Run 2 MySQL update queries in one PHP file

Question

I try to run

// Create connection
$conn = new mysqli($servername,$username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} 
$sql = "UPDATE pm_videos SET `description` = REPLACE(  `description` ,'Facebook',  '')";
$sql = "UPDATE pm_videos SET `description` = REPLACE(  `description` ,'Twitter',  '')";

if ($conn->query($sql) === TRUE) {
echo "Record updated successfully";
} else {
echo "Error updating record: " .  $conn->error;
}
$conn->close();

but only Twitter will be replaced. I'm new in PHP. Maybe someone could help me to fix this. Thank you.

Answer 1

The $conn->query($sql) statement runs your query. You are overwriting the first $sql (the one with Facebook) before you run it. You'll wan to do something like

$sql = "UPDATE pm_videos SET `description` = REPLACE(  `description` ,'Facebook',  '')";
if ($conn->query($sql) === TRUE) {
    echo "Record updated successfully with Facebook";
} else {
    echo "Error updating record: " .  $conn->error;
}
$sql = "UPDATE pm_videos SET `description` = REPLACE(  `description` ,'Twitter',  '')";
if ($conn->query($sql) === TRUE) {
    echo "Record updated successfully with Twitter";
} else {
    echo "Error updating record: " .  $conn->error;
}

Answer 2

Your first query won't be executed because the $sql string will be overwritten, and you end up executing only the second one.

Since you are executing the same query more than once, only changing a parameter, I would suggest you to use prepared statements and placeholders, something like this:

$sql = "UPDATE pm_videos SET `description` = REPLACE(`description`, ?,  '')";

$conn = new mysqli($servername,$username, $password, $dbname);
$stmt = $conn->prepare($sql);
$stmt->bind_param('s', $content);

$content='facebook';
$stmt->execute();

$content='twitter';
$stmt->execute();

$stmt->close();
$conn->close();

(I removed all error checks, you might want to add your own error checking routine).