multiple dynamic drop down submission on button click to mysql table - php

Question

hey guys i am working on a small project for my college. it consists of a table which is dynamic(linked to mysql table) and one cell in every record is a dynamic drop down(all are linked to mysql table). and the user has to choose from a list of drop down values for each drop down generated and it can also be left blank. here is the php code for that:
<form class="appnitro" method="post" action="">
<div class="form_description">
<center><h2>NOMINATE ENTRY</h2></center>
<p><center><font size='3'>
<?php
$con=mysqli_connect("localhost","user",pass","db");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$dept = $_POST['department'];
$class = $_POST['class'];
$result = mysqli_query($con,"SELECT * FROM prizemaster");
$result1 = mysqli_query($con,"SELECT * FROM studentmaster WHERE dept='$dept' and class='$class'");
echo "<table border='1'>
<tr>
<th>Prize ID &nbsp &nbsp &nbsp &nbsp</th>
<th>Prize Name &nbsp &nbsp &nbsp &nbsp </th>
<th>Name &nbsp &nbsp &nbsp &nbsp &nbsp &nbsp </th>
</tr>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['prizeid'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td><select name='name'>";
echo "<option></option>";
while($drop = mysqli_fetch_array($result1))
{
echo "<option value='".$drop['name']."'>" . $drop['name'] . "</option>";
}
mysqli_data_seek($result1, 0);
echo "</select></td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?></center></font></div>
<p>
<center><button type="submit" formaction="stnomins.php">Nominate</button></center>
</form>
the above is the form and when the nominate button is clicked i want this code to be executed for each and every drop down value selected:
<?php
$con=mysqli_connect("localhost","user","pass","db");
$myname = $_POST['name'];
$sql2="SELECT * FROM studentmaster WHERE name='$myname'";
$result = mysqli_query($con,$sql2)or die(mysqli_error());
$row = mysqli_fetch_array($result);
$mydept=$row['dept'];
$myclass=$row['class'];
$myregno=$row['regno'];
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql1="INSERT INTO studenttransaction (`transid`, `date`, `prizeid`, `regno`, `name`, `class`, `department`, `status`) VALUES ('' , CURRENT_TIMESTAMP() , '', '$myregno', '$myname', '$myclass', '$mydept', '1')";
if (!mysqli_query($con,$sql1))
{
die('Error: ' . mysqli_error($con));
}
mysqli_close($con);
?>
kindly help me out i need to submit it in 2 days. please explain with code because i am new to php and not fully known to it

Answer 1

I have updated both of your html & php code, please have a look :
In form file:
echo "<td><select name='name[]'>"; // added [] here to make it array
echo "<option value=''>Select student</option>"; // added value & option item
These above lines of code have been modified accordingly in the file below
<form class="appnitro" method="post" action="">
<div class="form_description">
<center><h2>NOMINATE ENTRY</h2></center>
<p><center><font size='3'>
<?php
$con=mysqli_connect("localhost","user",pass","db");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$dept = $_POST['department'];
$class = $_POST['class'];
$result = mysqli_query($con,"SELECT * FROM prizemaster");
$result1 = mysqli_query($con,"SELECT * FROM studentmaster WHERE dept='$dept' and class='$class'");
echo "<table border='1'>
<tr>
<th>Prize ID &nbsp &nbsp &nbsp &nbsp</th>
<th>Prize Name &nbsp &nbsp &nbsp &nbsp </th>
<th>Name &nbsp &nbsp &nbsp &nbsp &nbsp &nbsp </th>
</tr>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['prizeid'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td><select name='name[]'>";
echo "<option value=''>Select student</option>";
while($drop = mysqli_fetch_array($result1))
{
echo "<option value='".$drop['name']."'>" . $drop['name'] . "</option>";
}
mysqli_data_seek($result1, 0);
echo "</select></td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?></center></font></div>
<p>
<center><button type="submit" name="nominate" formaction="stnomins.php">Nominate</button></center>
</form>
Slightly updated your PHP code. Started a loop for each drop-down of names if button is clicked/submitted.
<?php
$con=mysqli_connect("localhost","user","pass","db");
if (mysqli_connect_errno()){
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if(isset($_POST['nominate'])){
foreach($_POST['name'] as $key => $myname){
//$myname = $_POST['name'];
$sql2="SELECT * FROM studentmaster WHERE name='$myname'";
$result = mysqli_query($con,$sql2)or die(mysqli_error());
if($row = mysqli_fetch_array($result)){
$mydept=$row['dept'];
$myclass=$row['class'];
$myregno=$row['regno'];
$sql1="INSERT INTO studenttransaction (`transid`, `date`, `prizeid`, `regno`, `name`, `class`, `department`, `status`) VALUES ('' , CURRENT_TIMESTAMP() , '', '$myregno', '$myname', '$myclass', '$mydept', '1')";
if (!mysqli_query($con,$sql1)){
die('Error: ' . mysqli_error($con));
}
}//if mysqli_fetch_array condition closed
}// for loop closed
}// if submit button(nominate) closed
mysqli_close($con);
?>
Hope this will work for you.

Answer 2

its working fine. I have another doubt is that how to echo the value of drop down when a drop down value is selected.Anybody who knows the answer of it, kindly provide the solution of it.