Question

这样可行，它以以下形式返回我的数组：

{"results":[{"cat_id":"8",
"cat_name":"dental hygienist"},
{"cat_id":"5","cat_name":"stocktaker"},
{"cat_id":"9","cat_name":"builder"}]}

我的PHP代码是：

$query2 = "SELECT DISTINCT cat_id,cat_name FROM review WHERE public_or_private = 2";
$result2 = mysqli_query($con,$query2);
$rows = array();
while($row = mysqli_fetch_assoc($result2)) {
    $rows['results'][] = $row;
}  
echo json_encode($rows);

但是我如何才能与参数化查询相似？

mysqli_query()，mysqli_store_result()或mysqli_use_result() ...我尝试了几件这样的事情，但我不断获得null。

我的PHP代码：

$query3 = "SELECT DISTINCT cat_id,cat_name FROM review WHERE user_id = ? AND public_or_private = 0";
$stmt3 = $con->prepare($query3) or die(mysqli_error($con));
$stmt3->bind_param('i', $user_id) or die ("MySQLi-stmt binding failed ".$stmt3->error);
$stmt3->execute() or die ("MySQLi-stmt execute failed ".$stmt3->error);
$result3 = $stmt3->get_result();
$rows = array();
while ($row = mysqli_fetch_assoc($result3)) {
    $rows['results'][] = $row;
} 
echo json_encode($rows);

Answer 1

我不知道为什么，但这对我有用：

$numbervar = 0;
                $query3 = "SELECT DISTINCT cat_id,cat_name FROM review WHERE user_id = ? AND public_or_private = ?";
                $stmt3 = $con->prepare($query3) or die(mysqli_error($con));
                $stmt3->bind_param('ii', $user_id,$numbervar) or die ("MySQLi-stmt binding failed ".$stmt3->error);
                $stmt3->execute() or die ("MySQLi-stmt execute failed ".$stmt3->error);
                $result3 = $stmt3->get_result();

                $rows = array();

                while ($row = mysqli_fetch_assoc($result3)) {

                    $rows['results'][] = $row;

                } 

echo json_encode($rows);

试图让mysqli_fetch_assoc与bind_param一起工作

1 个答案: