Member Avatar for aianne

Sorry I'm new to these stuff but I am having a hard time about this. So I'll trying to create a php and mysql program which can view records from database server. In my form I have 2 dropdown list and 2 submit button, the first one the the list of "section" and submit button, and the second one is a list of "year level" and submit button again. If I select the section "A" and click submit button, it will display only the student's information that belongs only on section "A" and same as on "year level". I can't figure it out whenever i click the submit button, i can't view the records. I hope it made sense. Please do you have any idea to make it work? here is what I have so far--->

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html>
<head>
<title>View</title>
</head>
<body>
<form action="view.php" method="POST">
<td> </td><label>By Section: </label>
<td> </td><select name="Section" value="<?php echo $Section; ?>">


<option value="a">A</option>


<option value="b">B</option>


<option value="c">C</option>


<option value="d">D</option>

</select>


<td> </td><input type="submit" value="GO" name="sec"/>




<td> </td><label>By Year: </label>
<select name="Year" value="<?php echo $Year; ?>">


<option value="1">First Year</option>


<option value="2">Second Year</option>


<option value="3">Third Year</option>


<option value="4">Fourth Year</option>

</select>


<td> </td><input type="submit" value="GO" name="year"/>




<?php

include('quiz1_connect.php');
mysql_connect("$server", "$user", "$pass")or die("cannot connect"); 
mysql_select_db("$db")or die("cannot select DB");

if (isset($_POST['sec']))
{
if (is_numeric($_POST['Section']))
{
$Section = $_POST['Section'];

$result = mysql_query("SELECT * FROM tblquiz1 WHERE SecName='$Section'") 
or die(mysql_error()); 



echo "<table border='1' cellpadding='10'>";
echo "<tr> <th>ID:</th> <th>First Name:</th> <th>Last Name:</th> <th>Year:</th> <th>Section:</th> </tr>";


while($row = mysql_fetch_array( $result )) {


echo "<tr>";
echo '<td>' . $row['IDNo'] . '</td>';
echo '<td>' . $row['FName'] . '</td>';
echo '<td>' . $row['LName'] . '</td>';
echo '<td>' . $row['YearLvl'] . '</td>';
echo '<td>' . $row['SecName'] . '</td>';
echo "</tr>"; 
} 


echo "</table>";

}
}
?>
</form>
</body>
</html>
  • 3 Contributors
  • 3 Replies
  • 871 Views
  • 20 Hours Discussion Span
  • Latest Post Latest Post by aianne
Member Avatar for GliderPilot

You need to seperate the two options into two forms than have if statements to display the results depending on what was selected:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html>
<head>
<title>View</title>
</head>
<body>
<form action="view.php" method="POST">
<td> </td><label>By Section: </label>
<td> </td><select name="Section" value="<?php echo $Section; ?>">


<option value="a">A</option>


<option value="b">B</option>


<option value="c">C</option>


<option value="d">D</option>

</select>


<td> </td><input type="submit" value="GO" name="sec"/>

</form>

<form action="view.php" method="POST">

<td> </td><label>By Year: </label>
<select name="Year" value="<?php echo $Year; ?>">


<option value="1">First Year</option>


<option value="2">Second Year</option>


<option value="3">Third Year</option>


<option value="4">Fourth Year</option>

</select>


<td> </td><input type="submit" value="GO" name="year"/>

</form>




<?php

  include('quiz1_connect.php');
  mysql_connect("$server", "$user", "$pass")or die("cannot connect"); 
  mysql_select_db("$db")or die("cannot select DB");


if (isset($_POST['sec']))
{

  $Section = $_POST['Section'];

  $query = "SELECT * FROM tblquiz1 WHERE SecName='$Section'";
}

if (isset($_POST['year']))
{

  $Year = $_POST['Year'];

  $query = "SELECT * FROM tblquiz1 WHERE YearLvl='$Year'";

}

if (isset($query))
{

  $result = mysql_query($query) or die(mysql_error()); 

  
  echo "<table border='1' cellpadding='10'>";
  echo "<tr> <th>ID:</th> <th>First Name:</th> <th>Last Name:</th> <th>Year:</th> <th>Section:</th> </tr>";


  while($row = mysql_fetch_array( $result )) {


    echo "<tr>";
    echo '<td>' . $row['IDNo'] . '</td>';
    echo '<td>' . $row['FName'] . '</td>';
    echo '<td>' . $row['LName'] . '</td>';
    echo '<td>' . $row['YearLvl'] . '</td>';
    echo '<td>' . $row['SecName'] . '</td>';
    echo "</tr>"; 
  } 


  echo "</table>";


}


?>

</body>
</html>
Edited by GliderPilot because: n/a
Member Avatar for ko ko

Only submit button should be identified for one form. No matter how much submit buttons are you using within the same form, they've only submission of that form.

Try what @Glider Pilot mentioned above.

Member Avatar for aianne

Oh. It works! Thank you so much guys! especially Sir GliderPilot, thank you. really. :))

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