无法从JavaScript到php获取参数

Question

Unable to get parameters passed from javascript to loginme.php

This is simple form in index.php

<form method="POST">

            <input type="text" id="userid" name="userid"></input>
            <input type="password" id="pass" name="pass"></input>
            <input type="button" value="Log in" onclick="letUserLogin()"/>
</form>

Javascript function : myscript.js

function letUserLogin() {
    var userid = document.getElementById("userid").value;
    var pass = document.getElementById("pass").value;

  if (window.XMLHttpRequest) {

    xmlhttp=new XMLHttpRequest();
  } else { 
    xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
  xmlhttp.onreadystatechange=function() {
    if (xmlhttp.readyState==4 && xmlhttp.status==200) {

      alert(xmlhttp.responseText); //only shows 'and'
    }
  }
  xmlhttp.open("POST","loginme.php?userid="+userid+"&pass="+pass,true);
  xmlhttp.send();
} 

Simple echo statement in loginme.php loginme.php

<?php

// username and password sent from form 
$username=$_POST['userid']; 
$password=$_POST['pass']; 

echo"$username and $password";

?>

Answer 1

You are passing GET parameters:

xmlhttp.open("POST","loginme.php?userid="+userid+"&pass="+pass,true);
                                ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

... thus you need to fetch them from $_GET , not $_POST :

$username=$_GET['userid']; 
$password=$_GET['pass']; 

You possibly want to use the send() method instead to send your data. Right now, your payload is empty:

xmlhttp.send();

Answer 2

You can resolve this with JQuery quite easily if you want: This method also allows you to put the URL within the action parameter of the form and uses POST which is more secured for transferring password information:

JQUERY:

$(document).on('submit', "form", function(e){ //We add a listener
   e.preventDefault();
   $.post($(this).attr('action'), $(this).serialize())
    .done( function( data ) {
     //Do something with response          
 });
});

Note that you can of course change the listener to only listen to a specific form. In this case all forms submits will be caught rather than a specific one in a page.

HTML:

<form action="/path/to/loginme.php">
    <input type="text" name="userid">
    <input type="password" name="pass">
</form>

PHP:

$username=$_POST['userid']; 
$password=$_POST['pass']; 
echo "$username and $password";

Answer 3

You are using POST but explicitly setting the values into the query string, GET style. So basically you are sending a blank POST .

You need to send the values like this:

xmlhttp.open("POST","loginme.php", true);
xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
xmlhttp.send("userid=" + userid + "&pass=" + pass);

Answer 4

try this:

var userid =encodeURIComponent(document.getElementById("userid").value)
var pass =encodeURIComponent(document.getElementById("pass").value)
var parameters="userid="+userid+"&pass="+pass
mypostrequest.open("POST", "loginme.php", true)
mypostrequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded")
mypostrequest.send(parameters)

use of encodeURIComponent() to encode any special characters within the parameter values. Call setRequestHeader() and set its content type to "application/x-www-form-urlencoded" . This is needed for any POST request made via Ajax.