PHP mysql查询返回错误结果
这是我的密码:PHP mysql查询返回错误结果,php,mysql,Php,Mysql,这是我的密码: <?php if(isset($_POST["stjel"])) { if($_POST["type"] == "Penger") { $luck = rand(1,4); $randuser = $sql->query("SELECT brukernavn FROM `brukere` WHERE `level`='1' AND `liv`>='0' ORDER BY RAND() LIM
<?php
if(isset($_POST["stjel"])) {
if($_POST["type"] == "Penger") {
$luck = rand(1,4);
$randuser = $sql->query("SELECT brukernavn FROM `brukere` WHERE `level`='1' AND `liv`>='0' ORDER BY RAND() LIMIT 1");
$randbruker = $sql->query("SELECT * FROM `brukere` WHERE `brukernavn`='$randuser'");
$target = mysql_fetch_object($randbruker);
$hvormyefra = $target->hand * 0.15;
$amount = $targed->hand - $hvormyefra;
if($luck == 1) {
echo"<div class=\"velykkett\">You stoled $amount from";
echo $randuser;
echo"</div>";
} else {
echo"You failed to steal money!";
}
}
}
?>
您应该获取结果并将其存储在另一个变量中$randuser将始终返回资源id。此外,您不需要编写两个单独的查询来获取随机用户的所有信息。
请尝试删除这两个查询,并仅写入此查询
SELECT * FROM `brukere` WHERE `level`='1' AND `liv`>='0' ORDER BY RAND() LIMIT 1
此查询将选择一个随机用户并返回与之相关的所有数据。query
函数返回资源。你可以从中得到你需要的东西。@u\u mulder我还有别的办法吗?/如何获取我尝试过的资源,但它只给我资源id#39您已经编写了$target=mysql\u fetch\u对象($randbruker)代码>。你不认为你应该对$randuser
也这样做吗?