how to insert data into server database with user input?
currently, i retrieve data from database is like that
private void getdatafromphp(){
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
//http post
try{
HttpClient httpclient = new DefaultHttpCl开发者_如何学Goient();
HttpPost httppost = new HttpPost("http://10.0.2.2/video.php");
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
is = entity.getContent();
}catch(Exception e){
Log.e("log_tag", "Error in http connection"+e.toString());
}
//convert response to string
try{
BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
sb = new StringBuilder();
sb.append(reader.readLine() + "\n");
String line="0";
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
result=sb.toString();
}catch(Exception e){
Log.e("log_tag", "Error converting result "+e.toString());
}
//paring data
try{
jArray = new JSONArray(result);
JSONObject json_data=null;
json_data = jArray.getJSONObject(jArray.length()-1);
url=json_data.getString("VideoUrl");
}catch(JSONException e1){
}catch(ParseException e1) {
e1.printStackTrace();
}
}
with this php
<?php
mysql_connect("localhost","root","");
mysql_select_db("imammuda");
$sql=mysql_query("select * from Video");
while($row=mysql_fetch_assoc($sql))
$output[]=$row;
print(json_encode($output));
mysql_close();
?>
now i want insert data into database. how to do that?
I had found the sql command which is "insert into table (column1, column2) values ('value1', 'value2')".
This is insert with constant values which is type in php.
What i want is from java there get input from user then copy this input into php 'value1' after that run the php to update the database.
Depending on whether you are using Get or Post
i will assume GET
$value = $_GET['value']; // this will retrieve the value from the url and save it in a variable
mysql_connect("localhost","root","");
// escape the value first
$value = mysql_real_escape_string($value);
mysql_select_db("imammuda");
$result = mysql_query("insert into Video (value) values ('$value')");
?>
learn more about working with the db here
UPDATE
to know the correct request method you can use this.
$req;
if ($_SERVER['REQUEST_METHOD'] == 'GET') {
$req = $_GET;
}else {
$req = $_POST;
}
now you can use $req as your request variable:
$value = $req['value'];
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