Data registration with Angularjs and PHP does not work

Question

Data registration with Angularjs and PHP does not work

Asked 8 years, 10 months ago

Viewed 1,211 times

-1

Hello,

I have a form, simple, with 3 fields, name, email and password in HTML and along with an Angularjs code. However, when submitting the form, I get the message that the data sent to the bank, but they are not there in the same.

<!DOCTYPE html>
<html ng-app="myApp">
<head>
<title>CRUD AngularJS</title>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/angularjs/1.4.3/angular.min.js></script>
</head>
<body ng-controller="MainCtrl as ctrl">
<form ng-submit="ctrl.submit()">
Nome<br>
<input type="text" ng-model="ctrl.user.nome"><br>
Email<br>
<input type="text" ng-model="ctrl.user.email"><br>
Senha<br>
<input type="password" ng-model="ctrl.user.senha"><br>
<input type="submit" id="submit" value="Submit" />
</form>

<script type="text/javascript">
angular.module('myApp', [])
    .controller('MainCtrl', ['$scope', '$http', function($scope, $http){
        $scope.list = [];
        var self = this;
        self.submit = function() {
          console.log('User clicked submit with ', self.user);
            $http.post('php/salvar.php', {'nome': $scope.nome, 'email': $scope.email, 'senha': $scope.senha})
            .then(function(response) {
                console.log(response.status);
                console.log(response.data.msg);

              }, function(response) {
                console.log(response.status);
                console.log(response.msg);

              });
        }
}]);
</script>
</body>
</html>

PHP:

?php
$user = 'root';
$password = 'root';
$db = 'angularDB';
$host = 'localhost';

$con = mysqli_connect("localhost", $user, $password, $db);

// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$nome = $_POST['nome']; 
$email = $_POST['email'];
$senha = $_POST['senha'];

$ins = mysqli_query($con, "INSERT INTO users VALUES (NULL, '$nome', '$email', '$senha')");

if($ins == 1){
echo json_encode( array('status' => 200, 'msg' => 'Cadastro efetuado com sucesso!'));
}else{
echo json_encode( array('status' => 0, 'msg' => 'Erro ao cadastrar no banco.'));
}
?>

What it returns in the request answer?

– henriquedpereira

2015/08/27 at 19:31
For security reasons, use Prepared statements in queries, especially those involving strings, helping to deliver SQL Injection. What is the value of the variable $ins after the query? Does the PHP/Apache/Nginx error log have any error messages? Try enabling error display to get a better sense of what’s going on.

– Vinícius Gobbo A. de Oliveira

2015/08/27 at 19:31
Ta very simple and visible the first problem, working or not

– Otto

2015/08/27 at 19:32
Sorry @Otto, how could I treat this?

– GustavoSevero

2015/08/27 at 20:22
@Viníciusgobboa.deOliveira, I don’t know how to use Prepared statement with mysqli. Can help me?

– GustavoSevero

2015/08/27 at 20:23
@Gustavosevero the way this in the answer

– Otto

2015/08/27 at 20:24
Well @Otto, I corrected the treatment. I put it in the post edition, check it out. And what happens is that only the code is inserted in the bank, that is, only the code field is filled, the other fields are blank... Very strange.

– GustavoSevero

2015/08/27 at 20:31
From what I’ve noticed, the problem is not in php but in Angular.

– GustavoSevero

2015/08/28 at 16:51
Note that in your angular include code is breaking.

– Ivan Ferrer

2015/08/28 at 17:36
Ready @Ivanferrer, include fixed. But here on my computer, there is no such break and yet, nothing.

– GustavoSevero

2015/08/29 at 02:44
Still broken... rrsrsrs, missing quotes

– Ivan Ferrer

2015/09/01 at 21:11

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3 answers

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by Cleber Azevedo • 31 points · Answer 1 · 2015-08-27T19:33:28+00:00

2

Check that no errors occurred in PHP at insertion time:

example:

if (mysqli_error($ins))
  echo json_encode( array('status' => 1, 'msg' => 'Ocorreu um problema'));
else
  echo json_encode( array('status' => 1, 'msg' => 'Cadastro efetuado com sucesso!'));

I use this code and run php only in the url, @Cleberazevedo?

– GustavoSevero

2015/08/27 at 20:24
Well, I just ran php in the url, as I asked you and the message that appeared on the screen was this, @Cleberazevedo. "{"status":1,"msg":"Successful registration!"}"

– GustavoSevero

2015/08/27 at 20:25
@Gustavosevero, this method $http.post({...}) the correct would not be $http({...}) ? you can use an equivalent: jQuery.post({...})or $.post({...}). Just don’t forget to insert the jquery library.

– Ivan Ferrer

2015/08/27 at 21:18
So the problem is not in inserting with php, make sure you are passing the ng parameters correctly?

– Cleber Azevedo

2015/08/27 at 23:21
Did you use mysqli_error($ins) ? It will return you if there was an error in the PHP query

– Cleber Azevedo

2015/08/27 at 23:30
@Ivanferrer, I’m sure it’s not because in the Angular documentation it shows how I put it. Check the link: [https://docs.angularjs.org/api/ng/service/$http]

– GustavoSevero

2015/08/28 at 00:51
Ah, okay, I wasn’t sure about that, but the exit wouldn’t have to be something like that: parse_str(file_get_contents("php://input"),$putData); is that I’m more used to defining the post as object value:$http({
 headers: {'Content-Type': 'application/json'},
 url: 'url.php', method: 'POST', data: $.param({param: param}) });

– Ivan Ferrer

2015/08/28 at 12:21
Here is an explanation of the problem of using $http.post(), in English, but I think I can understand: http://stackoverflow.com/questions/19254029/angularjs-http-post-does-not-send-data

– Ivan Ferrer

2015/08/28 at 12:23
@Ivanferrer, do you have any more hints, hint?

