在我的数据库中,我有我的用户,他们每个人都有自己的"黄金"数量,我制作了一个按钮,每次点击都会增加他们的黄金数量100。问题是,我不知道如何为特定的登录用户做这件事。这是我当前的更新代码
$addgold2 = mysqli_query($con,"UPDATE stats SET stats.gold = stats.gold + 100 WHERE stats.id=$retval2");
如果我删除"$retval2"并只输入ID,它就可以工作了。我想调用登录用户的ID,然后只更新这些统计数据。这是我的代码
$id2 = "SELECT users.id FROM users WHERE users.username=$username";
$retval2 = mysqli_query($con,$id2);
现在,如果我回显$id2,它显示ID是"1",但不是1!现在是7点(
我也得到这个错误
Catchable fatal error: Object of class mysqli_result could not be converted to string in C:'wamp'www'Game'addgold.php on line 19
我把放在代码的第一行
最后,这里是整个代码(表单在不同的页面上)
<?php error_reporting(E_ALL); ini_set('display_errors', 1);
include("connect.php");
include("header.php");
$username = $_SESSION['userlogin'];
$id2 = "SELECT users.id FROM users WHERE users.username=$username";
$retval2 = mysqli_query($con,$id2);
$sql = "SELECT stats.id, stats.gold, users.id, users.username FROM stats, users WHERE users.username = '$username' AND stats.id = users.id";
$retval = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($retval, MYSQL_ASSOC))
{
echo "Gold: {$row['gold']} <br> ".
"--------------------------------<br>";
}
$row2 = mysqli_fetch_row($retval2);
$id2 = mysqli_num_rows($retval2) or die(mysql_error()." ".$id2);
echo $id2;
if(isset($_SESSION['userlogin'])){
$addgold2 = mysqli_query($con,"UPDATE stats SET stats.gold = stats.gold + 100 WHERE stats.id=$retval2");
if(isset($addgold2['submit'])){
}
echo "You have earned 100 gold!";
mysqli_close($con);
}
?>
您在$id2
sql 中错过了一个引号
$id2 = "SELECT users.id FROM users WHERE users.username='$username'";
$id2 = mysqli_num_rows($retval2) or die(mysql_error()." ".$id2);
如果它有1行,它将返回1。。不是ID
$addgold2 = mysqli_query($con,"UPDATE stats SET stats.gold = stats.gold + 100 WHERE stats.id=$retval2");
$retval2是mysqli_query结果,而不是ID我认为$row2["id"]是您的id。