mysqli_affected_rows creates new connection (read

问题:

I am learning to use mysqli instead of mysql and facing some problems. This is my code.

require("classes/dbo.class.php");

$db->dml('update table set name="abc" where a_id=5);
echo "Rows Affected : ".mysqli_affected_rows($db->link());

Here's dbo.class.php

class dbo
{
    private $db = "dbname";
    private $user = "root";
    private $pass = "";
    private $host = "localhost";

    function link()
    {
        $link = mysqli_connect($this->host, $this->user, $this->pass) or die(mysqli_error($link));
        return $link;
    }

    function dml($q)
    {
        $link = mysqli_connect($this->host, $this->user, $this->pass) or die(mysqli_error($link));
        mysqli_select_db($link,$this->db) or die(mysqli_error($link));
        mysqli_query($link, $q) or die(mysqli_error($link));

    }

}

$db = new dbo();

Now the problem is I don't understand how to pass database link ($link) in mysqli_affected_rows() function. I tried above, but it seems to create a new database connection, so mysqli_affected_rows returns 0 instead of 1.

I thought of creating a new method dmlWithMysqliAffectedRow() in dbo.class.php that returns affected rows instead of true and false.

My solution looks stupid to me. Which is the better way to do this?

回答1:

First you have an error in this code, you have not closed the string for your query. Its also better to use a double quoted string and single quotes around you SQL data parameters.

require("classes/dbo.class.php");

//$db->dml('update table set name="abc" where a_id=5);
$db->dml("update table set name='abc' where a_id=5");
echo "Rows Affected : ".mysqli_affected_rows($db->link());

Second mysqli_ has a perfectly good OO implementation, use that rather than trying to re-invent the wheel, and getting a square one.

Documentation: for this can be found here, in many languages