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  • #1

    Trying to download table -edit- and upload

    I am fairly new here and learning. I'm making a menu for a sub shop. I want to allow the client to edit prices. 1)I know how to pull data date from my table and display it in a form (allowing the price fields to be "editable"). 2)I know how to create a form and submit the entered values to my table in the database (as variables).

    My problem is that I do not know how to connect steps 1 & 2. Is there a way to re-assign the form cells (that I am displaying), as variables so I can then re-upload them to the table?

    Am I going about this all wrong? Is there a more common way to build a simple interface for the client, or open-source code? I hope I explained myself and I would sincerely appreciate any help or direction.
    Code:
    <html> 
<head> 
        <title>View mymenu</title> 
</head> 
<body> 

<h2>View mymenu</h2>
<?php

//Displays all data 'mymenu' table 

//Connect to database
        require ('dbstuff.php');
        $db = connectDB();
 
        $result = mysql_query("SELECT * FROM mymenu")  
                or die(mysql_error());
				                  
// Display data from mysql table "mymenu"
?>		
  <form action="" form name="editform" method="post"> 	
<?	              
        echo "<table border='1' cellpadding='10'>"; 
        echo "<tr> <th>ID</th><th>type</th> <th>half</th><th>full</th></tr>"; 
// loop 
        $count= 0;
		while($row = mysql_fetch_array( $result )) { 
        ++$count;	  
// echo out the contents of each row into a table 
	             echo "<tr>"; 
				 echo '<td name="ID">' . $row['ID'] . '</td>';  
				 echo '<td>' . $row['type'] . '</td>' ;
				 echo '<td><input type="text" value= ' . $row['half'] . '></td>';
				 echo '<td><input type="text" value= ' . $row['full'] . '></td>';
 				 echo "</tr>";   
        }  
 
        // close table> 
        echo "</table>"; 
		?>
 <input type="submit" name="submit" value="Submit">
 </form>
</body> 
</html>
    Last edited by Buffmin; Sep 12, 2011, 02:15 PM.
    Tags: None
  • #2
    After these lines:
    Code:
    require ('dbstuff.php');
$db = connectDB();
    ADD this:
    Code:
    if(isset($_POST['submit']))
{

$half = $_POST["half"];
$full = $_POST["full"];
$type = $_POST["type"];
$result = mysql_query("UPDATE mymenu SET half = $half, full = $full WHERE type = $type");
}
    Evolution - The non-random survival of random variants.
    Physics is actually atoms trying to understand themselves.

    Comment

    • #3
      Umm...Sunfighter, that's not going to come close to working.

      Buffmin: For starters, you don't have any *NAMES* in your form fields in your <form>!!!

      So when you submit the <form> *NOTHING AT ALL* will be sent back to PHP for it to process.

      And Sunfighter's solution doesn't work, even if you give them names, because his answer will only handle *ONE SINGLE ITEM*. Well, it would, except he based the UPDATE on $type, whereas $type isn't a <form> field in your page. It should be based on the ID.

      So... With the understanding that I am *NOT* a PHP programmer, so well could be a typo or more here...
      Code:
      <html> 
<head> 
        <title>View mymenu</title> 
</head> 
<body> 

<h2>View mymenu</h2>
<?php

//Displays all data 'mymenu' table 

//Connect to database
require ('dbstuff.php');
$db = connectDB();

// if the form is posted back to this page, then process it:
if ( isset($_POST["submit"]) )
{
    $itemcount = $_POST["itemcount"];
    for ( $i = 1; $i <= $itemcount; ++$i )
    {
        $id = $_POST["id" . $i];
        $half = $_POST["half" . $i];
        $full = $_POST["full" . $i];
        $sql = "UPDATE mymenu SET half = $half, full = $full WHERE ID = $id";
        $result = mysql_query( $sql ) or die(mysql_error());
    }
}
// in any case, display items for update:
$sql = "SELECT id, half, full, type FROM mymenu ORDER BY id";
$result = mysql_query( $sql ) or die(mysql_error());
				                  
// Display data from mysql table "mymenu"
?>		
<form action="" name="editform" method="post"> 	
<table border='1' cellpadding='10'>
<tr><th>ID</th><th>type</th> <th>half</th><th>full</th></tr>
<?php
// loop 
$count= 0;
while($row = mysql_fetch_array( $result )) 
{ 
    ++$count;	  
    $id = $row["id"];
    echo "<tr>\n"; 
    echo '<td><input type="hidden" name="id'. $count . '" value=". $id . "/>' . $id . "</td>\n";  
    echo '<td>' . $row['type'] . "</td>\n";
    echo '<td><input name="$half' . $count . '" value="' . $row['half'] . '"/>' . "</td>\n";
    echo '<td><input name="$full' . $count . '" value="' . $row['full'] . '"/>' . "</td>\n";
    echo "</tr>\n";   
}  
echo "</table>\n";
echo "<input type="hidden" name="itemcount" value="' . $count . '"/>\n";
?>
<input type="submit" name="submit" value="Submit">
</form>
</body> 
</html>
      Be yourself. No one else is as qualified.

      Comment

      • #4
        Thank you guys . Works perfect. I added the names, and I added the "hidden" variable. Thank you very much.

        Comment

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