我们已经将其设置为用户可以选中一个框以指示内容是成人内容。未选中的上传照片将上传并存储在数据库中,但是,如果他们选中该框以将其标记为成人,它仍会上传到服务器,但不会在表中创建任何字段。
$mature=$HTTP_POST_VARS["mature"];
if($mature=="on") {
// $adult=1; // has no function, could be useful?
}
else {
// $adult=0; // has no function, could be useful?
// ALBUM CODE HERE ALBUM CODE HERE ALBUM CODE HERE ALBUM CODE HERE ALBUM CODE HERE
$sql="select photo_id as a from user_pics where user_id = $_SESSION[user_id]";
$result2 = mysql_query($sql);
$pic=mysql_num_rows($result2);
if($pic==0) {
// if no photos found within the database table user_pics
// Identify if adult or not.
$mature_content=$HTTP_POST_VARS["mature"];
if($mature_content=="on") {
$adult_rated=1;
}
else {
$adult_rated=0;
}
// Identify if adult or not.
$sql="insert into user_pics";
$sql.="(user_id";
$sql.=", caption";
$sql.=", mature";
$sql.=", photo_url";
$sql.=", medium_photos";
$sql.=", small_photos)";
$sql.=" values($_SESSION[user_id]";
$sql.=", '$caption'";
$sql.=", $adult_rated";
$sql.=", '$picture_url'";
$sql.=", '$med_picture_url'";
$sql.=", '$small_picture_url')";
$result=mysql_query($sql);
$photo_id = mysql_insert_id();
} else {
// if one or more than one photo
// Identify if adult or not.
$mature_content=$HTTP_POST_VARS["mature"];
if($mature_content=="on") {
$adult_rated=1;
}
else {
$adult_rated=0;
}
// Identify if adult or not.
$sql="insert into user_pics";
$sql.="(user_id";
$sql.=", caption";
$sql.=", mature";
$sql.=", photo_url";
$sql.=", medium_photos";
$sql.=", small_photos)";
$sql.=" values($_SESSION[user_id]";
$sql.=", '$caption'";
$sql.=", $adult_rated";
$sql.=", '$picture_url'";
$sql.=", '$med_picture_url'";
$sql.=", '$small_picture_url')";
$result=mysql_query($sql);
$photo_id = mysql_insert_id();
}
// ALBUM CODE HERE ALBUM CODE HERE ALBUM CODE HERE ALBUM CODE HERE ALBUM CODE HERE
}
如果是$mature=="on"
,那么您将跳过可能将某些内容插入数据库的整个代码块。因此,如果这是真的,则不会插入任何内容。除非我误解了你在问什么。