php - 为什么这个 get 函数会出现致命错误？

Question

我正在尝试根据 3 个要求选择我的帖子表中的所有行。

1. 帖子表中的用户 ID = 用户表中的用户 ID
2. Posts 表中的 hostid = users 表中的 userid
3. 帖子表中的评论 ID = 零

我想选择所有满足这些要求的行。这是错误...

Fatal error: Uncaught Error: Call to a member function fetch_assoc() on bool in C:\xampp\htdocs\loginsystem\includes\posts.inc.php:108 Stack trace: #0 C:\xampp\htdocs\loginsystem\home.php(38): getUploads(Object(mysqli)) #1 {main} thrown in C:\xampp\htdocs\loginsystem\includes\posts.inc.php on line 108

这是我的代码...

function getUploads($conn) {

        $userName = $_GET["user"];
        $sqluserid = "SELECT userid FROM users WHERE userName = $userName";
        $userid = mysqli_query($conn, $sqluserid);
        $sqlusercontent = "SELECT * FROM posts WHERE hostid = $userid AND userid = $userid AND commentid = 0";
        $usercontent = mysqli_query($conn, $sqlusercontent);
        $postid = 1;

    // This is the actual upload content
        while ($row = $usercontent->fetch_assoc()) {
            echo "<div class='postbox'><p>";
            echo $row['title']."<br>";
            echo $row['date']."<br>";
            echo "<div><img src='posts/".$userid."/".$postid.".*></div>";
            echo $row['description']."<br>";
            echo "</p>";
            echo "</div>";
            $postid++;
        }    
    }

Answer 1

您没有调用$userid->fetch_assoc()获取第一个查询的结果。您需要执行此操作，然后使用$row['userid']将用户标识移出行。

您还需要在第一个查询中为用户名加上引号。最好使用准备好的语句 and$stmt->bind_param()来防止 SQL 注入。

但是首先不需要两个查询，您可以将两个表连接起来。

SELECT p.*
FROM posts AS p
JOIN users AS u ON p.userid = u.userid AND p.hostid = u.userid
WHERE u.username = '$userName' AND p.commentid = 0

您应该检查查询的结果，然后您会在第一个查询中收到语法错误通知。

$result = mysqli_query($conn, $sql) or die($conn->error);

Answer 2

1）尝试添加这个

//This will prevent if you dont have any data
if ($usercontent = $conn->query($query)) {
    while ($row = $usercontent->fetch_assoc()) {
      //the rest of your code here
    }
/* free result set */
$result->free();
} else {
 echo "No Data";
}

2）您可以通过$_GET['user']这种方式传递as参数，您可以分开如何传递用户名。

function getUploads($conn, $username) {

}

3）您可以使用INNER JOIN只创建一个查询

你的代码最后应该是这样的

function db () {
    static $conn;
    if ($conn===NULL){ 
        $conn = mysqli_connect ("localhost", "root", "", "database");
    }
    return $conn;
}

function getUploads($userName)
{
    $conn = db();
    $sqlusercontent = "SELECT p.*
    FROM posts AS p
    JOIN users AS u ON p.userid = u.userid AND p.hostid = u.userid
    WHERE u.username = '$userName' AND p.commentid = 0";

   // not suere why hostid is equal to userid ?
    $postid = 1; //$postid is not suppose to come from the db ?

    // This is the actual upload content
    if ($usercontent = mysqli_query($conn, $sqlusercontent)) {
        while ($row = $usercontent->fetch_assoc()) {
            echo "<div class='postbox'><p>";
            echo $row['title'] . "<br>";
            echo $row['date'] . "<br>";
            echo "<div><img src='posts/" . $row['userid'] . "/" . $row['postid']  . "'></div>";
            echo $row['description'] . "<br>";
            echo "</p>";
            echo "</div>";
            $postid++; //$postid is not suppose to come from the db ?

        }
        /* free result set */
        $usercontent->free();
    } else {
        echo "No Data";
    }
}
    $userName = $_GET["user"];
    getUploads($username);