– GustavoSevero

2015/08/28 at 16:53
I’ll try to give you an answer, but I won’t test it, you tell me if it worked blz.

– Ivan Ferrer

2015/08/28 at 16:57

Show 5 more comments

by iLeonardo Carvalho • **561** points · Answer 2 · 2015-12-06T23:38:19+00:00

As you use ng-Ubmit on a specific location you can repurpose and summarize the code even in ng-model, I will pass on my way to create <form> with ng-submit with angular based on what you need:

HTML

<!DOCTYPE html>
<html ng-app="myApp">
<head>
    <title>CRUD AngularJS</title>
    <script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/angularjs/1.4.3/angular.min.js"></script>
</head>
<body ng-controller="MainCtrl">
    <form ng-submit="submit()">
        Nome<br>
            <input type="text" ng-model="nome"><br>
        Email<br>
            <input type="text" ng-model="email"><br>
        Senha<br>
            <input type="password" ng-model="senha"><br>
        <input type="submit" id="submit" value="Submit" />
    </form>
</body>
</html>

JS

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

Myapp.controller('MainCtrl', ['$scope', '$http', function($scope, $http) {

    $scope.submit = function () {

        var formData = { 'nome' : $scope.nome, 'email' : $scope.email, 'senha' : $scope.senha };
        var postData = 'myData='+JSON.stringify(formData);

        var request = $http({
            method: "POST",
            url: './php/salvar.php',
            data: postData,
            headers: { 'Content-Type': 'application/x-www-form-urlencoded; charset=UTF-8' }
        });

        request.success(function (data, status, headers, config) {
            $scope.success = true;
            console.log(status + ' - ' + data); //Captura de Dados
            if ( data.trim() === '[Protocolo] = #1') {
              alert("[INFO]: Cadastro efetuado com sucesso! ");
            }
            if ( data.trim() === '[Protocolo] = #2') {
              alert("[INFO]: Erro ao cadastrar no banco. ");
            }
        });
        request.error(function (data, status, headers, config) {
            $scope.error = true;
            console.log(error);
        });
    }
}]);

PHP

<?php

$user = 'root';
$password = 'root';
$db = 'angularDB';
$host = 'localhost';

$con = mysqli_connect("localhost", $user, $password, $db);

// Check connection
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

header('Content-Type:text/html;charset=UTF-8');

    include("../class/conexao.class.php");

    $myData = json_decode($_POST['myData'], true);

$nome = $myData['nome'];
$email = $myData['email'];
$senha = $myData['senha'];

$ins = mysqli_query($con, "INSERT INTO users VALUES (NULL, '$nome', '$email', '$senha')");

if($ins == 1) {
    echo ("[Protocolo] = #1");
} else {
    echo ("[Protocolo] = #2");
}

?>

by Ivan Ferrer • **12,096** points · Answer 3 · 2015-08-28T17:34:55+00:00

First we have HTML with angular:

  <!DOCTYPE html>
    <html ng-app="myApp">
    <head>
    <title>CRUD AngularJS</title>

    </head>
    <body ng-controller="MainCtrl as ctrl">
    <form ng-submit="ctrl.submit()">
    Nome<br>
    <input type="text" ng-model="ctrl.user.nome"><br>
    Email<br>
    <input type="text" ng-model="ctrl.user.email"><br>
    Senha<br>
    <input type="password" ng-model="ctrl.user.senha"><br>
    <input type="submit" id="submit" value="Submit" />
    </form>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/angularjs/1.4.3/angular.min.js"></script>
<script type="text/javascript">
angular.module('myApp', [])
    .controller('MainCtrl', ['$scope', '$http', function($scope, $http){

        $scope.status = null;
        $scope.msg = null;
        $scope.user = {};

        $scope.$watch('user', function (newValue, oldValue) {
             if (newValue !== oldValue) {
                 $scope.post(newValue);
             }
        });

        $scope.post = function (user) {
            try {

                $http({
                    url: "php/salvar.php",
                    method: 'POST',
                    data: {'user': user}

                }).success(function (response) {
                    if (angular.isDefined(response.status)) {
                         console.log(response.status);
                        $scope.status = response.status;
                         console.log(response.msg);
                        $scope.msg = response.msg;
                    }

                });

            } catch (e) {
                console.log(e);
            }

        };

}]);
</script>
</body>
</html>

And in PHP, I don’t remember whether it’s array or object, I treated for both cases, but you can give one var_dump($_POST); die(); to see the way out, but I think it’s something like that:

<?php

$user = 'root';
$password = 'root';
$db = 'angularDB';
$host = 'localhost';

 $con = mysqli_connect("localhost", $user, $password, $db);

// Check connection
if (mysqli_connect_errno()) {
   echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

 $user = $_POST['user'];

if (is_array( $user)) {
 $nome = $user['nome']; 
 $email = $user['email'];
 $senha = $user['senha'];
} else {
 $nome = $user->nome; 
 $email = $user->email;
 $senha = $user->senha;
}

 $ins = mysqli_query($con, "INSERT INTO users VALUES (NULL, '$nome', '$email', '$senha')");

if ($ins) {
    json_encode(array('status' => 1, 'msg' => 'Cadastro efetuado com sucesso!'));
} else {
   echo json_encode( array('status' => 0, 'msg' => 'Erro ao cadastrar no banco.'));
}
?>