Question

我可以通过编辑程序PHP（只是）来完成，但是OOP是另一回事。所以我对我在这里所做的事情并不熟悉，但我正在努力......

我有一个名为Quote.object.php的文件，其中包含以下内容：

        $Query = new DbQuery( "INSERT", "quotes", $array );
        $this->id = mysqli_insert_id();

mysqli_insert_id需要提供数据库连接参数，但我不知道该怎么做。还有一个名为Mysql.handler.php的文件包含数据库连接变量 - 有没有一种方法可以将$ con作为$ Query的参数提供？

class DbQuery extends DbConnectionInfo{
// file: includes/classes/MysqlQuery.php
// contains functions needed to perform queries on mysql database and functions for     necessary data processing for application
//      SELECT = new DbQuery("select", table,cols[$value]  ,where[$col=$value],order[$value],limit);
//      INSERT = new DbQuery("insert", table, data[$col=$value]);
//      UPDATE = new DbQuery("update", table, data[$col=$value],where[$col=$value],limit);

// set testing as true for SQL reports in page
var $results;
var $sql;

function __construct($mode,$table = '',$var1 = '',$var2 = '',$var3 = '',$var4='')
//connects to database according to info in DbConnectInfo, runs query, closes connection
{
$temp = '';
$con = mysqli_connect($this->host, $this->user, $this->pass) or die ('There was a problem connecting to the database ' . (ENVIRONMENT == 'Development' ? mysqli_error() . "$this->user, $this->pass, $this->host" : ''));
mysqli_select_db($con,$this->db) or die ('There was a problem connecting to the database'  . (ENVIRONMENT == 'Development' ? mysqli_error() : ''));

我正在尝试从DbQuery获取$ con，因此我可以将其放入mysqli_insert_id()。我认为这就是我需要的东西？有没有办法从DbQuery获得$ con并放入mysqli_insert_id()？或者您需要更多信息才能知道这一点吗？

NB 我试图简明扼要地展示相关信息，如果我错过了其他有用的信息，我会道歉。

Answer 1

您正在定义一个对象，因此请将$ con作为类变量，例如

function __construct() {
    $this->con = mysqli_connect(...);
    ^^^^^^^----store in object
}

function foo() {
    $result = mysqli_query($sql, $this->con);
                                   ^^^^^^^^---retrieve from object
}

如何将$ con作为$ Query的参数提供？

1 个答案: