Update form

Greenman

Hi,

I am having a bit of trouble getting the following edit form to work. I think this is because of something to do with retrieving id's from the DB??

Any ideas?

Regards,
G

<html>
<head>
<title>Update Link</title>

</head>
<?php

$db = mysql_connect("localhost", "user", "password")
or die ("cant connect to the DB");

print ("You have connected to database successfully");

?><br><br><?php

mysql_select_db("test",$db);

?>
<h2>Update Link</h2>

<?PHP

processed when form is submitted back onto itself

if ($_SERVER['REQUEST_METHOD'] == "POST") {

    $id = $_POST['id'];
    $first = addslashes($_POST["first"]);
    $last = addslashes($_POST["last"]);
    $address = addslashes($_POST["address"]);
    $dept = addslashes($_POST["dept"]);

    # setup SQL statement
    $sql  = " UPDATE employees SET";
    $sql .= " first = '$first', ";
    $sql .= " last = '$last', ";
    $sql .= " address = '$address', ";
    $sql .= " dept = '$dept' ";
    $sql .= " WHERE id = '$id' ";
    $sql .= " LIMIT 1 ";

    # execute SQL statement
    $result = mysql_query($sql,$db);

    # check for errors
    if (mysql_error()) { print "Database Error: $sql " . mysql_error(); }

    print "<p><b>Link Updated</b></p>\n";

}
else { # display edit form (not post method)

    # setup SQL statement to retrieve link that we want to edit
    $sql  = " SELECT * FROM employees ";
  	$sql .= " WHERE id = $id ";

    # execute SQL statement
$results = mysql_query($sql,$db);

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

    # retrieve values
    $row = mysql_fetch_array($results);
    $first = $row["first"];
    $last = $row["last"];
    $address = $row["address"];
    $dept = $row["dept"];

?>
<form name="fa" action="editform2.php" method="post">
<table>
<tr><td><b>First: </b> </td><td><input type="text" name="first" value="<?=$first?>" size=40></td>

</tr>
<tr><td><b>last:</b> </td><td><input type="text" name="last" value="<?=$last?>" size=40></td></tr>
<tr><td><b>address: </b> </td><td><input type="text" name="address" value="<?=$address?>" size=40></td></tr>
<tr><td valign=top><b>dept: </b> </td><td> <textarea name="dept" rows=5 cols=40><?=$dept?></textarea></td></tr>
<tr><th colspan=2><p><input type="submit" value="update link"></p></th></tr>
</table>
</form>

<?php } php?>
</body>
</html>

jpmoriarty

please use the php tags in the future: [php] and [/php]

<html> 
<head> 
<title>Update Link</title> 

</head> 
<?php 

$db = mysql_connect("localhost", "user", "password") 
or die ("cant connect to the DB"); 
print ("You have connected to database successfully"); 

?><br><br><?php 

mysql_select_db("test",$db); 

?> 
<h2>Update Link</h2> 

<?PHP 
# processed when form is submitted back onto itself 
if ($_SERVER['REQUEST_METHOD'] == "POST") { 

$id = $_POST["id"]; 
$first = addslashes($_POST["first"]); 
$last = addslashes($_POST["last"]); 
$address = addslashes($_POST["address"]); 
$dept = addslashes($_POST["dept"]); 

# setup SQL statement 
$sql = " UPDATE employees SET"; 
$sql .= " first = '$first', "; 
$sql .= " last = '$last', "; 
$sql .= " address = '$address', "; 
$sql .= " dept = '$dept' "; 
$sql .= " WHERE id = '$id' "; 
$sql .= " LIMIT 1 "; 

# execute SQL statement 
$result = mysql_query($sql,$db); 

# check for errors 
if (mysql_error()) { print "Database Error: $sql " . mysql_error(); } 

print "<p><b>Link Updated</b></p>\n"; 

} 
else { # display edit form (not post method) 

$id = $_GET = $myrow["id"]; 


# setup SQL statement to retrieve link that we want to edit 
$sql = " SELECT * FROM employees "; 
$sql .= " WHERE id = $id "; 

# execute SQL statement 
$results = mysql_query($sql,$db); 
// $result = mysql_query("SELECT * FROM employees",$db); 

# retrieve values 
$row = mysql_fetch_array($results); 
$first = $row["first"]; 
$last = $row["last"]; 
$address = $row["address"]; 
$dept = $row["dept"]; 
?> 
<form name="fa" action="editform2.php" method="post"> 
<table> 
<tr><td><b>First: </b> </td><td><input type="text" name="first" value="<?=$first?>" size=40></td> 

</tr> 
<tr><td><b>last:</b> </td><td><input type="text" name="last" value="<?=$last?>" size=40></td></tr> 
<tr><td><b>address: </b> </td><td><input type="text" name="address" value="<?=$address?>" size=40></td></tr> 
<tr><td valign=top><b>dept: </b> </td><td> <textarea name="dept" rows=5 cols=40><?=$ddept"?></textarea></td></tr> 
<tr><th colspan=2><p><input type="submit" value="update link"></p></th></tr> 
</table> 
</form> 

<?php } php?> 
</body> 
</html>

and we need more on what's not working? errors? nothing happening? what?

jpmoriarty

and by putting it into the tags you can spot very easily the problem at the bottom:

you've got a " inside the php tags which probably stop it from displaying,

Greenman

Hi jpmoriarty,

This is the error I am getting when I load the page,
Hopefully this will help.

Regards,
G

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/html/test1/editform2.php on line 63

line 63 ($row = mysql_fetch_array($results)😉

jpmoriarty

Originally posted by Greenman
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/html/test1/editform2.php on line 63

that means you're getting no results from your query.

what is this supposed to mean?
$id = $_GET = $myrow["id"];
two equals on one line?

I think you mean

to access a get variable (ie page.php?this=that)

you'd do $_GET['this'].

I'm not sure what you mean by $_GET = $myrow["id"], but it doesnt make sense.

Greenman

Hi jpmoriarty,

I have remoded this rogue line (now where did that come from 🙂)
but still no joy.
I have updated my posted code above

Regards,
G

jpmoriarty

well yes but

You've still not put in the php tags [php] and [/php]. Please use them as it makes the code so much easier to read.
you're still getting no results from the query. Try putting the $id in single quotes like you have previously.

Greenman

Hi John,

It still wont paly 🙁

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/html/test1/editform3.php on line

$row = mysql_fetch_array($results);

Regards,
G

jpmoriarty

okay, i know what the error message is and i know what line it happens on.

But i'm going to reiterate that the error is because you are getting no results from your query . So showing me what line the error happens on makes no difference, as that isnt the line that we have a problem with. your problem is with the sql that your code is generating.

Before you have that line echo out your SQL statement. if you cant see any glaringly obvious errors with it (and i'd bet good money that you will) then copy it and paste it directly into mysql, and then notice what it's problem with the line is.

And please, please, please put php tags around your code so that we can read it properly.