Inserting Data in mySql Database using AngularJs

Question

I am trying to insert some data in mysql database server using AngularJs with PHP/MySQL.But it's not working.Nothing is showing.Why it's not working ? My codes -

index.html

<!DOCTYPE html>
<html>

<head>

<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js">    </script>

</head>

<body ng-app="myApp" ng-controller="empcontroller">

 <form>

Employee Name:-<input type="text" ng-model="uname" />
Password:-<input type="Password" ng-model="pass" />
<input type="button" value="Submit" ng-click="postData()" />

</form>

<script src="app.js"></script>

 </body>

</html>

app.js

var app = angular.module('myApp', []);

app.controller('empcontroller', function ($scope, $http) {
$scope.postData = function (post) {
var data = {
uname:$scope.empcontroller.uname,
pass:$scope.empcontroller.pass
}
$http.post("http://localhost/query.php", { 'uname':     $scope.empcontroller.uname, 'pass': $scope.empcontroller.pass })
.success(function (data, status, headers, config) {
    console.log($scope.empcontroller.uname);
});
};   
});

query.php

<?php 
header('Access-Control-Allow-Origin: *');
header("Content-Type: application/json; charset=UTF-8");

$mysql_host = "localhost";
$mysql_database = "FirstDB";
$mysql_user = "root";
$mysql_password = "";
// Create connection
$conn = new mysqli($mysql_host, $mysql_user,    $mysql_password,$mysql_database);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} 
$data = json_decode(file_get_contents("php://input"));
$uname = mysql_real_escape_string($data->uname);
$pass = mysql_real_escape_string($data->pass);

$sql = "INSERT INTO Employee (Username,Password) VALUES ('$uname','$pass')";
$result = $conn->query($sql);
echo($uname);
error_log($uname);
$conn->close();

?>

When I am trying to run the php file (query.php) in my localhost server - it gives me the following error -

Fatal error: Uncaught Error: Call to undefined function >mysql_real_escape_string() in I:\xampp\htdocs\query.php:15 Stack trace:

0 {main}

thrown in I:\xampp\htdocs\query.php on line 15

the error is obvious, google that !!!!

Rathma
– Rathma

2016-11-19 15:31:14 +00:00
Commented Nov 19, 2016 at 15:31 — Rathma
– Rathma, Commented Nov 19, 2016 at 15:31

Tirthraj Barot · Accepted Answer · 2016-11-19 16:42:45Z

1

It should be mysqli_real_escape_string

`i` is missing.

UPDATE

According to me, your approach can be modified a bit. The error you are getting about the uname not found, that could be corrected but according to me, when you are using AngularJS, you can go more json based.

The form could be like,

<form ng-init="obj={}">

Employee Name:-<input type="text" ng-model="obj.uname" />
Password:-<input type="Password" ng-model="obj.pass" />
<input type="button" value="Submit" ng-click="postData(obj)" />

</form>

Hence now you have your object to be passed from view to controller as a function parameter.

Now in the controller,

app.controller('empcontroller', function ($scope, $http) {
    $scope.postData = function (obj) {
//NOTE THAT WE DON'T HAVE TO MAKE A NEW OBJECT TO PASS IN POST IF THIS APPROACH IS FOLLOWED.
 //HENCE, OUR OBJECT IS THE SAME WHICH WE PASSED FROM VIEW.
        $http.post("http://localhost/query.php",obj).then(function (response) {
            console.log(response.data);
        },function(error){
            console.error("Error : ",error);
        });
    };   
});

So you will get your $data in php script exactly as you want.

Moreover, you can make a factory or service for such http calls.

This is the approach I suggest, following the best practices.

edited Nov 19, 2016 at 16:42

answered Nov 19, 2016 at 15:30

Tirthraj Barot

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10 Comments

Asif Al Noman Over a year ago

Still my code is not working . when i am clicking submit button , nothing is showing .

Tirthraj Barot Over a year ago

Turn on the error reporting. @AsifAlNoman and tell me the error

Asif Al Noman Over a year ago

Actually i am talking about index.html file . when i am clicking the submit button of index.html file nothing is showing.

Tirthraj Barot Over a year ago

Put an alert and check of the function os being called. See the network tab in developers console in browser.. And see the response obtained in that network tab. @AsifAlNoman

Tirthraj Barot Over a year ago

And frankly I had been facing this error in the past... I found an alternative function of mysqli_real_escape_string and used it on javascript side and avoided using that function in php.. Things are running as smooth as ever...! :D @AsifAlNoman

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