Question

我有这个查询在动态表格中显示我的“页面”，但是它显示了所有页面，我想限制“页面”仅在其“campaignid”列与当前“广告系列”列匹配时显示“，任何人都有任何想法。

这是显示数据的查询：

  $counter=1;               $userID=PageDB::getInstance()->get_user_id_by_name($_SESSION['user']);              $result=PageDB::getInstance()->get_pages_by_campaign_id($campaignID);
            $i=0;
                while ($row = mysqli_fetch_array($result)):
                 $style = "";
                    if($i%2==0)
    {
        $style = ' style="background-color: #EFEFEF"';
    }
    echo "<tr".$style.">";
                echo "<td>" . htmlentities($row['pid']) . "</td>";
                echo "<td>". htmlentities($row['id']) . "</td>";
                ?>
                <td>
                <form style="display:none;"></form>
                    <form name="editPage" action="editPage.php" method="GET">
                        <input type="hidden" name="pageID" value="<?php echo $pageID = $row['pid']; ?>" />
                        <input type="submit" name="editPage"  value="<?php echo "Page "  . $counter; ?>" style="padding-top:4px; padding-bottom:6px;background:none;border:none; cursor:pointer"/>
                    </form> 
    </td>           
                <?php
                $pageID = $row['pid'];  
                $counter++;
                $i++;

                echo "</tr>\n";
            endwhile;
            mysqli_free_result($result);              
?>

以下是我查询数据库的方法。

public function get_pages_by_campaign_id($campaignID) {
    $campaignID = $this->real_escape_string($campaignID);
    return $this->query("SELECT pid,campaignid,id,campaign_name FROM pages,campaigns WHERE pages.campaignid = campaigns.id ORDER BY campaigns.id LIMIT 1,10");
    }

正如我所说它正在返回正确的数据，但我只想显示数据： pages.campaignid = campaigns.id [1]或pages.campaignid = campaigns.id [r]等...

Answer 1

您需要if-else循环内的while条件。如果当前行等于campaigns.id

，则此if-else语句将进行比较

显示正确的结果，但我想将它们限制为相同的值

1 个答案: