将数据从javascript发布到php

Question

I have two dropdowns in my web-page where I select manager and project.

If I choose any manager then I want to get the project which are assigned to particular manager.

I think I can't do this without the help of javascript and ajax.

So I have passed the selected value of manager to javascript file and again I have posted the value using ajax. But it seems the code is not working.

Here is my php code.

<form method="post" action="<?php $_PHP_SELF ?>">
    Select Manager <select id='managed' name="managed" onchange="getManager()">
                       <option value="">---select---</option>
<?php 
    $conn=mysqli_connect('localhost','root','root','projmanagement'); 
    $result=mysqli_query($conn,'SELECT manager_id,manager_name FROM manager'); 
    while($row=mysqli_fetch_assoc($result)) { 
        echo "<option value='$row[manager_id]'>$row[manager_name]</option>"; 
    } 
?>
                    </select>
            Select Project <select name="projectsd">
                               <option value="">---select---</option>
<?php          
    $temp = $_POST['managed'];
    var_dump($temp);
    die();
    $result1=mysqli_query($conn,'SELECT project_id,project_name FROM project inner join manager on project.m_id=$temp'); 
    while($row1=mysqli_fetch_assoc($result1)) { 
        echo "<option value='$row1[project_id]'>$row1[project_name]</option>"; 
    } 
?> 

This is my javascript code.

function getManager() {
    var myvar=document.getElementById('managed').value;

    var xmlhttp;
    if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari
        xmlhttp=new XMLHttpRequest();
    } else {// code for IE6, IE5
        xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
    }
    xmlhttp.onreadystatechange=function() {
        if (xmlhttp.readyState==4 && xmlhttp.status==200) {
            document.getElementById("managed").innerHTML = xhr.responseText;
        }
    }
    xmlhttp.open("POST","TaskMaster.php",true);
    xhr.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
    xhr.send("managed=" + myvar);
}

I'am a beginner to php and javascript.

Please give me an idea to solve my problem.

Answer 1

Try This :

<form method="post" action="<?php $_PHP_SELF ?>">
    Select Manager <select id='managed' name="managed" onchange="getManager()">

        <option value="">---select---</option>
<?php 
        $conn=mysqli_connect('localhost','root','root','projmanagement'); 
        $result=mysqli_query($conn,'SELECT manager_id,manager_name FROM manager'); 
        while($row=mysqli_fetch_assoc($result)) { 
            echo "<option value='$row[manager_id]'>$row[manager_name]</option>"; 
        } 
        ?> </select>
        Select Project <div id="project_container"><select name="projectsd"></div>

JS :

function getManager()
{
var myvar=document.getElementById('managed').value;

   var xmlhttp;
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
 xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
   xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
 if (xmlhttp.readyState==4 && xmlhttp.status==200)
  {

   document.getElementById("project_container").innerHTML = xhr.responseText;
 }
 }
 xmlhttp.open("POST","TaskMaster.php",true);
 xhr.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
 xhr.send("managed=" + myvar);
   }