My previous problem got solved but know i want to use html select value as input to another mysql query. My code so far -
<?php
$con=mysqli_connect("server","user","pwd","db");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"select eid from tbl1 order by eid;");
?>
<select id="eid" size="1">
<option selected="true" disabled="disabled">Choose eid</option>
<?
while($row = mysqli_fetch_array($result))
{
?>
<option value="<? echo $row_list['eid']; ?>"<? if($row['eid']==$select){ echo "selected"; } ?>>
<?echo $row['eid'];?>
</option>
<?
}
?>
</select>
<select name="tagging">
<option selected="true" disabled="disabled">Choose tagging</option>
<option value="option1">option1</option>
<option value="option2">option2</option>
<option value="option3">option3</option>
</select>
<?
$result = mysqli_query($con,"select distinct val from tbl2 where val!='' order by val;");
?>
<select id='val'>
<option selected="true" disabled="disabled">Choose val</option>
<?
while($row = mysqli_fetch_array($result))
{
?>
<option value="<? echo $row_list['val']; ?>"<? if($row['val']==$select){ echo "selected"; } ?>>
<?echo $row['val'];?>
</option>
<?
}
?>
</select>
<?
mysqli_close($con);
?>
Now the output window have three html select, i want to pass this three value in mysql select query to show some result. But don't know how to do it. All kind of advice are welcome.