Reputation: 92
I am trying a simple post method to insert into a database.Below is the code to point me out my mistake.
This is android code to send a post request.I don't have any idea whats wrong but the parameter that is needed to be insert is not passed and an empty string is being inserted in database.
java
String add = "myName";
BufferedReader reader = null;
String text = "";
Log.d("K2NK", "Connecting....");
URL url;
String response = "";
try {
url = new URL("http://10.13.210.114/myDatabase/insert.php");
URLConnection conn = url.openConnection();
conn.setDoOutput(true);
OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
wr.write(add);
wr.flush();
//get the server response
reader = new BufferedReader((new InputStreamReader(conn.getInputStream())));
StringBuilder sb = new StringBuilder();
String line = null;
//read server response
while ((line = reader.readLine()) != null) {
//append server response in string
sb.append(line + "\n");
}
text = sb.toString();
Log.d("K2NK", text);
}catch (IOException e) {
e.printStackTrace();
} finally {
try {
if(reader!=null){
reader.close();}
} catch (IOException e) {
e.printStackTrace();
}
}
PHP script is as below to insert data into database
PHP
<?php
$name = ($_POST['name']);
$con = new mysqli("localhost", "root", "","syncadapter");
if($con->connect_error){
die("Connection failed: ".$con->connect_error);
}
//sql to delete record
$sql="INSERT INTO table_student (name) values ('$name')";
if($con->query($sql)===TRUE){
echo json_encode("insert successful: name:".$name);
} else {
echo json_encode("insert failed");
}
?>
Upvotes: 0
Views: 57
Reputation: 1033
The string you passed should be:
String add = "name=myname";
It is trying to find a Post variable by name name which does not exist.
Upvotes: 2