Php 从数据库获取值不工作

我试图从本地数据库中获取特定值，但似乎不起作用，以下是我的代码：

$acceptance = " SELECT acceptance_status FROM teams_0517 WHERE team_id=$team_id"; 
$acceptancex = mysql_query($acceptance);
mysql_close($connection);

if($acceptancex == 0)
{
    ob_clean();
    header('Location: index.php');
    ob_flush();
}
else
{
    ob_clean();
    header('Location: signup.php');
    ob_flush();
}

---

当我运行代码时，php将跳过if条件并继续执行其他语句，因此

$acceptance

似乎没有值。。有什么帮助吗？

您需要在执行查询后获取结果。 看

试着这样做：

$acceptance = " SELECT acceptance_status FROM teams_0517 WHERE team_id=$team_id limit 1"; 
$result = mysql_query($acceptance);
$acceptancex = mysql_fetch_object($result); // fetch object


if($acceptancex->acceptance_status == 0)
{
    ob_clean();
    mysql_close($connection);
    header('Location: index.php');
    ob_flush();
}
else
{
    ob_clean();
    mysql_close($connection);
    header('Location: signup.php');
    ob_flush();
}

试试这个：-

$acceptance = " SELECT acceptance_status FROM teams_0517 WHERE team_id=$team_id limit 1"; 
$result = mysql_query($acceptance);
$acceptancex = mysql_fetch_object($result); // fetch object

if($acceptancex->acceptance_status == 0)
{
    $redirect = "index.php";
}
else
{
    $redirect = "signup.php";
}
mysql_close($connection);

ob_clean();
header('Location: '.$redirect);
ob_flush();

---

设置出口的良好做法；使用标题（“位置…”后<代码>打印（$acceptancex）在这一行之后：

$acceptancex=mysql\u fetch\u object（$result）

和die以查看结果，并确保

$team\u id

正确打印此值，以便始终跳过条件