$rows = $statement->fetchAll(PDO::FETCH_ASSOC);

得到我桌子的所有行.

模式定义为“id INTEGER PRIMARY KEY,title TEXT,year INTEGER,price REAL”

来自fetchAll结果的一行是

array(4) {
    [0]=>
    string(1) "1"
    [1]=>
    string(14) "The Dark Night"
    [2]=>
    string(4) "2008"
    [3]=>
    string(5) "19.95"
  }

为什么所有数据类型都以字符串形式返回？我希望它们按模式中的定义返回.我理解sqlite的“无类型”特性,但它们确实将有限的数据类型定义为TEXT,INTEGER和REAL.如何使用指定的数据类型返回数据？我不想遍历每一行,并用PHP转换它 – 这似乎太慢了.

完整的测试代码如下：

<?
class sqlite_test {
    function __construct()
    {
        $this->dbase_filename = "test.sqlite";
        $this->init_dbase();

    }

    function init_dbase()
    {
        $this->pdo = new PDO("sqlite:".$this->dbase_filename);
    }

    function open_table()
    {
        $table = "dvds";

        if ( ! ($test_to_see_if_table_exists = $this->pdo->query("SELECT 1 from $table")) ) {
            $schema = "id INTEGER PRIMARY KEY,price REAL";
            $query = "CREATE TABLE $table ($schema)";

            $this->pdo->exec($query);
        }

        return $test_to_see_if_table_exists;
    }

    function add_test_records()
    {
        $query[]='INSERT INTO dvds (id,title,year,price) VALUES (null,"The Dark Night",2008,19.95)';
        $query[]='INSERT INTO dvds (id,"The Wizard of Oz",1939,9.95)';
        $query[]='INSERT INTO dvds (id,"Jaws",1977,6.95)';

        $this->pdo->exec( join(';',$query) );
    }

    function dump_test_records()
    {
        $query = "SELECT * FROM dvds";

        if ($statement = $this->pdo->prepare($query)) {
            $statement->execute();

            $rows = $statement->fetchAll(PDO::FETCH_ASSOC);

            echo "<pre>";
            var_dump($rows);
            echo "</pre>";
        }
    }

    function main()
    {
        if ( ! $this->open_table() ) {
            $this->add_test_records();
        }

        $this->dump_test_records();
    }
}

$test = new sqlite_test();
$test->main();