php - Post variable via ajax to mysql db - phonegap -

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i want post simple variable phonegap app mysql database. dont work. db show me empty results.

mysql database looks this, tablename user

rows:

user (varchar 25),vip (varchar 1),call (varchar 1)

at top of project inserted:

            <script type="text/javascript" src="js/tcplugin.js"></script>             <script type="text/javascript" src="js/phoneapp.js"></script>             <script type="text/javascript" src="js/start.js"></script>

start.js file ajax request - other 2 working fine guess there not problem

my start.js file looks this:

$(document).ready( function() {                   var email = "john@web.de";                     $.ajax({                          type: 'post',                          url: 'http://www.blablabla.de/phone/action.php',                          data: {                          data: {"email" : email },                          //data:email,                          },                          success: function(result) {                          console.log(result);                          }                          });

and action.php looks this:

<?php         $dbhost = "bla";         $dbuser = "bla";         $dbpass = "bla";         $dbname = "bla";    $user = $_post['data']['email'];  $conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname);  if ($conn->connect_error) {     die("connection failed: " . $conn->connect_error); }   $sql = "insert user (email) values ('" . $user . "')";   if ($conn->query($sql) === true) {     echo "new record created successfully"; } else {     echo "error: " . $sql . "<br>" . $conn->error; }  $conn->close();  ?>

but everytime start app db empty. debugger concole not giving me errors. db connection working, tried echo connection result. , other js working fine.

action.php

<?php         $dbhost = "bla";         $dbuser = "bla";         $dbpass = "bla";         $dbname = "bla";  print_r($_post); // see full contents of post  $user = $_post['data']['email'];  print php_eol . $user . php_eol; // see full contents of $user var  $conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname);  if ($conn->connect_error) {     die("connection failed: " . $conn->connect_error); }   $sql = "insert user (email) values ('" . $user . "')";  print php_eol . $sql . php_eol; // see actual sql query, can test later in sql editor ex. phpmyadmin   if ($conn->query($sql) === true) {     echo "new record created successfully"; } else {     echo "error: " . $sql . "<br>" . $conn->error; }  $conn->close();  ?>

make these edits, run ajax , check response tab in debugger.


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