Loading data from database with php and pure ajax in a textbox

Question

i have one dropdown list which has data for \"subjects\" loaded from database. when i clicked on one subject what it should do is load related \"subject_id\" value inside te

Answer 1

Your subject_id is not displaying because you have not printed your book_id after fetching results from database in getbook.php

After this $result=mysql_query($sql);

Write echo $result['your_book_id_field_name'];

Answer 2

The mistake was done at ajax part document.getElementById("bookid").innerHTML and must be replaced with document.getElementById().value since I had to put data to Html Element that cotain value i.e Textbox(as textbox contain value attribute).

InnerHTML is used to manipulate the html elements that does not contain value, ** div,h1, ** etc. for details see below link.

http://www.verious.com/qa/what-39-s-the-difference-between-document-get-element-by-id-quot-test-quot-value-and-document-get-element-by-id-quot-tes/

ajax code

  function show_bookid(str)
{
    var xmlhttp;
    if(str.length==0)
    {
        document.getElementById("bookid").value="";
        return;
    }
    if(window.XMLHttpRequest)
    {
        xmlhttp= new XMLHttpRequest();
    }
    else
    {
        xmlhttp=new ActiveXOjbject("Microsoft.XMLHttpRequest"); 
    }
    xmlhttp.onreadystatechange=function()
    {
        if (xmlhttp.readyState==4 && xmlhttp.status==200)
        {
            document.getElementById("bookid").value=xmlhttp.responseText;
        }
    }
        xmlhttp.open("GET","getbook.php?q="+str,true);
        xmlhttp.send();

}

getbook.php

<?php
$b=$_GET['q'];
include('includes/security.php');
 include('includes/dbconnect.php'); 
$database=new MySQLDatabase();

$sql="select * from tbl_bkcat where book_id='".$b."'";
$result=mysql_query($sql);
$row=mysql_fetch_array($result);
echo $row['Book_id'];
?>

addbook.php

            <form name="bookadd" action="" class="jNice" method="post">
            <fieldset>

                <p><label>Subject</label>
                 <select name="subject" onChange="show_bookid(this.value);">

                 <?php

                    while($sel_rows=mysql_fetch_array($subresult))
                    {
                 ?>
                    <option value="<?php echo $sel_rows['book_id'];?>">
                    <?php echo $sel_rows['subject']?>
                    </option> 
                    <?php 
                        }
                    ?>
                </select>
                </p>

                <p>
                <label >Book ID</label>                 
               <input type="text" id="bookid" name="book"/>
                </p>