php - PHP中的mysql_query参数错误

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Closed 7 years ago.

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mysql_fetch_array() expects parameter 1 to be resource, boolean given in select

$connection = mysql_connect('localhost', 'username', 'password') or die('Database connection failed:' . mysql_error());                                

mysql_select_db('db_test', $connection) or die('Database connection failed: ' . mysql_error());

$var = '';

switch ($var) {
    case 'dosomething':
        break;
    default:
        default_func();
}

function default_func() {
    if (isset($_POST['submit'])) {
        $query = "INSERT INTO `table_test` (`some_field`) VALUES ('test')";
        $result = mysql_query($query, $connection) or die(mysql_error());

    header('Location: index.php?submit=success');
    }
}

mysql_close($connection);

我已经收到“ mysql_query（）期望参数2在给定错误的情况下为资源null”。

我确保将$ connection分配给数据库连接资源id的值，因此这似乎不是问题。不知道为什么$ connection的值给我一个NULL。关于为什么会发生此错误的任何解释？

Answer 1

$ connection变量在default_func中不可见。如果要成为它，只需在此处将其声明为global：

function default_func() {
    global $connection;
  ...
}

但是，当代码变得稍微复杂一些时，全局变量可能会导致难以诊断的错误。隐藏的全局状态（如果将第二个参数完全保留为 mysql_query，则使用的是全局状态）更糟。您最好将连接传递给函数：

switch ($var) {
    case 'dosomething':
        break;
    default:
        default_func($connection);
}

function default_func($conn) {
    if (isset($_POST['submit'])) {
        $query = "INSERT INTO `table_test` (`some_field`) VALUES ('test')";
        $result = mysql_query($query, $conn) or die(mysql_error());

    header('Location: index.php?submit=success');
    }
}