PHP/MySQL Database/sqlite query

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Версия от 07:06, 26 мая 2010; Admin (обсуждение | вклад) (1 версия)
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Advanced Functions

 
$dbconn = sqlite_open("phpdb");
    if ($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";
    }



sqlite_query.php

 
<?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: $empid) <br />";
    }
    sqlite_close($sqldb);
?>



Using the sqlite_query() Function

 
<?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);
?>