Php code issue 
Author: Errol W.Here is the code I am using.
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT order_id, label, value FROM cartinvoice_addresses";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo " " . $row["label"]. " " . $row["value"] ;
if( $row["label"] = "address"){ "<br>"}
}
} else {
echo "0 results";
}
$conn->close();
?>
I added this to the original code "if( $row["label"] = "address"){ "<br>"}" I want the output to skip a line when it finds the word "address" in the table column. When I run the process I get "This Page not working" I suspect the syntax is wrong. Help me. Without that piece of code, there is no problem.
Try this:
if( $row["label"] = "address"){ echo "<br>"; }
Author
Hello Axel,
I am not getting the page not working after using your advice. However I am not getting the line break.
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT order_id, label, value FROM cartinvoice_addresses";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo " " . $row["label"]. " " . $row["value"] ;
if( $row["label"] = "address"){ echo "<br>"; }
}
} else {
echo "0 results";
}
$conn->close();
?>
Try to modify your script with :
$conn = mysqli_connect("my_server", "user", "password", "dbname");
and to save it in PHP file like test.php
and try tu use your test.php without WSX5. Directly from your web server like https://myserver.com/test.php
It's better to debug the PHP script outside of WSX5.
When it's run, you can add it to use it...
Hope this helps
Axel
Author
Thanks, I'll give it a go.
Author
Oh, Yeah that much better for fixing code. Thanks
Hello Errol,
The comparison operator for equality in PHP is a double equals sign as opposed to a single.
Thus:
if( $row["label"] == "address")
This is why your code is failing the if statement.
Also, you might want to capitalise "address" to match what is in the MySQL database.
if( $row["label"] == "Address")
Kind regards,
Paul
Search the WebSite X5 Help Center
Author
Paul, that did the trick.
Great, thanks for letting us know!