联系表格使用php，大查询和angularjs发送到数据库和电子邮件

Question

我正在尝试使用AngularJS和PHP创建一个简单的联系表单，该表单将发送到电子邮件和AngularJS Google - big query database ，但没有成功。

而且似乎到处都没有它的例子...充满爱心寻求任何帮助...

我输入的代码显示了我要使用MySQL ，我需要将其调整为bigquery 。

   <body ng-app="myapp" ng-controller="formcontroller">
    <div class="contact-us-page">
        <div class="container">
            <div class="row">
                <div class="col-sm-4 col-xs-12">
                    <h1>CONTACT US</h1>
                    <form name="userForm" novalidate ng-submit="insertData()">
                       <label class="text-success" ng-show="successInsert">{{successInsert}}</label>
                        <div>
                            <label for="username" >Name</label>
                            <input type="text" name="userName" ng-model="insert.user_name" class="form-control" ng-class="{ 'has-error' : userForm.userName.$invalid && userForm.$submitted && userForm.userName.$pristine }" required>
                            <!--<span class="text-danger" ng-show="errorUsername">{{errorUsername}}</span>-->
                        </div> 
                        <div>
                            <label for="companyName">Company name</label>
                            <input type="text" name="companyName" ng-model="insert.company_name" class="form-control" ng-class="{ 'has-error' : userForm.companyName.$invalid && userForm.$submitted && userForm.companyName.$pristine }" required>
                            <!--<span class="text-danger" ng-show="errorCompanyname">{{errorCompanyname}}</span>-->
                        </div> 
                        <div>
                            <label for="email">Email</label>
                            <input type="email" name="email" ng-model="insert.email" class="form-control" ng-class="{ 'has-error' : userForm.email.$invalid && userForm.$submitted && userForm.email.$pristine }" required>
                            <!--<span class="text-danger" ng-show="errorEmail">{{errorEmail}}</span>-->
                        </div> 
                        <div>
                            <label for="phone">Phone</label>
                            <input type="text" name="phone" ng-model="insert.phone" class="form-control" ng-class="{ 'has-error' : userForm.phone.$invalid && userForm.$submitted && userForm.phone.$pristine }" required>
                            <!--<span class="text-danger" ng-show="errorPhone">{{errorPhone}}</span>-->
                        </div> 
                        <!--<div>
                            <input type="file" name="upload_cv" value="Upload CV" class="upload-cv">
                        </div> -->   
                        <input type="submit" name="insert" class="button-style" value="Send">
                    </form>   
                </div>  
            </div>  
        </div>  
    </div> 
    <script src="app.js"></script>    
</body>

app.js：

 var application = angular.module("myapp", []);
application.controller("formcontroller", function($scope, $http){
 $scope.insert = {};
 $scope.insertData = function(){
  $http({
   method:"POST",
   url:"insert.php",
   data:$scope.insert,
  }).success(function(data){
   if(data.error)
   {
    $scope.errorUsername = data.error.user_name;
    $scope.errorCompanyname = data.error.company_name;
    $scope.errorEmail = data.error.email;
    $scope.errorPhone = data.error.phone;
    $scope.successInsert = null;
   }
   else
   {
    $scope.insert = null;
    $scope.errorUsername = null;
    $scope.errorCompanyname = null;
    $scope.errorEmail = null;
    $scope.errorPhone = null;
    $scope.successInsert = data.message;
   }
  });
 }
});

insert.php：

 <?php
//insert.php
$connect = mysqli_connect("localhost", "root", "", "resonai");
$form_data = json_decode(file_get_contents("php://input"));
$data = array();
$error = array();

if(empty($form_data->user_name))
{
 $error["user_name"] = "User Name is required";
}

if(empty($form_data->company_name))
{
 $error["company_name"] = "Company Name is required";
}

if(empty($form_data->email))
{
 $error["email"] = "Email Name is required";
}

if(empty($form_data->phone))
{
 $error["phone"] = "Phone Name is required";
}

if(!empty($error))
{
 $data["error"] = $error;
}
else
{
 $user_name = mysqli_real_escape_string($connect, $form_data->user_name); 
 $company_name = mysqli_real_escape_string($connect, $form_data->company_name);
 $email = mysqli_real_escape_string($connect, $form_data->email);
 $phone = mysqli_real_escape_string($connect, $form_data->phone);
 $query = "
  INSERT INTO contactus_page(user_name, company_name, email, phone) VALUES ('$user_name', '$company_name', '$email', '$phone')
 ";
 if(mysqli_query($connect, $query))
 {
  $data["message"] = "Thank you. We will contact you soon!";
 }
}

echo json_encode($data);

?>

具有bigquery的insert.php-无法正常工作...

<?php
//insert.php
composer require google/cloud-bigquery;

require __DIR__ . '/vendor/autoload.php';

use Google\Cloud\BigQuery\BigQueryClient;

$projectId = 'burnished-data-183409';

$form_data = json_decode(file_get_contents("php://input"));
$data = array();
$error = array();

if(empty($form_data->user_name))
{
 $error["user_name"] = "User Name is required";
}

if(empty($form_data->company_name))
{
 $error["company_name"] = "Company Name is required";
}

if(empty($form_data->email))
{
 $error["email"] = "Email Name is required";
}

if(empty($form_data->phone))
{
 $error["phone"] = "Phone Name is required";
}

if(!empty($error))
{
 $data["error"] = $error;
}
else
{
# [START build_service]
$bigquery = new BigQueryClient([
'projectId' => $projectId
]);
# [END build_service]
# [START run_query]
$user_name = $form_data->user_name; 
$company_name = $form_data->company_name;
$email = $form_data->email;
$phone = $form_data->phone;

$query = 'insert into test1.cf (user_name, company_name, email, phone) values ('$user_name', '$company_name', '$email', '$phone')';
$options = ['useLegacySql' => true];
$queryResults = $bigQuery->runQuery($query, $options); 
# [END run_query]

# [START print_results]
if ($queryResults->isComplete()) {
    $data["message"] = "Thank you. We will contact you soon!";
    /*$i = 0;
    $rows = $queryResults->rows();
    foreach ($rows as $row) {
        printf('--- Row %s ---' . PHP_EOL, ++$i);
        foreach ($row as $column => $value) {
            printf('%s: %s' . PHP_EOL, $column, $value);
        }
    }
    printf('Found %s row(s)' . PHP_EOL, $i);*/
} else {
    throw new Exception('The query failed to complete');
}
# [END print_results]
# [END all]
}

echo json_encode($data);

?>

Answer 1

我想您正在尝试在BigQuery上执行MySQL当前正在执行的查询。 对于该用例，您应该使用BigQuery PHP客户端库。 有关其安装和使用的更多信息，请参见https://cloud.google.com/bigquery/create-simple-app-api [单击PHP选项卡]

联系表格使用php，大查询和angularjs发送到数据库和电子邮件

问题描述

1 个解决方案

解决方案1
0 2017-11-27 04:55:50

联系表格使用php，大查询和angularjs发送到数据库和电子邮件

问题描述

1 个解决方案

解决方案1 0 2017-11-27 04:55:50

解决方案1
0 2017-11-27 04:55:50