PHP/MySQL Database/sqlite query

Advanced Functions

   <source lang="html4strict">

$dbconn = sqlite_open("phpdb");

   if ($dbconn) {
           script!
           sqlite_query($dbconn, "INSERT INTO animal VALUES("a", 14)");
           sqlite_query($dbconn, "INSERT INTO animal VALUES("b", 16)");
           sqlite_query($dbconn, "INSERT INTO animal VALUES("c", 13)");
           var_dump(sqlite_array_query($dbconn, "SELECT * FROM animal", SQLITE_ASSOC));
   } else {
           print "Connection to database failed!\n";
   }
 
 </source>
   
  

sqlite_query.php

   <source lang="html4strict">

<?php

   $sqldb = sqlite_open("mydatabase.db");
   $results = sqlite_query($sqldb, "SELECT * FROM employee");
   while (list($empid, $name) = sqlite_fetch_array($results)) {
       echo "Name: $name (Employee ID: $empid) 
";
   }
   sqlite_close($sqldb);

?>

 </source>
   
  

Using the sqlite_query() Function

   <source lang="html4strict">

<?php

    $db = sqlite_open(":memory:");
    if(!$db) die("Could not create the temporary database");
    $query = "CREATE TABLE cities(name VARCHAR(255), state VARCHAR(2))";
    sqlite_query($db, $query);
    $cities[] = array("name" => "Chicago","state"=> "IL");
    foreach($cities as $city) {
        $query = "INSERT INTO cities VALUES(" .""{$city["name"]}", "{$city["state"]}")";
        if(!sqlite_query($db, $query)) {
         trigger_error("Could not insert city " . ""{$city["name"]}, {$city["state"]}");
        }
    }
    sqlite_close($db);

?>

 </source>