Right now I have working a DB connection to mysql. The html -> PHP -> query -> data reception works. I show the relevant code:
From the html file matters:
d3.json("http://path/file.php",   function(error, data) {
    console.log (data); 
});
file.php:
    <?php
    $username = "myusername"; 
    $password = "mypassword";   
    $host = "myhost";
    $database="myDB";
    $server = mysql_connect($host, $username, $password);
    $connection = mysql_select_db($database, $server);
    $myquery = "select * from `mytable`";
    $query = mysql_query($myquery);
    if ( ! $query ) {
        echo mysql_error();
        die;
    }
    $data = array();
    for ($x = 0; $x < mysql_num_rows($query); $x++) {
       $data[] = mysql_fetch_assoc($query);
    }
    echo json_encode($data);     
    mysql_close($server);
    ?>
What I want is to have only 1 .php file instead of 1 php file for every query. That means I need to send from the html a variable inputquery to the php file. I've tried several things such as changing:
`$myquery = "select * from `mytable`; into `$myquery = inputquery`;
And I think that the wrong point is the definition of the function that requests the data from the DB. What I tried (wrong, the following code does not work as expected):
var inputquery = "select * from `mytable`"
d3.json("http://serverhost/path/file.php",  function(error, data) {
    console.log (data); 
});
Maybe this is not working because I am not telling the function I want as an input to the .php file the variable inputquery. I tried to put it inside the function, but got "data is not defined" errors, so I think it is not worth it to show the wrong code.
How can I input that var inputquery to the .php file? It could not be the way I planned it.
Thank you
 
     
     
    