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.
There are few things you need to include in your web script pages.
<form object.name="value" defined.action="url" defined, to submit to a different web page.name=value pairs and based on your condition, should pass to database engine to store in a table.Why dont you wrap those 3 <select></select> into a <form> tag and add action and method attributes such as <form action="file_to_handle_select_data.php" method="POST"> and than in file_to_handle_select_data.php run the queries from $_POST array.