Question

Edit database record with forms

Asked by: bingie

I am using the code in this question (http:Q_22129709.html) to send data to my database.

I can view each record in the database by specifing the id (i.e. http://members.lycos.co.uk/bookit/view.php?id=26 ).

So, can I make this an editing page? Where as the data is shown in the form (same as the page that would used to create it) but I would use an update statement instead of insert. The sql is the easy part - I am unsure how to retreive the data from the database and have it populate a form, that I can then update to the database.

Thanks

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Asked On
2007-01-21 at 16:56:22ID22130950
Tags

database

,

edit

,

record

Topic

PHP Scripting Language

Participating Experts
1
Points
500
Comments
16

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Answers

 

by: gamebitsPosted on 2007-01-21 at 17:03:25ID: 18361830

You put the result from the db insite the form as value

<input type=text name=firstname value=$row['firstname']>

 

by: bingiePosted on 2007-01-21 at 17:13:49ID: 18361858

Hey gamebits :)

The code for my view page is

<html>
<head>

<?php
function mysql_die($error = "unknown")
{
echo "<H1>Error:". $error."</H1>";
   exit;
}

$db = mysql_connect("localhost","bookit","") or mysql_die("Unable to connect");
mysql_select_db("bookit_uk_db",$db);

$requete = "SELECT * FROM tblProgramControl WHERE id=".$id;
$result = mysql_query ($requete,$db)or mysql_die("No record");
$Used =mysql_fetch_object($result);
  mysql_free_result($result);
 ?>

</head>

<body>
<H2 align=center>Database Test Page</H2><br><br>
    <?php echo $Used->id ?>
    <BR>
    <?php echo $Used->password ?>
    <BR>
    <?php echo $Used->errString ?>
</body>

</html>


So , I combined that with the other code for the from and get

<html>
<head>

<?php
function mysql_die($error = "unknown")
{
echo "<H1>Error:". $error."</H1>";
   exit;
}

$db = mysql_connect("localhost","bookit","") or mysql_die("Unable to connect");
mysql_select_db("bookit_uk_db",$db);

$requete = "SELECT * FROM tblProgramControl WHERE id=".$id;
$result = mysql_query ($requete,$db)or mysql_die("No record");
$Used =mysql_fetch_object($result);
  mysql_free_result($result);
 ?>

</head>

<body>
<h1>Book It Program Control Interface</h1>
<br>

<?

$db = mysql_connect("localhost","bookingapp","") or mysql_die("Unable to connect");
mysql_select_db("bookingapp",$db);
?>

<?php

if(isset($post)){

//get values from the form
$isValid = $_POST['isValid'];
$errString = $_POST['errString'];
$allowPassword = $_POST['allowPassword'];
$password = $_POST['password'];
$lockedOutText = $_POST['lockedOutText'];

$qry = mysql_query("insert into `tblProgramControl` (`isValid`, `errString`, `allowPassword`, `password`, `lockedOutText`) values ('$isValid', '$errString', '$allowPassword', '$password', '$lockedOutText')");
$result = mysql_query ($qry,$db);

               }
                 else
               {

 echo "<form method=\"post\" action=\"programControl.php\">
 Program is active?
 <input type=checkbox name=isValid value=$Used['isValid']><br>
 Message to display to program user:
 <input type=text name=errString value=$Used['errString']><br>
 Allow password to unlock program?
 <input type=checkbox name=allowPassword value=$Used['allowPassword']><br>
 Password:
 <input type=text name=password value=$Used['password']><br>
 Text to display when locked out:
 <input type=text name=lockedOutText value=$Used['lockedOutText']><br>
 <input type=submit name=post value=\"Submit\">
</form>";

         }

?>

<a href="http://members.lycos.co.uk/bookit/output.php">View Data</a>


</body>
</html>

which has a parse error

 

by: bingiePosted on 2007-01-21 at 17:14:47ID: 18361862

oops!

Any ideas? thx again..

