Question

<?php
$tid = $_GET['tid'];
$id = $_SESSION['userid'];
$sql1 = "SELECT * FROM topics WHERE id='$tid' LIMIT 1";
$res1 = mysqli_query($connect, $sql1) or die(mysqli_error($connect));
while ($row = mysqli_fetch_array($res1, MYSQLI_ASSOC)) {
    $title = $row['topic_title'];
    $creator = $row['topic_creator'];
}
$sql = "SELECT * FROM users WHERE id='$creator' LIMIT 1";
$user_query = mysqli_query($connect, $sql) or die(mysqli_error($connect));
while ($row = mysqli_fetch_array($user_query, MYSQLI_ASSOC)) {
    $name = $row["first"].$row["last"];
}
echo $name;
?>

我对PHP有点新，但我做的事情与此完全相同，但这次我收到了错误。这里的一切都有效，除了$ name。我检查了我的SQL表并确保用户存在，并且有第一个和最后一个区域。我不知道还有什么可能是错的。

注意：未定义的变量：第**行 * 中的名称

谢谢。

Answer 1

请尝试使用以下代码：

<?php
$tid = $_GET['tid'];
$id = $_SESSION['userid'];
$tid = mysqli_escape_string($connect, $tid);
$sql1 = "SELECT * FROM topics WHERE id='{$tid}' LIMIT 1";
$res1 = mysqli_query($connect, $sql1) or die(mysqli_error($connect));
// Check for rows first.
if($res1 and mysqli_num_rows($res1)){
    // Use if as while is pointless on LIMIT 1
    if($row = mysqli_fetch_array($res1, MYSQLI_ASSOC)) {
        $title = $row['topic_title'];
        $creator = $row['topic_creator'];

        $creator = mysqli_escape_string($connect, $creator);
        $sql = "SELECT * FROM users WHERE id='{$creator}' LIMIT 1";
        $user_query = mysqli_query($connect, $sql) or die(mysqli_error($connect));
        // Check for rows first.
        if($user_query and mysqli_num_rows($user_query)){
            // Use if as while is pointless on LIMIT 1
            if ($row = mysqli_fetch_array($user_query, MYSQLI_ASSOC)) {
                $name = $row["first"].$row["last"]; // NO HIT!
            }
            echo $name;
        }else{
            echo 'no rows found (query 2).';
        }
    }
}else{
    echo 'no rows found (query 1).';
}
?>

变量$name未定义，因为未达到$name = ...;行。因此，请确保$sql查询实际返回结果。必须定义$name。

未定义的变量，不确定原因

1 个答案: