Display data retrieved from the drop down menu

Question

I need help. I want to display three data values ​​that the user will select from three drop-down menus. I created three drop-down lists in which the user will select 3 different parameters for each of the database, and then everything that the user will select should appear in the table. For example, if the user selects 111 for selection_id and 222 for fixture_id, then Jesty for the name should look like 111 222 Jesty in the table. At the moment, I only have drop-down lists that retrieve information from the database and then output it down. Wat. I need a way to display a custom selection or javascript that will print to the user what he selected.

Here is my drop down list 3

require "config.php"; //Database connection $resource_selections = mysql_query("SELECT DISTINCT selection_id FROM selections ORDER BY selection_id ASC"); $selections = array(); while($row = mysql_fetch_row($resource_selections)){ $selections[] = $row[0]; } $resource_fixtures = mysql_query("SELECT DISTINCT fixture_id FROM selections ORDER BY selection_id ASC"); $fixtures = array(); while($row = mysql_fetch_row($resource_fixtures)){ $fixtures[] = $row[0]; } $resource_names = mysql_query("SELECT DISTINCT name FROM selections ORDER BY selection_id ASC"); $names = array(); while($row = mysql_fetch_row($resource_names)){ $names[] = $row[0]; } if(count($selections) <= 0 || count($fixtures) <= 0 || count($names) <= 0){ echo 'No results have been found.'; } else { // Display form echo '<form name="form" method="post" action="selection.php">'; //SelectionID dropdown: echo "<select name='selection_id' id='selections' >"; foreach($selections as $selection) echo "<option id='$selection'>$selection</option>"; echo '</select>'; //FixtureID dropdown echo "<select name='fixture_id' id='fixtures' >"; foreach($fixtures as $fixture) echo "<option id='$fixture'>$fixture</option>"; echo '</select>'; //Name dropdown echo "<select name='name' id='names' >"; foreach($names as $name) echo "<option id='$name'>$name</option>"; echo '</select>'; echo "<input type='submit' name='submit' value='Submit' />"; echo '</form>'; } ?> 

Answer 1

You can try like this

  <script type="text/javascript"> function mergeval(val,flag) { if(flag==1) document.getElementById('mvalue1').innerHTML = val ; else if(flag==2) document.getElementById('mvalue2').innerHTML = val; else if(flag==3) document.getElementById('mvalue3').innerHTML = val; } </script> <?php session_start(); $_SESSION['values']=!empty($_SESSION['values']) ? $_SESSION['values']:array(); if(!empty($_REQUEST['selection_id'])&&!empty($_REQUEST['fixture_id'])&&!empty($_REQUEST['name'])) { $_SESSION['values'][] = array('id'=>$_REQUEST['selection_id'],'fid'=>$_REQUEST['fixture_id'],'name'=>$_REQUEST['name']); } $dispval ="<table>"; foreach(@$_SESSION['values'] as $key=>$val) { $dispval .="<tr><td>".$val['id']."&nbsp;".$val['fid']."&nbsp;".$val['name']."</td></tr>"; } $dispval .="</table>"; echo $dispval; ?> <?php require "config.php"; //Database connection $resource_selections = mysql_query("SELECT DISTINCT selection_id FROM selections ORDER BY selection_id ASC"); $selections = array(); while($row = mysql_fetch_row($resource_selections)){ $selections[] = $row[0]; } $resource_fixtures = mysql_query("SELECT DISTINCT fixture_id FROM selections ORDER BY selection_id ASC"); $fixtures = array(); while($row = mysql_fetch_row($resource_fixtures)){ $fixtures[] = $row[0]; } $resource_names = mysql_query("SELECT DISTINCT name FROM selections ORDER BY selection_id ASC"); $names = array(); while($row = mysql_fetch_row($resource_names)){ $names[] = $row[0]; } if(count($selections) <= 0 || count($fixtures) <= 0 || count($names) <= 0){ echo 'No results have been found.'; } else { // Display form echo '<form name="form" method="post" action="selection.php">'; //SelectionID dropdown: echo "<select name='selection_id' id='selections' onchange='javascript:mergeval(this.value,1)'>"; foreach($selections as $selection) echo "<option id='$selection'>$selection</option>"; echo '</select>'; //FixtureID dropdown echo "<select name='fixture_id' id='fixtures' onchange='javascript:mergeval(this.value,2)' >"; foreach($fixtures as $fixture) echo "<option id='$fixture'>$fixture</option>"; echo '</select>'; //Name dropdown echo "<select name='name' id='names' onchange='javascript:mergeval(this.value,3)'>"; foreach($names as $name) echo "<option id='$name'>$name</option>"; echo '</select>'; echo "<input type='submit' name='submit' value='Submit' />"; echo '</form>'; } ?> <table> <tr><td >&nbsp;<span id="mvalue1"></span>&nbsp;<span id="mvalue2"></span>&nbsp;<span id="mvalue3"></span></td></tr> </table> 

Answer 2

you can achieve this by using jquery, im functions to insert the code for you. Here is the script:

  <script language="JavaScript" type="text/javascript"> $(document).ready(function(){ $(".testval1").live('change',function(){ var x = $(this).val(); if(x!='') { $(".displaydata").append("<td>"+x+"</td>"); } }); }); </script> 

here is the html

  <body> <form> <select name='testval1' class='testval1'> <option value=''>select data</option> <option value="110">110</option> <option value="111">111</option> </select> <select name='testval2' class='testval1'> <option value=''>select data</option> <option value="2223">2223</option> <option value="222">222</option> </select> <select name='testval3' class='testval1'> <option value=''>select data</option> <option value="Jestysd">Jestysd</option> <option value="Jesty">Jesty</option> </select> <table class="displaydata"> </table> </form> 

you need to add jquery-1.9.1.min.js for this .. all the best