所以我的查询就在这里我现在正在使用php进行分页现在每当我在我的查询中设置限制就会出现此错误警告:mysqli_fetch_array()期望参数1为mysqli_result,在C:\ xampp \ htdocs中给出布尔值第292行\ project \ Deshboard.php
$query=mysqli_query($conn,"SELECT *
FROM
notes LIMIT $start,$limit
INNER JOIN
subjects
ON
subjects.subject_id = notes.subject_id
WHERE by_teacher='".$_SESSION['admin_id']."'," );
答案 0 :(得分:1)
首先,欢迎来到Stack Overflow!
我现在多次阅读查询以为您创建一个可靠的答案,并且必须首先在接受参数时忘记prepared queries的重要性。这样可以避免对您的数据库进行SQL injection /恶意攻击。
完成此操作后,根据SQL SELECT-syntax,LIMIT
应位于WHERE
声明之后。
PHP代码应该看起来像这样:
// prepare a mysqli query with notes.by_teacher requiring
// a parameter (indicated by the question mark)
$stmt = mysqli_prepare(
$conn, 'SELECT * FROM notes
INNER JOIN subjects ON subjects.subject_id = notes.note_id
WHERE notes.by_teacher=?
LIMIT ?,?'
);
// set the first parameter as type string (s)
// to the value of $_SESSION['admin_id']
mysqli_stmt_bind_param($stmt, 'sii', $_SESSION['admin_id'], $start, $limit);