Search Criteria in PHP - php

Question

JAVASCRIPT
function ShowContent(id) {
if(id.length < 1) { return; }
document.getElementById(id).style.display = "block";
}
PHP
$test = mysql_query('SELECT userid,testid,testname FROM test_table WHERE userid = 9 ORDER BY name');
TestName : <select>
while($row=mysql_fetch_array($test))
{
$testname = $row['testname'];
$testid = $row['testid'];
echo '<option value='.$testid.'>'.$testname.'</option>
}
</select>
echo '<div id="testid" style="display:none;">
echo '<table>
<tr><th>USERID</th</tr>
<tr><th>TESTNAME</th></tr>
<tr><th>MARKS</th></tr>
publishtest = mysql_query('SELECT id,userid,publishtest,marks FROM publish_table WHERE userid= 9 GROUP BY id');
while($row=mysql_fetch_array($publishtest))
{
$userid = $row['userid'];
$testid = $row['id']; /*This is test id*/
$testname = $row['publishtest']; /* This is test name*/
$marks = $row['marks'];
echo '<tr><td>'.$userid.'</td>
<td>'.testid.'</td>
<td>'.testname.'</td>
<td>'.$marks.'</td>
</tr>
}
echo '</table>';
echo '</div>';
I need when I select test from select tag, then div should be displayed of that testid. I want either in javascript, jquery

Answer 1

You can try this:
JavaScript
$('#showDivSelect').change(function (e) {
var divId = $(this).children('option:selected').val();
$('#' + divId).css('display', 'block');
});​
HTML
<select id="showDivSelect">
<option id="o1">Div1</option>
<option id="o2">Div2</option>
</select>
<div id="Div1"></div>
<div id="Div2"></div>​
CSS
#Div1 {
width: 30px;
height: 30px;
background-color: red;
display: none;
}
#Div2 {
width: 30px;
height: 30px;
background-color: blue;
display: none;
}​
Here is working example: http://jsfiddle.net/ePSDu/