[英]PHP: Send form data to php script which then posts data to another form processor like zapier.com?
I use a 3rd party form processor which is zapier.com. 我使用的是zapier.com的第三方表单处理程序。 The issue is, I need a way to redirect the user after submitting data to the 3rd party form processor.
问题是,在将数据提交给第三方表单处理器之后,我需要一种重定向用户的方法。 Zapier.com accepts post, get, and put submissions.
Zapier.com接受发布,获取和放置提交。
I was thinking of: 我在想:
I might be over thinking this, but the only way I see of doing this is making a form that posts data that has been posted to it. 我可能会对此有所考虑,但是我看到的唯一方法是制作一个表单,该表单发布已发布到其上的数据。 Otherwise the form will just send my user to the 3rd party processor without redirecting them to whatever page I choose.
否则,该表单只会将我的用户发送到第三方处理器,而无需将他们重定向到我选择的任何页面。 The 3rd party form processor doesn't have a way of me using custom redirects.
第三方表单处理器无法使用自定义重定向。
Using javascript and Ajax it can be done like this (with jQuery): 使用javascript和Ajax可以做到这一点(使用jQuery):
$("#idOfTheForm").submit(function(e) {
e.preventDefault();
$.ajax({
method : "post",
url : this.action,
data : $(this).serialize(),
success : function() {
window.location = "yourUrlOfThanks.html";
},
error : function() {
alert("Something went bad");
}
});
});
So basically it is: sent a post request to the action url of this form, and once it throws an 200 code (found and everything went ok) then redirect to my page. 基本上就是这样:向此表单的操作网址发送一个发布请求,一旦它抛出200个代码(找到并一切正常),然后重定向到我的页面。 If soemthing went wrong, then the server will throw an 40* status code and the script will go into error function.
如果发生错误,则服务器将抛出40 *状态代码,并且脚本将进入错误功能。
You can use Guzzle to proxy the user request. 您可以使用Guzzle 代理用户请求。
$client = new Guzzle\Http\Client();
$request = $client->post('/3rd.party.php')
->addPostField('user_field_1', $_POST['user_field_1'])
->addPostField('user_field_2', $_POST['user_field_2']);
$response = $request->send();
if ($response->isSuccessful()) {
//show message
}
The downside is that you can't be 100% sure that the form submission was indeed successful. 缺点是您不能100%确定表单提交确实成功。 In order to achieve that you could scrap the
$request->getBody()
and check if a known success message is present. 为了实现此目的,您可以
$request->getBody()
并检查是否存在已知的成功消息。
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