I am struggling to view some data via PHP. I have a SQL database with Google. I am struggling to extract the data and desperately need some help!
The PHP keeps saying there is 0 records, even though there is a number of records within the 'timing' database and 'events' table.
If anyone has any ideas why this is not working I would be very grateful!
<?php
$link = new mysqli('IP_ADDRESS:3306','root','PASSWORD',timing);
if (!$link) {
die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully ';
$sql = "SELECT event_id FROM events";
$result = $link->query($sql);
if ($result->num_rows > 0) {
echo "<table><tr><th>Event ID</th><th>Event Name</th></tr>";
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["event_id"];
}
echo "</table>";
}
else {
echo "0 results";
}
mysql_close($link);
?>
Unless timing is a defined constant, and it looks like that is not the case, being that a string you should wrap it in quotes or double-quotes. A non-defined constant should cast to a string in any case, but since i can't see any other error in your code that might be what is causing your issue. Try changing
$link = new mysqli('IP_ADDRESS:3306','root','PASSWORD',timing);
to
$link = new mysqli('IP_ADDRESS:3306', 'root', 'PASSWORD', 'timing');
Ok i think i got it. From PHP.net:
OO syntax only: If a connection fails an object is still returned. To check if the connection failed then use either the mysqli_connect_error() function or the mysqli->connect_error property as in the preceding examples.
This means $link in your code will always be a MySQLi resource, and that it will always cast to true. What you need to do is to check your resource object for errors properly, like this:
if ($link->connect_error)
die('Connect Error: ' . $link->connect_error);