Updating the DB from a form

Anon

I can script a PHP form to updatethe database, but I want to create something like this:

Once a user logs in with an id/password, a form appears with all of their account information inside of it.

They can then update the information and update the database.

An e-mail is then sent to me telling me that they have updated the information.

I can do most everything, but I need help with getting their information to show up in the form.

Thanks
Jon
jon@jonkowalski.com

Anon

What you need to do is query your source and echo your results to fields in your form

Mark
hdwt.com

kidestranged

just build the form in php and use the info from the selected record as the value.

echo "<input type='text' value='$mysql_field_variable' name='formname'>\n";

Anon

Now... when I try the following, first name gets filled in but not last name... any ideas why?

<?php
$dbh=mysql_connect("64.23.60.54", "jon", "jdkldl");
mysql_select_db("test", $dbh);
$query="SELECT * FROM customers";
$res=mysql_query($query, $dbh);
?>
<form method="POST" action="customer.php">
<p>

Update Your Account Information<p>

First Name <input type="text" name="fname" size="20" value="
<?php while($row=mysql_fetch_array($res))
{echo$row[fname];}
?>">
<p>
Last Name <input type="text" name="lname" size="20" value="
<?php while($row=mysql_fetch_array($res))
{echo$row[lname];}
?>">
<p>

Anon

I tried the following code:

<?php
$dbh=mysql_connect("server", "test", "test");
mysql_select_db("test", $dbh);
$query="SELECT * FROM customers";
$res=mysql_query($query, $dbh);
?>
<form method="POST" action="customer.php">
<p>
Update Your Account Information<p>

First Name <?php echo "<input type='text' value='$fname' name='fname'>
";
?>

<BR>

and nothing was filled into the first field. It was filled in the second, but not the third. Any ideas?

Anon

Of course it won't work; your code does not make sense.

To begin with, your SQL query looks for ALL customers, not just the one who might be using the form. You need to add a WHERE clause to the SQL to narrow it down.

Further, mysql_query() returns something known as a result set, which might contain zero to many rows of data.

You can't address that data directly; you have to use one of the fetch functions. The function mysql_fetch_array() is one of them. You're using it. That's fine, but ...

When you call mysql_fetch_array, you advance an internal pointer in the result set.

Notice how your code contains two while() loops. Why? If your SQL has returned one, and only one, record, then there should not be a while() loop at all.

Naturally, your second call to mysql_fetch_array() won't return anything, because the result set is exhausted. Your first call has run through the entire result set, returning the last object found.

Anon

See Steve Yelvington's reply further down the tree for the third missing field; for the first it's simply because you never said what $fname was.

khaine

Try changing you code to

<?php
$dbh=mysql_connect("64.23.60.54", "jon", "jdkldl");
mysql_select_db("test", $dbh);
$query="SELECT * FROM customers";
$res=mysql_query($query, $dbh);
$row=mysql_fetch_array($res)
?>
<form method="POST" action="customer.php">
<p>

Update Your Account Information<p>

First Name <input type="text" name="fname" size="20" value="
<?php echo$row[fname];} ?>">
<p>
Last Name <input type="text" name="lname" size="20" value="
<?php echo$row[lname]; ?>">
<p>

Basically what every one has already said but I though it might help.

Mark.

Anon

Ok, I can now pull the values from the database into the form, but I cannot get the database to update based on the values in the form. I know I am screwing something very easy up... any pointers?

Anon

Ok, I can now pull the values from the database into the form, but I cannot get the database to update based on the values in the form. I know I am screwing something very easy up... any pointers?

Anon

where is your link identifier

mysql_query($query);

It seems to be missing

Mark

Anon

The link identifier argument is optional, and generally not necessary unless you are connecting to two databases, or to one database as two separate users (example: one reads, one writes).

As the manual says:

mysql_query() sends a query to the currently active database on the server that's associated with the specified link identifier. If link_identifier isn't specified, the last opened link is assumed. If no link is open, the function tries to establish a link as if mysql_connect() was called with no arguments, and use it.

Anon

Still think it's good practice to supply the link identifier

It makes double sure that a link was opened for use in the first pace

Mark

Anon

Now... I am authenticating users coming into this site. Is there any way to authenticate passwords that have already been encrypted by the database? Thank you so much for your help!

Anon

If you store the passwords in encrypted form in MySQL, you have to encrypt the test value in order to perform the comparison.

Something like this:

select * from table where userid='$userid' and password=PASSWORD('$password')