我正在尝试使用此
更新数据库表的列$url = mysqli_connect($servername, $dbusername, $usrpassword, $dbname);
// Check connection
if (!$url) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT adv_val FROM current_advert";
if($result = mysqli_query($url, $sql)){
while($row = mysqli_fetch_assoc($result)){
$adv_val = $row['adv_val'];
}
}
echo "current advert is:" . $adv_val;
// Attempt select query execution
$sql = "SELECT advert_text FROM advertisements WHERE advert_id = $adv_val";
if($result = mysqli_query($url, $sql)){
while($row = mysqli_fetch_assoc($result)){
$advert = $row['advert_text'];
}
}
echo "<br>current advert text is:" . $advert;
if ($adv_val == 1 OR $adv_val == 2 OR $adv_val == 3) {
$adv_val = $adv_val + 1;
} else {
$adv_val = 1;
}
$sql = "UPDATE current_advert SET adv_val='$adv_val'";
// Close connection
echo "<br>next advert id is:" . $adv_val;
mysqli_close($url);
与数据库的连接是可以的,因为我能够在脚本的开头读取数据。这让我很伤心!
答案 0 :(得分:1)
$sql = "UPDATE current_advert SET adv_val='$adv_val'";
mysqli_query($url, $sql);
你错过了第二行。