Update a Forecast Record

[chefmars]

I am currently inserting records into a MYSQL table using a php form.
I use the INSERT command in the form to capture the variables in the table.

The form is for Restaurateurs to use to forecast sales for a future date.
That future date is one of the fields in the record. Each record has a unique number ID.
Two of the fields in the record (table) are not on the initial form because (at a later time) I want the user to be able to modify the record and insert the actual number of customers and also any mitigating circumstances (into those two empty fields) that could account for the difference.

Could someone help me with the code that could allow the user to call up from the table of records, the full record for viewing and allow them to add that (actual) information and then"update" the record? (I think it requires another form but I am not sure)

Thanks...
Chefmars

[Roger_Ramjet]

No you can use the same form for this quite easily, modifying the presentation to suit.

However, that adds a layer of complexity and if you are not sure about creating general purpose scripts and forms then do it in 2. Here is what the update would look like.

edit.php

// edit.php get and displays record for update
if (isset($_POST['id'])) {
$rid=$_POST['id'];
// query for the data
$sql = "SELECT * FROM table WHERE rid=$rid";
$result = mysql_query($sql) or die (mysql_error());
if (!$result or mysql_num_rows($result) != 1) {
   echo 'database error, record not found';
   break;
}

$row=mysql_fetch_array($result);
echo '<form name="update" method="post" action="update.php">
    <input type="hidden" name="rid" value="' . $row['rid'] . '" />
    <input type="text" name="forcast" value="' . $row['forcast'] . '" readonly /><br />
    <input type="text" name="actual" /><br />
    <input type="text" name="circumstance" />
    <input type="submit" name="submit" /></form>';

then update.php

if (isset($_POST['submit'])) {
   $act = $_POST['actual'];
   $circ = $_POST['circumstance'];
   $rid = $_POST['rid'];
   $sql = "UPDATE table SET actual=$act, circumstance='$circ' WHERE rid=$rid";
   $result = mysql_query($sql);
}

[chefmars]

Roger,

Thank you very much for you help.
The issue has been resolved.
I did have an initial problem with line 10 of your code.

echo 'database error, reocrd not found';

Regards and thanks again,
Chef Mars

[Roger_Ramjet]

Fixed the typo in the original post.