Webservice code:
function login($uname)
{
$id=1;
$link = mysql_pconnect("localhost", "root", "root") or die("Could not connect");
mysql_select_db("sparq",$link) or die("Could not select database");
$sql=mysql_query("select username,password from user_login where user_id=1");
//$result = mysql_query($query);
$arr = array();
while($obj = mysql_fetch_object($sql))
{
$arr[] = $obj;
}
// $obj = mysql_fetch_object($sql);
header("Content-type: application/json");
echo json_encode($arr);
}
客户端代码:
$url="http://localhost/web.php";
if (isset($_POST['Login']))
{
$ch = curl_init($url); // Initialize a CURL session
curl_setopt($ch, CURLOPT_HEADER, 0); // options for a CURL transfer
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_POSTFIELDS,"username=".$username );
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 0);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
$data = curl_exec($ch); // Perform a CURL session
curl_close($ch);
$arr =array();
$arr=json_decode($data,true);
echo 'I am here'; //echo1
echo $data; //echo2
echo $arr[0]->username; //echo3
我得到下面的输出:
I am here //echo1
[{"username":"akhilnk@gmail.com","password":"asdf123"}] //echo2
Notice: Trying to get property of non-object in F:'xampp'htdocs'webtest'login.php on line 38 //echo3
This:
echo $arr[0]->username;
应:echo $arr[0]['username'];
:
$arr = array();
while($obj = mysql_fetch_object($sql)) {
$arr[] = $obj;
}
应:$arr = array();
while($row = mysql_fetch_assoc($sql)) {
$arr[] = $row;
}
不能通过json发送php对象。javascript(和json)调用的对象是php中的关联数组。
Replace: echo $arr[0]->username;美元的arr[0]("用户名");