 

by: gamebitsPosted on 2007-01-21 at 17:16:40ID: 18361871

Sorry was quite a short answer here is an example of what I mean

$resultID = mysql_query ("SELECT sec_id, section_id, street
                           FROM Sections
                           WHERE city = '$city'
                           LIMIT $offset, $rowsPerPage");

                              echo "<form action=\"index.php\" method=POST>";
                              echo "<input type=hidden name=open value=updsect>";
                              echo "<input type=hidden name=city value=\"$city\">";
                              echo "<input type=hidden name=display_edit_sect_form>";

                                $num_results = mysql_num_rows($resultID);

                              echo "<p align=left><font color=red><b>$numrows</font> Sections Total.</b></p>";

                             print "<table border=\"1\" cellpadding=2 cellspacing=0>
                <THEAD>
                     <tr>

                           <th width=\"5\" class=admin_grid_heading>&nbsp;</th>
                           <th width=\"75\" class=admin_grid_heading>Sec ID</th>
                           <th width=\"125\" class=admin_grid_heading>Street</th>
                         
                       echo "</THEAD>";
                       echo "</tr>";

             // Define your colors for the alternating rows

                $color1 = "#ECECEC";
                $color2 = "#ffffff";
                $datarow_count = 0;

                        while ($datarow=mysql_fetch_assoc($resultID)) {
                             $datarow_color = ($datarow_count % 2) ? $color1 : $color2;

                          echo '<TBODY><tr bgcolor='. $datarow_color .'><td><input type="checkbox" name="Record_id[]" value='. $datarow['sec_id'] .'></td>';
                          echo '<input type="hidden" name="id" value='. $datarow['sec_id'] .'>';
                          echo '<td align=\"left\" class=admin_grid_cell>'. $datarow['section_id'] .'</td>';
                                                                            echo '</tr>';


    $datarow_count++;
                          }
                           echo "<tr><td colspan=8 class=dark_gray_row align=center><span class=member_menu_text> $first  $prev  <font color=yellow>$nav</font>  $next  $last </span></td></tr>";
                          echo "</td></tr><TBODY></table>
                          <br><input type=\"submit\" name=\"submit\" value=\"Edit >>\"></form>";

There is some extra information in there because I use different row colors as well as paging but you should be able to figure it all, feel free to ask if you have any questions.

Gamebits

 

by: gamebitsPosted on 2007-01-21 at 17:32:37ID: 18361928

OK, Try this code make sure you do your db connect first

<?php
$requete = "SELECT * FROM tblProgramControl WHERE id=".$id;

 $num=mysql_numrows($requete);

echo "<b>Your query return <font color=\"red\">$num</font> results.</b><br><br>";

$i=0;
while ($i < $num) {

$isValid = mysql_result($requete,$i,"isValid");
$errString = mysql_result($requete,$i,"errString");
$allowPassword = mysql_result($requete,$i,"allowPassword");
$password = mysql_result($requete,$i,"password");
$lockedOutText = mysql_result($requete,$i,"lockedOutText");

 echo "<form method=\"post\" action=\"programControlEDIT.php\">
           Program is active?
               <input type=checkbox name=isValid value=$isValid><br>
           Message to display to program user:
               <input type=text name=errString value=$errString><br>
           Allow password to unlock program?
               <input type=checkbox name=allowPassword value=$allowPassword><br>
           Password:
               <input type=text name=password value=$password><br>
           Text to display when locked out:
               <input type=text name=lockedOutText value=$lockedOutText><br>
               <input type=submit name=post value=\"Submit\">
</form>";

$i++;
}

?>

If it works the way it is suppose to you will see your form filled with the data from the db.

Gamebits

 

by: bingiePosted on 2007-01-21 at 17:40:46ID: 18361960

ok, before i try that - should i incorperate that into the other code you wrote? or make it a new page?

 

by: gamebitsPosted on 2007-01-21 at 17:51:29ID: 18362006

I got to go but just a quick note the form you will see will not change anything in the db if you submited, you will need to make another file name

programControlEDIT.php as per the form action name to actually do the changes in the db tha file should look like this

$edit = mysql_query("UPDATE `tblProgramControl`
                                    SET `isValid` = $_POST['isValid'],
                                           `errString` = $_POST['errString'],
                                           `allowPassword` = $_POST['allowPassword'],
                                           `password` = $_POST['password'],
                                           `lockedOutText` = $_POST['lockedOutText']
                                   WHERE `id` = '$id'");

Now that I see it you should adda hidden field in the form to keep track of the id. Thw WHERE clause is very important otherwise you will change all the rows in the db.

Gamebits

 

by: gamebitsPosted on 2007-01-21 at 17:54:25ID: 18362011

To answer your question when I build an application I like to make different files for each things, made it easier to debug, when everything is working the way i want it to be than I look if it's worth it to set everything on the same page or not.

The last one I did I had about 10 pages, all in the same page I have 3000 lines of code.

