Question

在这种形式中，我试图将用户重定向到给定的URL。在那里我只能使用get方法传递参数。但是当我使用get方法时，我的表单数据不会保存在数据库中。我想在数据库中同时保存数据。

这是我的代码：

<?php

include_once 'CryptoUtils.php';
include_once 'dbconnect.php';

if(isset($_POST['btn-signup']))
{
    $mobile = mysql_real_escape_string($_POST['Mnumber']);
    $email = mysql_real_escape_string($_POST['email']);
    $fname = mysql_real_escape_string($_POST['fname']);
    $address = mysql_real_escape_string($_POST['address']);
    $sitename = mysql_real_escape_string($_POST['sitename']);


    $q = ("INSERT INTO template_users(Mnumber,email,fname,address,sitename) VALUES('$mobile','$email','$fname','$address','$sitename')");
    mysql_query ($q) or die("Problem with the query: $q<br>" . mysql_error());
}

?>
<!DOCTYPE>
<html>
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8"/>
    <title>Ipayy </title>
    <link rel="stylesheet" href="style.css" type="text/css"/>

</head>
<body>
<center>
    <div id="login-form">
        <form action="http://api.ipayy.com/v001/c/oc/dopayment" method="get" >

            <table align="left" width="40%" border="0">
                <h1 align="left">Create Your Web Site</h1></br></br>




                <input type="hidden" name="gh" value="<?php echo $encrypted_string; ?>" />

                <tr>
                    <td><input type="text" name="Mnumber" value=""  placeholder="Your Mobile Number" required /></td>
                </tr>

                <tr>
                    <td><input type="email" name="email"  value=""  placeholder="Your Email" required /></td>
                </tr>

                <tr>
                    <td><input type="text" name="fname"   value=""  placeholder="Your First Name" required /></td>
                </tr>

                <tr>
                    <td><input type="text" name="address" value=""  placeholder="Your Address" required /></td>
                </tr>

                <tr>
                    <td><input type="text" name="sitename" value=""  placeholder="Your Site name" required /></td>
                </tr>

                <tr>
                    <td><button type="submit" name="btn-signup" onclick="myFunction()">Create My gomobi website</button></td>
                </tr>

            </table>
        </form>

    </div>
</center>
</body>
</html>

这是我的数据库连接：

<?php
if(!mysql_connect("localhost","root",""))
{
    die('oops connection problem ! --> '.mysql_error());
}
if(!mysql_select_db("ipay"))
{
    die('oops database selection problem ! --> '.mysql_error());
}
?>

Answer 1

当您向外部链接发送请求并且还希望在您自己的服务器中处理相同的请求时，您必须使用AJAX发送两个请求，或者您必须使用curl将一个请求发送到外部链接。希望以下内容可能有所帮助。

<button type="button" name="btn-signup" onclick="myFunction()">Create My gomobi website</button>

检查上面的按钮代码，我将其更改为键入按钮，因为我更喜欢使用myFunction（）和AJAX发布表单数据。现在，下面是myFunction（）的可能代码。 [注意：你必须已经在myFunction（）中有一些代码，我不知道，如果需要你只需保留它们并添加以下代码]如果ajax表单发布你不需要在表单中使用方法或操作标签

function myFunction()
{
  var gh = $('input[name=gh]').val();
  var Mnumber = $('input[name=Mnumber]').val();
  var email = $('input[name=email]').val();
  var fname= $('input[name=fname]').val();
  var address= $('input[name=address]').val();
  var sitename= $('input[name=sitename]').val();
  $.get("http://api.ipayy.com/v001/c/oc/dopayment",
    { 

        gh : gh,
        Mnumber : Mnumber,
        email : email,
        fname : fname,
        address : address,
        sitename : sitename
    },
    function(data,status){
        //console.log(data);
      if(status == 'success')
       {
         //another POST or GET ajax method can be initiated with same data              
          //but differant url (the url of your php code which is inserting    data to database)Here is an example below
        $.post("storedatatodb.php",
        {
           gh : gh,
        Mnumber : Mnumber,
        email : email,
        fname : fname,
        address : address,
        sitename : sitename

