update fields in a html form

timmybuck

hi guys

im getting this error message

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\Apache Group\Apache2\htdocs\m82\updatethree.php on line 29

below is my code

<html>

<body>

<?php

$db = mysql_connect("localhost", "phpuser", "******");
mysql_select_db("booking",$db);

if ($submit) {

$sql = "UPDATE event_categories SET categoryID='$categoryID', category='$category'";

$result = mysql_query($sql);

echo "Thank you! Information updated.\n";
}

else {

// query the DB

$sql = "SELECT * FROM event_categories WHERE categoryID=$categoryID";

$result = mysql_query($sql);

$myrow = mysql_fetch_array($result);

categoryID:<input type="Text" name="categoryID" value="<?php echo $myrow["categoryID"]?>"><br>

category:<input type="Text" name="category" value="<?php echo $myrow["category"]?>"><br>

</form>

<?php

$result = mysql_query("SELECT * FROM event_categories",$db);

while ($myrow = mysql_fetch_array($result)) {

printf("<a href=\"%s?categoryID=%s\">%s</a><br>\n", $php_SELF, $myrow["categoryID"], $myrow["category"]);

}

</body>
</html>

dartcol

Check out superglobals at php.net. It might help you out.
HTH

timmybuck

the data from my drop down keeps disappearing once i select it and choose submit. Also it dissapears from my database, but categoryID remains, its just the category that disappears.

can anyone see why

<html>

<body>
<form action="updatefour.php" method="post" target="_self">
<?php

$db = mysql_connect("localhost", "phpuser", "******");
mysql_select_db("booking",$db);

$event_categories=mysql_query("SELECT * FROM event_categories ")

OR DIE("Unable to find events. Error: ".mysql_error());

print("<select name=\"categoryID\" size=1>\n");

print("<option value=\"b\">Select");
print("<option value=\"b\">-----");
while($row = mysql_fetch_array ($event_categories))
{
print("<option value=\"$row[categoryID]\">$row[category] \n");
}
print("</select>\n");

</form>

<form method="post" action="<?php echo $php_SELF?>">
<?php

$categoryID = $POST['categoryID'];
$category = $POST['category'];
$sql = "SELECT * FROM event_categories WHERE categoryID=$categoryID";
$result = mysql_query($sql);
if ( $result == true){
$myrow = mysql_fetch_array($result);
}
?>
<input type=hidden name="categoryID" value="<?php echo $myrow["categoryID"]?>">
categoryID:<input type="Text" name="categoryID" value="<?php echo $myrow["categoryID"]?>"><br>
category:<input type="Text" name="category" value="<?php echo $myrow["category"]?>"><br>
<?php
$sql = "UPDATE event_categories SET categoryID=$categoryID, category='$category' WHERE categoryID=$categoryID";
$result = mysql_query($sql);
?>

</form>

</body>
</html>

Weedpacket

UPDATE event_categories SET categoryID=$categoryID, category='$category' WHERE categoryID=$categoryID

Why update the value of something if you know it already has that value?

Hint: to make your posts legible to others, put your code in [php]...[/php] tags.

It's spelled $PHP_SELF (or, better, $_SERVER['PHP_SELF']), not $php_SELF. There is a difference.

Which form field in the first form does $_POST['category'] get its value from?

Errr.... and because of where it is, the UPDATE will happen while the form to "Update information" is still being written out. And that's well before the form is even seen by the user, let alone before they hit the submit button. Are you sure you want it to happen then, and not after they've submitted the form back? Maybe I'm just confused by how you've designed the program.

timmybuck

im confused

can u show me how u would do it