if order_status = 'S', set is_placement to 1 (this is a bit) and set date_updated to curdate()"

I can do that with an update, but the trick is I need to store the order_id of those with the status of 'S' in an array to use in another query.

I'm not sure how I would loop and store those in an array for that.

Here's the select:

$orderCheck = "
    SELECT 
        order_id,
        order_status
    FROM order_status
 ";

EDIT (based on answers below): So I can make an array of order IDs like so:

$result = mysqli_query($connection,$orderCheck);
while ($row = mysqli_fetch_array($result))
    {
        $array = $row['order_id'}
        if($row['order_status'] == 'S'){
        Store ORder IDs here}
    }

But with that array and stored order IDs, I need to update the records for each ID.

How can I preform that update and store the necessary IDs as well?

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Answer 1

The following will store the data the way you needed:

$result = mysql_query($orderCheck);
$order_ids = array();

while ($row = mysql_fetch_array($result))
    {
    $order_id = $row['order_id'];
    if ($row['order_status'] == "S")
        {
        array_push($order_ids, $order_id);
        }
    }