        },
        function(data,status)
       {
           if(status == 'success')
             {
               // do something - redirect to some page
             }
       });
       }
    else
    {
       //show error message
       // or do nothing
    }
    });
}

在上面的例子中，storedatatodb.php将如下所示

<?php
  include_once 'dbconnect.php';
$mobile = mysql_real_escape_string($_POST['Mnumber']);
$email = mysql_real_escape_string($_POST['email']);
$fname = mysql_real_escape_string($_POST['fname']);
$address = mysql_real_escape_string($_POST['address']);
$sitename = mysql_real_escape_string($_POST['sitename']);


$q = ("INSERT INTO template_users(Mnumber,email,fname,address,sitename) VALUES('$mobile','$email','$fname','$address','$sitename')");
mysql_query ($q) or die("Problem with the query: $q<br>" . mysql_error());
?>

另一种方法是使用PHP curl。

在这种情况下，您可以在表单标记和操作中使用任何GET或POST方法，使用处理数据的php文件的url插入db。假设文件存储在atatodb.php中。代码如下：

<?php
  include_once 'dbconnect.php';
$gh = $_POST['gh'];
$mobile = mysql_real_escape_string($_POST['Mnumber']);
$email = mysql_real_escape_string($_POST['email']);
$fname = mysql_real_escape_string($_POST['fname']);
$address = mysql_real_escape_string($_POST['address']);
$sitename = mysql_real_escape_string($_POST['sitename']);

$q = ("INSERT INTO template_users(Mnumber,email,fname,address,sitename) VALUES('$mobile','$email','$fname','$address','$sitename')");
mysql_query ($q) or die("Problem with the query: $q<br>" . mysql_error());

  //curl to send data to external URL 
  // In case of GET request
  $geturl = "http://api.ipayy.com/v001/c/oc/dopayment?gh=".$gh."&mobile=".$mobile."&email=".$email."&fname=".$fname."&address=".$address."&sitename=".$sitename;
 // Get cURL resource
  $curl = curl_init();
 // Set some options
  curl_setopt_array($curl, array(
  CURLOPT_RETURNTRANSFER => 1,
  CURLOPT_URL => $geturl
));
// Send the request & save response to $resp
 $resp = curl_exec($curl);
//Print error if any
if(curl_errno($curl))
{
echo 'error:' . curl_error($curl);
}
 // Close request to clear up some resources
 curl_close($curl);
?>

Answer 2

如果您要将数据提交到同一页面然后改变

<form action="http://api.ipayy.com/v001/c/oc/dopayment" method="get" >

到

<form action="">

也从

改变

<?php

include_once 'CryptoUtils.php';
include_once 'dbconnect.php';

if(isset($_POST['btn-signup']))
{
    $mobile = mysql_real_escape_string($_POST['Mnumber']);
    $email = mysql_real_escape_string($_POST['email']);
    $fname = mysql_real_escape_string($_POST['fname']);
    $address = mysql_real_escape_string($_POST['address']);
    $sitename = mysql_real_escape_string($_POST['sitename']);


    $q = ("INSERT INTO template_users(Mnumber,email,fname,address,sitename) VALUES('$mobile','$email','$fname','$address','$sitename')");
    mysql_query ($q) or die("Problem with the query: $q<br>" . mysql_error());
}

?>

到

<?php

include_once 'CryptoUtils.php';
include_once 'dbconnect.php';

if(isset($_POST['btn-signup']))
{
    $mobile = mysql_real_escape_string($_GET['Mnumber']);
    $email = mysql_real_escape_string($_GET['email']);
    $fname = mysql_real_escape_string($_GET['fname']);
    $address = mysql_real_escape_string($_GET['address']);
    $sitename = mysql_real_escape_string($_GET['sitename']);


    $q = ("INSERT INTO template_users(Mnumber,email,fname,address,sitename) VALUES('$mobile','$email','$fname','$address','$sitename')");
    mysql_query ($q) or die("Problem with the query: $q<br>" . mysql_error());
}

?>

如何将数据传递给数据库？

2 个答案: