php中,从数据库读取数据,并以json格式返回数据。具体方法如下:
第一步,定义相关变量
$servername = "localhost"; $username = "root"; $password = "root"; $mysqlname = "datatest"; $json = ''; $data = array(); class User { public $id; public $fname; public $lname; public $email; public $password; } |
第二步,链接数据库,代码如下:
// 创建连接 $conn = mysqli_connect($servername, $username, $password, $mysqlname); |
第三步,定义查询语句,并执行,代码如下:
$sql = "SELECT * FROM userinfo"; $result = $conn->query($sql); |
第四步,获取查询出来的数据,并将其放在事先声明的类中,最后以json格式输出。代码如下:
if($result){ //echo "查询成功"; while ($row = mysqli_fetch_array($result,MYSQL_ASSOC)) { $user = new User(); $user->id = $row["id"]; $user->fname = $row["fname"]; $user->lname = $row["lname"]; $user->email = $row["email"]; $user->password = $row["password"]; $data[]=$user; } $json = json_encode($data);//把数据转换为JSON数据. echo "{".'"user"'.":".$json."}"; }else{ echo "查询失败"; } |
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