Php 为什么这个get函数会出现致命错误？

我试图根据3个要求选择posts表中的所有行

posts表中的userid=users表中的userid posts表中的hostid=users表中的userid posts表中的commentid=零 我想选择所有符合这些要求的行。 这就是错误

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

这是我的密码

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++;
        }    
    }

您没有调用$userid->fetch\u assoc来获取第一个查询的结果。您需要这样做，然后使用$row['userid']将userid从行中取出

在第一个查询中，还需要在用户名周围加引号。最好使用预先准备好的语句和$stmt->bind_param来防止SQL注入

但是首先不需要两个查询，您可以连接两个表

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，然后在第一个查询中就会收到语法错误的通知

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

我试着加上这个

//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您可以将$_GET['user']作为参数传递，通过这种方式，您可以单独说明如何传递用户名

function getUploads($conn, $username) {

}

3您可以使用内部联接创建一个查询

您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);

---

你需要调用$userid->fetch\u assoc，就像你要得到第二个查询的结果一样。我不知道“p.”是什么。。postid不是来自数据库hostid和userid必须都是相等的userid，因为信息在数据库中的放置方式p是posts作为p的别名