Fetching the current location of the browser and send the data in db using ajax and php

Question

I am doing an silly mistake in getting the location of the browser and storing using an external ip , ajax and php file.I am getting an error as

Uncaught SyntaxError: Unexpected token :

I think i am not sending the data properly in the php file, Here is my AJAX function:

$(document).ready(function() {
  $.get("https://ipinfo.io", function(response) {
      var city_data = val(response.city);
      var region_data = val(response.region);
        data: {city: city_data, region: region_data},
        type: "post",
        url: "test3.php",
        success: function(data){
             alert("Data Saved: " + data);
        }
}, "jsonp");

});

and my php file as :

<?php
 if(isset($_REQUEST))
 {
 mysql_connect("localhost","root","");
 mysql_select_db("practice");
 error_reporting(E_ALL && ~E_NOTICE); 
 print_r($_POST);
 return;
  $city=$_POST['city'];
  $region=$_POST['region'];
  $sql="INSERT INTO location(area,country) VALUES ('$city','$region')";
  $result=mysql_query($sql);
  if($result){
  echo "Work Done.";
   } 
   }
   ?>

I used mysql as this is for testing, Please have a look on my codes and let me know where i went wrong. Thanks in advance.

Please check your syntax. You have the body of your callback function mingled with AJAX properties. — DarthJDG
– DarthJDG, Commented Aug 1, 2017 at 16:28
since you are practicing, why dont you adopt PDO right now ? will save you tons if issues later, not to mention that mysql_* is deprecated right now. Read about little bobby and SQL injection. — YvesLeBorg
– YvesLeBorg, Commented Aug 1, 2017 at 16:35
@DarthJDG Can you write in the answer field where i did wrong as i am not getting it. Thanks — Piyush Seofaceup
– Piyush Seofaceup, Commented Aug 1, 2017 at 16:36
@YvesLeBorg Sure ,I will practice with PDO here after , Can you tell me where i went wrong — Piyush Seofaceup
– Piyush Seofaceup, Commented Aug 1, 2017 at 16:37
Well the return; saves you from SQL injections because the SQL will never get executed. If called from the global scope, then execution of the current script file is ended. — chris85
– chris85, Commented Aug 1, 2017 at 16:40

Guillermo Andres Fuentes Moral · Accepted Answer · 2017-08-01 16:37:14Z

1

You have error in syntax, who say @DarthJDG Add $.ajax to your request

Example that will serve you for your problem

jQuery(document).ready(function($){
  $.getJSON('https://ipinfo.io', function(data){
      var city_data = data.city;
      var region_data = data.region;
      $.ajax({
        data: {city: city_data, region: region_data},
        type: "post",
        url: "test3.php",
        success: function(data){
             alert("Data Saved: " + data);
        }
      })
  })

});

answered Aug 1, 2017 at 16:37

Guillermo Andres Fuentes Moral

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2 Comments

Piyush Seofaceup Over a year ago

It does solved my error but the data is not getting stored in the database

Guillermo Andres Fuentes Moral Over a year ago

You remove print_r($_POST); return; ? and out print $sql execute in mysql client or command line mysql and see which output delivers the command line

Dan · Accepted Answer · 2017-08-01 17:39:08Z

1

You mixed up the data object and the success callback. According to the documentation of jQuery.get(), the first parameter is the url, the second the data-object, the third the callback if the request succeeded and the last one the data type (like json, html or xml).

$(document).ready(function() {
    $.get("https://ipinfo.io", function(response) {
        console.log(response);

        var city_data = response.city;
        var region_data = response.region;

        $.post("test3.php",
            {
                // data
                city: city_data,
                region: region_data
            },
            function (data) {
                // success
                console.log("data successfully saved.");
            }
        )
    }, "json")
        .fail(function(jqXHR, textStatus, errorThrown) {
            console.log(textStatus, errorThrown)
        });
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

edited Aug 1, 2017 at 17:39

answered Aug 1, 2017 at 16:38

Dan

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5 Comments

Piyush Seofaceup Over a year ago

Hello, I am not getting any error but the data is also not getting stored in the database

Dan Over a year ago

Take a look into the developer tools console. Do you get an error?

Dan Over a year ago

Add the error controller or die(mysqli_error()) after the mysqli_query() function to see if something's wrong with the query.

Dan Over a year ago

Is the script inside the if(isset($_REQUEST)) reachable? Otherwise try to replace it with if ($_SERVER["REQUEST_METHOD"] == "POST").

Piyush Seofaceup Over a year ago

I got it what i am looking for . Thank you for the help

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