如何测试以查看是否通过Java正确发送了POST请求？

Question

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 ;)

Answer 1

使用firebug的Firefox。

Answer 2

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:

xr.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');

...after your open call. Some examples here and here .