Question

我正在用PHP开发一个小项目。我试图获取与数据库中类别相对应的数字以显示，但在显示页面上多次出现相同的错误：

警告：mysqli_num_rows（）期望参数1为mysqli_result，在第34行的C：\ xampp \ htdocs（A）Book 2.0 \ Bootstrap \ content_function.php中给出布尔值

以下是该错误针对的代码：

function getnumtopics($cat_id, $subcat_id){
include ('../db.php');
$select = mysqli_query($mysqli, "SELECT category_id, subcategory_id FROM topics WHERE".$cat_id." = category_id
                    AND ".$subcat_id." = subcategory_id");

return mysqli_num_rows($select);

}

第34行是return语句。我的语法似乎有问题吗？

Answer 1

您应先检查$select(result)的结果（即mysqli_query），然后再将其传递给mysqli_num_rows，请参阅documentation

Answer 2

问题在这里：

"SELECT category_id, subcategory_id FROM topics WHERE".$cat_id." = category_id AND ".$subcat_id." = subcategory_id"

查询中有两个问题，首先是您忘记在WHERE之后加上空格，其次是在与字符串比较时，必须将其与单引号连接起来，例如：

"SELECT category_id, subcategory_id FROM topics WHERE ".$cat_id." = 'category_id' AND ".$subcat_id." = 'subcategory_id'";

尝试上述查询。

Answer 3

尝试一下：

function getnumtopics($cat_id, $subcat_id){
include ('../db.php');
$select = mysqli_query($mysqli, "SELECT category_id, subcategory_id FROM topics WHERE ".$cat_id." = category_id
                    AND ".$subcat_id." = subcategory_id");
$get = mysqli_num_rows($select);

return $get;

必须通过mysqli_num_rows传递哪些参数？

3 个答案: