如何测试我是否通过 Javascript 正确发送 POST 请求？
这是特殊情况：我使用一个书签来调用一个 .js，该 .js 向我的服务器上的 PHP 文件发送 POST 请求。这是 .js 文件中的 POST 请求：
var snd = ("qu=" + encodeURIComponent(t) + "&dl=" + encodeURIComponent(dl) + "&dt=" + encodeURIComponent(dt));
xr = new XMLHttpRequest();
xr.open("POST", "http://quotebook.us/s/process2.php",true);
xr.onreadystatechange=function() {
if (xr.readyState==4) {
var xmldoc = xr.responseText;
window.alert(xr.responseText);
}
}
xr.send(snd);
下面是我在 PHP 中所做的事情。但尽我所能，我无法弄清楚如何将某些内容返回到 .js 文件，以便它可以在警报中显示它（因此，我可以确认它首先发送了数据）。
<?php
if ($_SERVER['REQUEST_METHOD'] != 'POST') {
echo "This page is not for viewing";
exit;
}
$qo = $_POST["qu"];
$dl = $_POST["dl"];
$dt = $_POST["dt"];
echo "First parm: $qo, second param: $dl, third param: $dt";
?>
最终，我想获取这些变量并将它们写入 MySQL 数据库，但我至少需要一天的时间才能学习如何做到这一点...
对此过程的任何帮助将非常受欢迎，我已经有很多了一次查找有关处理不是由用户表单发送的 POST 请求的任何内容。显然，编写将数据发送到 MySQL 的书签是一门黑术；）
Here's the particular situation: I'm using a bookmarklet to call a .js that sends a POST request to a PHP file on my server. Here's the POST request in the .js file:
var snd = ("qu=" + encodeURIComponent(t) + "&dl=" + encodeURIComponent(dl) + "&dt=" + encodeURIComponent(dt));
xr = new XMLHttpRequest();
xr.open("POST", "http://quotebook.us/s/process2.php",true);
xr.onreadystatechange=function() {
if (xr.readyState==4) {
var xmldoc = xr.responseText;
window.alert(xr.responseText);
}
}
xr.send(snd);
And below is what I'm doing in PHP. But try as I might, I can't figure out how to get something BACK to the .js file so it can display it in an alert (and consequently, so I can confirm that it's sending the data in the first place).
<?php
if ($_SERVER['REQUEST_METHOD'] != 'POST') {
echo "This page is not for viewing";
exit;
}
$qo = $_POST["qu"];
$dl = $_POST["dl"];
$dt = $_POST["dt"];
echo "First parm: $qo, second param: $dl, third param: $dt";
?>
Ultimately I want to take these variables and write them to a MySQL database, but I'm at least a day away from learning how to do that...
Any help on this process would be very welcome, I've had a heck of a time finding anything about processing POST requests that AREN'T sent by a user form. Apparently writing bookmarklets that send data to MySQL is a black art ;)
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对于 Firefox，请使用 firebug。
Use firebug for firefox.
为了测试您是否正确执行，我可能会在 Firefox 上使用 Firebug 或在 Chrome 上使用开发工具；无论使用哪种方式，您都可以看到发送或接收的实际 HTTP 数据。但我认为你真正的问题是，为什么 POST 不起作用？ （您可以考虑更新问题标题。）
答案可能是您没有设置内容类型。 POST 是通用的，您可以发布任何内容。在您的情况下，您要发布 URL 编码的数据，因此请尝试
在
open调用之后添加： ...。 此处和此处。To test that you're doing it correctly, I'd probably use Firebug on Firefox or Dev Tools on Chrome; with either, you can see the actual HTTP data sent or received. But I think your real question is, why isn't the POST working? (You might consider updating your question title.)
And the answer may be that you're not setting the content type. POST is generic, you can post anything. In your case, you're posting URL-encoded data, so try adding:
...after your
opencall. Some examples here and here.