php - Fetch db data depending on selected drop down value -

is possible fetch data db upon changing value of drop down, want pull data corresponds value of drop down , show text area, got codes i've tried create don't know how make work want.

i want page show drop down after selecting value show data in text area without refreshing page, here code:

<div id="maincontent"> <table width="619" border="0" align="center"> <td align="center"><form id="form1" name="form1" method="post" action="" > <fieldset> <legend><strong>ea</strong></legend> <p>  <select name="ea_name" id="ea_name"> <option value="" selected="selected">please select...</option> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> </select> </p> </fieldset> </form></td> </table>  <div id="results"></div> </div> 

i thinking "onchange" dont know how implement it, want echo resulting data text area within fieldset.

here code i've tried pull data , show text area:

<?php require 'include/db_open.php';  $ea_name = $_post['ea_name'];  $sql="select * ea_error ea_name = '" . $ea_name . "'";  $mydata = mysql_query($sql);  //to count if there results $numrow = mysql_num_rows($mydata);  if($numrow == 0) {     echo "no results found."; } else { echo "<fieldset><legend><strong>information</strong></legend><p> <table width="auto" height="172" border="0"> <tr><th scope="row">error</th></tr> <tr><th scope="row">resolution</th></tr> <tr><th scope="row">contact/s</th></tr>;"  while($info = mysql_fetch_array($mydata))  { echo "<form action='retrieve.php' method='post'>";  echo"<tr>";  echo  "<td align='center'>" . $info['error'] . "<input type=hidden name=error value=" . $info['error'] . " </td>"; echo  "<td align='center'>" . $info['resolution'] . "<input type=hidden name=resolution value=" . $info['resolution'] . " size='11' maxlength='11' /> </td>";  echo  "<td align='center'>" . $info['contacts'] . "<input type=hidden name=contacts value=" . $info['contacts'] . "' /> </td>";  echo "</tr>";  echo "</form>"; } } echo "</fieldset>";   include 'include/db_close.php'; ?> 

updated code:

<?php require 'include/db_open.php';  $ea_name = $_post['ea_name'];  $sql="select * ea_error ea_name = '" . $ea_name . "'";  $mydata = mysql_query($sql);  //to count if there results $numrow = mysql_num_rows($mydata);  if($numrow == 0) {     echo "no results found."; } else { echo '<fieldset><legend><strong>information</strong></legend><p> <table width="auto" height="172" border="0"> <tr><th>error</th></tr> <tr><th>resolution</th></tr> <tr><th>contact/s</th></tr>';  while($info = mysql_fetch_array($mydata))  { echo "<form action='retrieve.php' method='post'>";  echo"<tr>";  echo  "<td align='center'>" . $info['error'] . "<input type=hidden name=error value=" . $info['error'] . " </td>"; echo  "<td align='center'>" . $info['resolution'] . "<input type=hidden name=resolution value=" . $info['resolution'] . " size='11' maxlength='11' /> </td>";  echo  "<td align='center'>" . $info['contacts'] . "<input type=hidden name=contacts value=" . $info['contacts'] . "' /> </td>";  echo "</tr>";  echo "</form>"; } } echo "</fieldset>";   include 'include/db_close.php'; ?> 

this should trick. put javascript either in script tags in head of page, after jquery include, or in js file , include after jquery...

$(function() {  //  document.ready     $("#ea_name").on("change", function() {         $.ajax({             url: "phpfile.php",             type: "post",             data: {                 ea_name: $(this).val()             },             success: function(data) {                 $("#results").html(data);             }         });     }); }); 

it assigns event handler change event of drop down. when triggered sends ajax request php file (don't forget put correct filename url!) returns html. pushed results div.

note: fixed typo may or may not in php file. @ end of line create query, you'd missed closing ". in case copy , paste real file.