Want to display Inserted row inside a table in php form

saravanataee · January 5, 2013

Dear all,

I am having the following php form, which has a table in that, i ill fill up the details and when i click Place order it will store the details into database. My query is that instead of place order it to add item, and for each time i click add item, the store data should be shown below the page where i am working inside a table serially. I mean basically displaying the rows in a grid..

Once it reaches the maximum screen size i should be able to scroll down to see all the items i have added. After all finished adding items i should then want to Place order which finally takes all my items into a single table value..

Any suggestion or idea how to start would be greatful.

Thanks.

<?php
include("conndb.php");
function createoptions($table , $id , $field)
{
   $sql = "select * from $table ORDER BY $field";
   $res = mysql_query($sql) or die(mysql_error());
   while ($a = mysql_fetch_assoc($res))
   echo "<option value=\"{$a[$id]}\">$a[$field]</option>";
}

// mysql_connect('localhost', 'root', '');
// mysql_select_db('inventorydb');
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Place an Order</title>
<script language="javascript src="code.js">
</script>
<link href="loginmodule.css" rel="stylesheet" type="text/css" />


<style>



 #entrydt1 { margin-left: 100px; }
 #orderid1 { margin-left: 107px; }
 #orderdt1 { margin-left: 94px; }
 #itemid { margin-left: 142px; }
 #qtyid { margin-left: 122px; }
 #packid { margin-left: 120px; }
 #qtynoid { margin-left: 80px; }
 #Poid { margin-left: 80px; }
 #resetid { margin-left: 80px; }
 #clearid { margin-left: 80px; }
 #Name { margin-left: 50px;  width:100px;}


</style>

<script type="text/javascript" src="jquery.js"></script>
	    <script type="text/javascript" charset="utf-8">
	    $(function(){
		  $("select#category").change(function(){
		    $.getJSON("select.php",{category: $(this).val(), ajax: 'true'}, function(j){
			  var options = '';
			  for (var i = 0; i < j.length; i++) {
			    options += '<option value="' + j[i].optionValue + '">' + j[i].optionDisplay + '</option>';
			  }
			  $("select#subcategory").html(options);
		    })
		  })


		  $("select#subcategory").change(function(){
		    $.getJSON("select.php",{subcategory: $(this).val(), ajax: 'true'}, function(j){
			  var options = '';
			  for (var i = 0; i < j.length; i++) {
			    // options += '<option value="' + j[i].optionValue + '">' + j[i].optionDisplay + '</option>';
                   options = '<input type="text" name="subcat2" value="' + j[i].optionDisplay + '" />';
			  }
			  $("div#subcategory2").html(options);
		    })
		  })

	    })
	    </script>
           <script>function calculate() {
document.getElementById("t8").value =
parseFloat(document.getElementById("subcat2").value) *
(parseFloat(document.getElementById("MRP").value)*(parseFloat(document.getElementById("t7").value)/100));
}
</script>
</head>
<body>
<h1>Place an Order</h1>
<form method="post">




    <table border='1' >

		    <tr>
				    <td><label for="t1">Item</label></td>
				    <td><label for="t2">Quantity</label></td>
                       <td><label for="t2">Packing</label></td>
				    <td><label for="t2">Quantity in NO's</label></td>
				    <td><label for="t2">MRP/RATE</label></td>
				    <td><label for="t2">Discount</label></td>
				    <td><label for="t2">Amount</label></td>





		    </tr>
		    <tr>
				    <td>    <select name="cat" id="category">
    <option value="-1">--Select--</option>
       <?php
    createoptions("category", "cat_id", "category");
    ?>
    </select>
       </td>

                       <td><input name="t3" id="t3" size="10" value=""></td>
                       <td>
                        <select name="subcat"  id="subcategory">
    </select>
                       </td>
                       <td>
                       <!-- <select name="subcat2" id="subcategory2">
    </select> <input type="text" name="subcat2" value="" id="subcategory2" /> -->
       <div id="subcategory2"> </div>

                       </td>
                       <td><select name="MRP" id="MRP">
                       <option value=""><--Select--></option>
                           <?php
                           $sql = "SELECT Mrprate FROM mrp";
                           $result = mysql_query($sql);
                           while ($row = mysql_fetch_array($result)) {
                           echo "<option value='" . $row['Mrprate'] . "' >" . $row['Mrprate'] . "</option>";
                           }
                           ?></td>
                       <td><input name="t7" id="t7" size="10" placeholder="40%" value=""></td>
                       <td><input name="t8" id="t8" size="10" value="" onchange="calculate()"></td>


                       </tr>
                       <tr>
				    <td><input type="submit" name="submit" value="Place Order"></td>
				    <td><input type="submit" name="submit1" value="Reset"></td>

                       </tr>
                       </table>
                       </form>
                       </body>
                       </html>

<?php
if (isset($_POST['submit']))
  {		  
    include 'db.php';

								    $item=$_POST['cat'] ;
								    $qty= $_POST['t3'] ;
								    $pack= $_POST['subcat'] ;
								    $qtyno= $_POST['subcat2'] ;
								    $mrp= $_POST['MRP'] ;
								    $dis= $_POST['t7'] ;
								    $amt= $_POST['t8'] ;








		   $query=  mysql_query("INSERT INTO `po`(Item,Quantity,Packing, QuantityNO, MRP, Discount, Amount) VALUES ('$item','$qty','$pack','$qtyno','$mrp','$dis','$amt')");
			 if(!$query)
                {
die('Query Failed with exception'.mysql_error());
}
else{
echo"Thanks for the input";
}


		    }



?>

50r · January 5, 2013

There are two problem actually why no one has replyed to this post. 1 your question is quite not clear, 2 the code that you post is not well indented which makes it hard to read.

Try by explaining what the problem is. do you want to retrieve data from a database? after you insert it or what?

saravanataee · January 6, 2013

Hi 50r,

Yes, presently i have a table with one row in which i fill the five fields and when i click add item i store them into database. What i look for is, whenever i insert an item i want to display it in the same page below my present table. No mater how many items i insert, all should be displayed and i should be able to scroll and see them..!

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Want to display Inserted row inside a table in php form

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