谢谢
答案 0 :(得分:0)
$banners = $db->QueryFetchArrayAll("SELECT site_url FROM `banners` );
foreach($banners as $banner){
If ($banner['site_url']=='google.com or www.google.com'){
echo "this";
} else{
this"}
Site_url不是域名。它的网址是google.com/translate/etc/etc。我想提取域名并与google.com进行比较然后结果。
由于