java我无法在php中解码json对象
这是我的java代码,它发出http请求并将json对象发送到php脚本
登录。爪哇
String username = jTextField1.getText();
String password = jPasswordField1.getText();
JSONObject obj = new JSONObject();
obj.put("username", username);
obj.put("password", password);
//JSONArray list = new JSONArray();
//list.add(username);
//list.add(password);
//obj.put("logindata", list);
try {
HttpClient httpclient = new DefaultHttpClient();
HttpResponse response;
HttpPost httppost = new HttpPost("http://localhost/kolitha/json_test/index.php");
StringEntity se = new StringEntity("myjson" + obj.toString());
httppost.setEntity(se);
System.out.print(se);
httppost.setHeader("Accept", "application/json");
httppost.setHeader("Content-type", "application/json");
System.out.println(obj.toString());
//response = httpclient.execute(httppost);
HttpGet httpget = new HttpGet("http://localhost/kolitha/json_test/index.php");
response = httpclient.execute(httpget);
HttpEntity entity = response.getEntity();
System.out.println(EntityUtils.toString(entity));
} catch (Exception e) {
e.printStackTrace();
System.out.print("Cannot establish connection!");
}
这是我的索引。php脚本,我无法从json对象提取用户名和密码
索引。php
$obj = json_decode(file_get_contents('php://input'));
$username=$obj->{'username'};
$password=$obj->{'password'};
$connect=mysql_connect('localhost', 'root', '');
IF (!$connect){
die ('Failed Connecting to Database: ' . mysql_error());}
$d = mysql_select_db("kolitha_json_test");
if(!$d){ echo "db not selected";}
$sql="SELECT * FROM login WHERE username='$username' AND password='$password' ";
$result=mysql_query($sql) or die (mysql_error());
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row
if($count==1)
{
echo "true";
}
else
{
echo "false";
}
?>
# 1 楼答案
好的,我留下了一条评论,但似乎没有人注意到一条评论。错误在Java代码中
它向名为
myjson
的json字符串添加前缀# 2 楼答案
为什么同时使用HttpPost和HttpGet您正在将字符串实体添加到HttpPost请求,而未使用它
而您使用的是Httpget,其中未设置您的用户名和密码(因此显然没有与Httpget请求相关联的json数据)
在此进行更改
如果您使用的是httppost,请在此处添加注释