我如何回顯mysql select as json的結果？當前它將回顯響應：ID: 1 - Name: John DoeID: 2 - Name: John Deo此致，$conn = new mysqli($servername, $username, $password, $dbname);// Check connectionif ($conn->connect_error) { die("Connection failed: " . $conn->connect_error);} $sql = "SELECT id, firstname, lastname FROM MyGuests";$result = $conn->query($sql);if ($result->num_rows > 0) { // output data of each row while($row = $result->fetch_assoc()) { echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>"; }} else { echo "0 results";}$conn->close();
1 回答
FFIVE
TA貢獻1797條經驗 獲得超6個贊
首先將所有行添加到一個數組中：
$data = [];
while($row = $result->fetch_assoc()) {
$data[] = $row;
}
然后將數組轉換為json并輸出：
echo json_encode($data);
- 1 回答
- 0 關注
- 119 瀏覽
添加回答
舉報
0/150
提交
取消