Php while和for循环问题

Question

我在下面描述的while循环..My问题中有问题。

$sql = mysql_query($query, $this->db);
        if(mysql_num_rows($sql) > 0)
        {
            $result = array();
            while($rlt = mysql_fetch_array($sql,MYSQL_ASSOC))
            {

                $theature = explode(",",$rlt['mw_movie_theature']);
                //echo 'count'.count($theature).'<br/>'; print_r($theature);

                for($i = 0; $i<count($theature); $i++ )
                {
                    $sqls = mysql_query("SELECT * FROM mw_theatres WHERE status = 1 AND id='".$theature[$i]."'", $this->db);
                    $rlts = array();
                    while($rlts = mysql_fetch_array($sqls,MYSQL_ASSOC))
                    {
                        $rlt['movie'] = $rlts;
                    }
                }
                $rlt['value'] = 'true';
                $result[] = $rlt;

            }

            echo '<pre>';print_r($result);die;

具有2,3,4值的$theature变量。但是$rlt‘’movie‘的值只给出了最后4个id结果。我想要2,3,4个id值。

Answer 1

你应该试试这个：--

$sqls = mysql_query("SELECT * FROM mw_theatres WHERE status = 1 AND id='".$theature[$i]."'", $this->db);
$rlts = array();
$j=0;
while($rlts = mysql_fetch_array($sqls,MYSQL_ASSOC))
{
     $rlt['movie'][$j] = $rlts;
     $j++;
}

Answer 2

因为每次您将$rlts赋给相同的变量$rlt‘’movie‘。尝试$rlt‘’movie‘

Answer 3

$sqls = mysql_query("SELECT * FROM mw_theatres WHERE status = 1 AND id IN (".$theature.")", $this->db);
$rlts = array();
$k = 0;
while($rlts = mysql_fetch_array($sqls,MYSQL_ASSOC))
{
    $rlt['movie'][$k] = $rlts;
    $k++;
}

在where类中，我们使用id In，它只检查变量中作为字符串的id，因此没有使用额外的for循环和explod函数