There are two tables in the database:
_url with type field
_splash with id and name fields
In the script, the data from the first table is displayed as: $url->type .
How to print name instead of $url->type if the type from the first table matches the id from the second, i.e. the data in the type field is taken from id ?
as far as I understood $url -> type makes a query to the database and simply displays the result. Then you just need to change the function:
$sql = "SELECT name FROM __splash JOIN __url ON __splash.id = __url.type;" //for MySQL $rs = mysqli_query($sql); foreach($row = mysqli_fetch_array($rs) as $key => $val) { echo $key . " :" . $val . "</br>"; } if you need to change on a specific id then you need to transfer its value to $id in the function and finally add WHERE __url.name = {$id} to the request. Well, or just make requests to the __splash table instead of __url .
In general, this is done through SQL, a view table is created that combines several tables into one (use join), and then output from this your table.
Source: https://ru.stackoverflow.com/questions/544941/
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