在MySQL中使用PHP时和if语句

Question

I have been stuck with this problem for awhile now. I'm sure the answer is on easily find able on Google but I have been unable to find the words to search so my last resort is here. Thanks in advanced!

I have this php code:

if(mysqli_num_rows($query) > 0) {
    while($row = mysqli_fetch_array($query)) {
        if ($row['status'] == "Opened") {
            $test = "1";
        } else {
            $test = "0";
        }
    }
}
echo $test;

The problem occurs when there are multiple entries in the database and one of those entries does not have status = "Opened" therefore $test returns "0". What I am trying to accomplish is if ANY of the entries have status = "Opened", $test will return "1".

Once again thanks in advance for all your help!

Answer 1

All you need to do is the following:

$query = "SELECT `status` FROM `table` WHERE `status` = 'Opened'";
$result = mysqli_query($conn, $query);
$test = (mysqli_num_rows($result) > 0) ? 1 : 0;

No need to fetch, it will be faster too.

Answer 2

It might be better to directly select the data you need from the database instead of filtering it out with PHP.

SELECT id, message, etc FROM tickets WHERE status = 'Opened'
// if it fetches data, you have open tickets and have directly access to the data of the open tickets.

SELECT count(*) AS total FROM tickets WHERE status = 'Opened'
// the result of this is a number of the total amount of opened tickets, perhaps better for this specific use.

But on the subject on how to fix your loop you can do the following:

$open = false;

while($row = mysqli_fetch_array($query)) {
    if ($row['status'] == "Opened") {
        $open = true;
        break;
    } 
}

As this will quit the while loop, setting $open for futher use:

if($open){
  // do your stuff;
}