我努力寻找解决方案，但我仍然没有设法解决它。请帮助。下面是我的java代码:-
public class MainActivity extends Activity {
String project_id;
String id;
InputStream is=null;
String result=null;
String line=null;
int code = 0;
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
final EditText e_id =(EditText) findViewById(R.id.editText1);
final EditText e_prjId =(EditText) findViewById(R.id.editText2);
Button insert =(Button) findViewById(R.id.button1);
id = e_id.getText().toString();
project_id = e_prjId.getText().toString();
insert.setOnClickListener(new View.OnClickListener() {
public void onClick(View v) {
insert();
}
});
}
public void insert() {
final ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("id",id));
nameValuePairs.add(new BasicNameValuePair("Project_Id",project_id));
new Thread(new Runnable() {
public void run() {
try {
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://192.168.0.111/insert.php");
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
is = entity.getContent();
Log.e("pass 1", "connection success ");
}
catch(Exception e){
Log.e("Fail 1", e.toString());
Toast.makeText(getApplicationContext(), "Invalid IP Address",
Toast.LENGTH_LONG).show();
}
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
StringBuilder sb = new StringBuilder();
while ((line = reader.readLine()) != null){
sb.append(line + "'n");
}
is.close();
result = sb.toString();
Log.e("pass 2", "connection success ");
}
catch(Exception e){
Log.e("Fail 2", e.toString());
}
try {
Log.i("tagconvertstr", "["+result+"]");
JSONObject json_data = new JSONObject(result);
code=(json_data.getInt("code"));
} catch (JSONException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
if(code==1)
{
Toast.makeText(getBaseContext(), "Inserted Successfully",Toast.LENGTH_SHORT).show();
}
else
{
Toast.makeText(getBaseContext(), "Sorry, Try Again",Toast.LENGTH_LONG).show();
}
}
}).start();
}
: -
<?php
$uname='root';
$pwd='';
$con = new PDO("mysql:host=192.168.0.111;dbname=wktask", $uname, $pwd);
$ID=$_REQUEST['ID'];
$Project_Id=$_REQUEST['Project_Id'];
$flag['code']=0;
if($r= $con->query("insert into task(ID,Project_Id) values('$ID','$Project_Id')"))
{
$flag['code']=1;
}
echo(json_encode($flag));
?>
我真的不知道是什么原因，我一直收到错误消息从JSON异常错误。真的很感激有人能帮助我。
谢谢
注意，PHP关联数组是区分大小写的
您正在发送id:
nameValuePairs.add(new BasicNameValuePair("id",id));
不等于ID
除了这个错误之外，你没有检查php脚本中的数据，我为你重写了:
$data = array();
if(isset($_POST['id'], $_POST['Project_Id']){
$id=$_POST['id'];
$project_id=$_POST['Project_Id'];
$uname='root';
$pwd='';
$con = new PDO("mysql:host=192.168.0.111;dbname=wktask", $uname, $pwd);
$con->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION );
$stmt = $con->prepare('INSERT INTO task (`ID`, `Project_Id`) values(:id, :project_id)'))
$success = $stmt->execute(array(':id'=>$id, ':project_id'=>$project_id));
if($success){
$data['code'] = 1;
$data['msg'] = 'INSERT successful';
}else{
$data['code'] = 0;
$data['msg'] = 'INSERT Failed';
}
}else{
$data['code'] = 0;
$data['msg'] = 'values are not set';
}
echo(json_encode($data));