為什麼只顯示第一個資料庫查詢的結果，而忽略了第二個查詢的功能？

Question

以下輸出顯示了第一個查詢的結果，但沒有顯示第二個查詢的結果。當我不使用函數進行查詢時，它可以正常工作。

function builtFooter($catactive, $catlink)
{
    global $conn;
    $sql = "SELECT catid, catname_de AS name, catlink as link, extern, extlink FROM categories WHERE catactive=? AND catlink!=? ORDER BY catsort";
    $stmt = $conn->prepare($sql);
    $stmt->bind_param("ss", $catactive, $catlink);
    $stmt->execute();
    $result = $stmt->get_result();
    while ($cat = $result->fetch_assoc()) {
        $resArr[] = $cat;
        echo '
' . $cat['name'] . '
';

        $cities = "SELECT COUNT(a.ad_id) AS ANZ,a.region, b.city, b.citylink FROM ads a, neueorte b WHERE a.zeigen=? AND a.gesperrt=? AND FIND_IN_SET(?, a.portale) AND a.catid=? AND b.po_id=a.region GROUP BY a.region ORDER BY ANZ DESC LIMIT 0,50";
        $stmt = $conn->prepare($cities);
        $zeigen = 'ja';
        $gesperrt = 'no';
        $category = $cat['catid'];
        $stmt->bind_param("ssss", $zeigen, $gesperrt, $portalnumber, $category);
        $stmt->execute();
        $result = $stmt->get_result();
        while ($city = $result->fetch_assoc()) {
            $resArr[] = $city;
            echo ' ' . $city['city'] . ' 
';
        }
    }
}

builtFooter('yes', 'markt');

Answer 1

這些查詢都正常工作，我得到了想要的結果：

$catactive = 'yes';
$catlink = 'markt';
$sql = "SELECT catid, catname_de AS name, catlink as link, extern, extlink FROM categories WHERE catactive=? AND catlink!=? ORDER BY catsort";
        $stmt = $conn->prepare($sql); 
        $stmt->bind_param("ss", $catactive, $catlink);
        $stmt->execute();
        $result = $stmt->get_result();
        while ($cat = $result->fetch_assoc()) {    
            $resArr[] = $cat;
           echo '<h3><a href="'.$cat['link'].'" title="'.$cat['name'].' bei $portalname">'.$cat['name'].'</a></h3>';
        
        $cities = "SELECT COUNT(a.ad_id) AS ANZ,a.region, b.city, b.citylink FROM ads a, neueorte b WHERE a.zeigen=? AND a.gesperrt=? AND FIND_IN_SET(?, a.portale) AND a.catid=? AND b.po_id=a.region GROUP BY a.region ORDER BY ANZ DESC LIMIT 0,50";
        $stmt = $conn->prepare($cities); 
        $zeigen = 'ja';
        $gesperrt = 'no';
        $category = $cat['catid'];
        $stmt->bind_param("ssss", $zeigen, $gesperrt, $portalnumber, $category);
        $stmt->execute();
        $r = $stmt->get_result();
        while ($city = $r->fetch_assoc()) {    
            $resArr[] = $city;
            echo' '.$city['city'].' <br />';
        }
       }

但是當我建立一個函數時，我無法得到第二個查詢的結果：

function builtFooter($catactive,$catlink){
        global $conn;
            $sql = "SELECT catid, catname_de AS name, catlink as link, extern, extlink FROM categories WHERE catactive=? AND catlink!=? ORDER BY catsort";
            $stmt = $conn->prepare($sql); 
            $stmt->bind_param("ss", $catactive, $catlink);
            $stmt->execute();
            $result = $stmt->get_result();
            while ($cat = $result->fetch_assoc()) {    
                $resArr[] = $cat;
               echo '<h3><a href="/sexkontakte/'.$cat['link'].'" title="'.$cat['name'].' bei $portalname">'.$cat['name'].'</a></h3>';
            
            $cities = "SELECT COUNT(a.ad_id) AS ANZ,a.region, b.city, b.citylink FROM ads a, neueorte b WHERE a.zeigen=? AND a.gesperrt=? AND FIND_IN_SET(?, a.portale) AND a.catid=? AND b.po_id=a.region GROUP BY a.region ORDER BY ANZ DESC LIMIT 0,50";
            $stmt = $conn->prepare($cities); 
            $zeigen = 'ja';
            $gesperrt = 'no';
            $category = $cat['catid'];
            $stmt->bind_param("ssss", $zeigen, $gesperrt, $portalnumber, $category);
            $stmt->execute();
            $r = $stmt->get_result();
            while ($city = $r->fetch_assoc()) {    
                $resArr[] = $city;
                echo' '.$city['city'].' <br />';
    }
    }
       }
       builtFooter('yes','markt');

我有categories、ads和neueorte這些表。我想建立頁腳鏈接，所以我想將類別作為h3標籤，然後按城市分組，其中有放置廣告的城市。希望我做得對，我是這個頁面的新手。