Gamebits

 

by: bingiePosted on 2007-01-22 at 07:57:53ID: 18366140

Ok so here is where I am - no errors, just not returning any records.

<html>
<head>

<?php

function mysql_die($error = "unknown")
{
echo "<H1>Error:". $error."</H1>";
   exit;
}


?>

</head>

<body>
<h1>Book It Program Control Interface</h1>
<br>

<?php

$db = mysql_connect("localhost","bookingapp","") or mysql_die("Unable to connect");
mysql_select_db("bookingapp",$db);


$requete = "SELECT * FROM tblProgramControl WHERE id=".$id;

$num=mysql_numrows($requete);

echo "<b>Your query return <font color=\"red\">$num</font> results.</b><br><br>";

$i=0;
while ($i < $num) {

$isValid = mysql_result($requete,$i,"isValid");
$errString = mysql_result($requete,$i,"errString");
$allowPassword = mysql_result($requete,$i,"allowPassword");
$password = mysql_result($requete,$i,"password");
$lockedOutText = mysql_result($requete,$i,"lockedOutText");

 echo "<form method=\"post\" action=\"programControlEDIT.php\">
           Program is active?
               <input type=checkbox name=isValid value=$isValid><br>
           Message to display to program user:
               <input type=text name=errString value=$errString><br>
           Allow password to unlock program?
               <input type=checkbox name=allowPassword value=$allowPassword><br>
           Password:
               <input type=text name=password value=$password><br>
           Text to display when locked out:
               <input type=text name=lockedOutText value=$lockedOutText><br>
               <input type=submit name=post value=\"Submit\">
</form>";

$i++;
}

?>


</body>
</html>

 

by: gamebitsPosted on 2007-01-22 at 13:51:46ID: 18369825

Make the variable .$id in the where clause as a value, as is the code would not return anything, replace the variable with a value you know exist in the db.

Gamebits

 

by: bingiePosted on 2007-01-22 at 15:13:44ID: 18370569

Done. Still no records being returned:

<html>
<head>

<?php

function mysql_die($error = "unknown")
{
echo "<H1>Error:". $error."</H1>";
   exit;
}

$db = mysql_connect("localhost","bookingapp","") or mysql_die("Unable to connect");
mysql_select_db("bookingapp",$db);


$requete = "SELECT * FROM tblProgramControl WHERE id=27";

$num=mysql_numrows($requete);

echo "<b>Your query return <font color=\"red\">$num</font> results.</b><br><br>";

$i=0;
while ($i < $num) {

$isValid = mysql_result($requete,$i,"isValid");
$errString = mysql_result($requete,$i,"errString");
$allowPassword = mysql_result($requete,$i,"allowPassword");
$password = mysql_result($requete,$i,"password");
$lockedOutText = mysql_result($requete,$i,"lockedOutText");

?>

</head>

<body>
<h1>Book It Program Control Interface</h1>
<br>

<?php

 echo "<form method=\"post\" action=\"programControlEDIT.php\">
           Program is active?
               <input type=checkbox name=isValid value=$isValid><br>
           Message to display to program user:
               <input type=text name=errString value=$errString><br>
           Allow password to unlock program?
               <input type=checkbox name=allowPassword value=$allowPassword><br>
           Password:
               <input type=text name=password value=$password><br>
           Text to display when locked out:
               <input type=text name=lockedOutText value=$lockedOutText><br>
               <input type=submit name=post value=\"Submit\">
</form>";

$i++;
}

?>


</body>
</html>

 

by: gamebitsPosted on 2007-01-22 at 15:37:50ID: 18370695

Can you try something else?

replace

$requete = "SELECT * FROM tblProgramControl WHERE id=27";

by

$requete = mysql_query("SELECT * FROM tblProgramControl WHERE id=27");

Gamebits

 

by: bingiePosted on 2007-01-22 at 16:23:48ID: 18371181

OK perfect. I used

$req = "SELECT * FROM tblProgramControl WHERE id=".$id;
$requete = mysql_query($req);

and the form is populated correctly.

Now...where do I put the code for the update? Is there an event for the form's submit click?

 

by: gamebitsPosted on 2007-01-22 at 16:40:18ID: 18371353

We will keep it simple for now

create a file and name it programControlEDIT.php

put the code to do the update in that file and upload the file in the same directory where is the pdate form.

Now open the form again make some changes and click submit

Let me know if you get an error or if the db is getting updated.

Gamebits

20120131-EE-VQP-002

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