Newbie: Mysql Example Edit Table....

**nasos** · 06:11 AM

Hi webmasters,

I would like to start PHP with mysql with a simple example by my own , so i made the following that tries to edit and remove a row from a table.

the table is named: shop
the database: test2
fields: article,dealer,price

So, I want to have the row displayed and beside every row to have a checkbox (not checked).
There will be two additional buttons below the rows that will display EDIT and DELETE (row).

I started with this:

<html>
<head>
<title>::: EDIT DATABASE :::</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body bgcolor="#999999">

PHP Code:


			
<?php



 $connection = mysql_connect("localhost", "", ""); 



   mysql_select_db("test2"); 



   { 

      $result = mysql_query("SELECT article,dealer,price FROM shop"); 

      

      $num_rows = mysql_num_rows($result);

      $itemchoice = 0;       



      if (($row  = mysql_fetch_array($result)) != 0) 

      { 

         echo "<table border=0>"; 



         echo "<tr bgcolor=#cccccc> 

            <td align=center>Article</td> 

            <td align=center>Dealer</td> 

            <td align=center>Price</td> 

            </tr>"; 



         do 

         { 

            $name = $row['article']; 

            $owner = $row['dealer']; 

            $species = $row['price']; 

            $itemchoice = $itemchoice + 1;    

             

            echo "<form method='post' action='<? $PHP_SELF ?>'> 

               <tr> 

               <td align='center'> 

               <input disabled type='text' name='article' value='" . $name . "' size='30' maxlength='30'></td> 

               <td align='center'> 

               <input disabled type='text' name='dealer' value='" . $owner . "'></td> 

               <td align='center'> 

               <input disabled type='text' name='price' value='" . $species . "' size='10' maxlength='10'></td> 

               <td align='center'> 

               <input style='background-color:#CCFF33' type=checkbox name='item' id='$itemchoice'>

               </tr></font>"; 

           } 

              

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



         echo "<form method='post' action='<? $_POST[$PHP_SELF] ?>'>"; 

         echo "<tr><td><input type='submit' name='delete' value='Delete'>"; 

         echo "<input type='submit' name='edit' value='Edit'></td></tr></table></form>";          

      } 



      else 

      {    

      echo "No records found!"; 

      }    

   } 

      

  if (isset($_REQUEST["delete"]);

$sql_del = mysql_query("DELETE FROM shop WHERE &itemchoice = ".$HTTP_POST_VARS['$item']); 

      }

if (!mysql_query ($sql_del)) { 

    $success = 0; 

    $msg = 'Record could not be deleted'; 

    } 

else { 

       $msg = 'Record for item '.$HTTP_POST_VARS['item'].' was deleted. Refreshing display...'; 

     } 



?>

</body>
</html>

This of course doesnt work ... so any help will be appreciated.!

**RonnieTheDodger** · 05-06-2004, 07:28 AM

You might want to try getting rid of the second <Form> tag. I have not ran your code yet, but that stuck out like a sore thumb.

Remove the echo statement following the while statement.

PHP Code:


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



         echo "<form method='post' action='<? $_POST[$PHP_SELF] ?>'>";

**nasos** · 05-06-2004, 09:33 AM

Thanks for replying,

well i followed your suggestion ...
but i d like to add this.

I know that I dont know much, but should I use an array for check buttons and use a variable for each when checked? It will be validated only when the DELETE button is pressed.

i removed the part:

PHP Code:


			
 if (isset($_POST["delete"]);

//update the database (corresonding to your select-query)

     $sql_del = mysql_query("DELETE FROM shop WHERE &itemchoice = " .$HTTP_POST_VARS['$item']);

      

      }

if (!mysql_query ($sql_del)) { 

    $success = 0; 

    $msg = 'Record could not be deleted'; 

    } 

else { 

       $msg = 'Record for item '.$HTTP_POST_VARS['item'].' was deleted. Refreshing display...'; 

     }

and added the array in javascript:

PHP Code:


			
 echo "<form method='post' action='<? $PHP_POST ?>'> 

               <tr> 

               <td align='center'> 

               <input disabled type='text' name='article' value='" . $name . "' size='30' maxlength='30'></td> 

               <td align='center'> 

               <input disabled type='text' name='dealer' value='" . $owner . "'></td> 

               <td align='center'> 

               <input disabled type='text' name='price' value='" . $species . "' size='10' maxlength='10'></td> 

               <td align='center'> 

               <input style='background-color:#CCFF33' type=checkbox value='$message'>"

               ?>

               <SCRIPT language=JavaScript><!--

               var foo=1;function CheckAll(){

                var i=0;for(i=0;i<document.list.elements.length;i++){

                     if(document.list.elements[i].name=='message'){

                             document.list.elements[i].checked=foo  }

                }; 

                 if(foo == 1){

                    foo=0}

                       else {

                          foo=1}

                  }

                //--></SCRIPT>

               <?php

               echo "</tr></font>"; 

           }

and it is working ... no the buttons arent working !

**RonnieTheDodger** · 05-08-2004, 09:27 PM

I am not very good at javascript. But I do know that you need to call the function CheckAll() at some point to run it.

You might want to go to some of the scripting sites and look for a routine that you can adapt for your purposes....something that is already written and known to work. If you are just starting to learn php, there is no better way than learning by example.

Also, I do not know why you were mixing javascript into your php like that. Try to stick with the php.