Screen Value?

jamesh

I have a form where users are going to be retrieving information, and then may change that information. I want to pull the "screen value" of that change (i.e., what they change the value to be) and check it against the db before committing the change. Is there a way to assign the screen value of a text box to a variable?

For example: the form field is "Street Address". The user does a find for a particular company and gets the street address of "1212 Elm St.". He changes the value on the form to read "1214 Oak St." and wants to save it. I want to be able to read that new value to pass it to the db, to see if it already exists. Can I capture that screen value?

Thanks--

theslug

So, if I understand correctly, you want to make sure that 1214 Oak Street does not exist in your db correct?

If that's the case, and the address is held in a column somewhere you could do something like:

$query = "SELECT address from table where address=\"$forminput\";
$result_query = mysql_query($query);

if ($result_query == 0)
{
$update = INSERT INTO table ('address') VALUES ('$forminput');
$result_update = mysql_query($update);
}
else
{
print("$forminput already exists!\n"):
}

jamesh

Well, that's most of it. However, the user is going to be doing a "find" to bring up what information is already in the db. Thus, the box I want to check the value of will already have a value that I know is in the db. If the user types over that value, the original value still shows. My problem with this is how to read in the user's change in order to test it. The text box's attributes look something like: <inpt type="text" nme="address" value=<php echo "\"$dbvalpulledhere\"" ?>. (yeah, I know it is missing letters, etc....I didn't want to actually put a box up, sorry).

A user will have the "dbvalpulledhere" value and types over it to change it. How do I read his changes? If I spit back the value of the box, I get the original value, not the changed one.....that is what I am looking for--the "trapped" value that the user puts in.

Sorry to be confusing....it might be hard to explain.

Thanks--
--Jim

theslug

Okay, so the user does a search.
You process the search and print a result page.

On the result page, you are allowing the users to change some of the information.
So in essence your input line will look somewhat like this.

The user then erases the output of $row[0] on the webpage and inputs their own value and then clicks SUBMIT (or whatever)
now in theory address should be updated with the new value?

Am I right on this one?

Just wanted to do a sanity check just to make sure I'm on the same page as you 😛

jamesh

Yep, that looks about right.

What I am trying to do is test against that user-input value (where they erased the $row[0] value) and check it against the db. (i.e., check for a duplicate before committing).

Thanks--

     --Jim

theslug

k, I'm going to have to see your code then..

if you are checking your DB with data contained in the variable from <INPUT NAME=var> it should work.

I have a series of pages doing exactly what you have described they are working.

basically (after the search is done)

$query = "SELECT address from table where address=\"$var\";
$result = mysql_query($query);

if($result == 0)
{
-- insert information into db --
}
else
{
echo "This info already exists"
}

Anon

I have had to do a lot of this exact type of thing recently. Here's what I came up with.

When I query the database, I use that data in the form AND I also put the data into a session variable. (I used an array (something like $db_data) where the key is the same as the table field name, value is the data for that field.)

Client sees a form, with the data (from database) already filled in. When they submit the form, I step through the post variables and compare them with the appropriate elements of $db_data array. If there are differences, you'll have to update the database.

One note of caution: Only checked checkboxes are passed as post variables. If you have 10 checkboxes, but only two are checked . . . only two variables will be sent. What this means is that when you use checkboxes, you'll also have to step through $db_data and compare all the checkbox data with the post variables to see if the client has removed any checkmarks in the form.

You don't have to use session variables to do this, but I think that's probably easiest. It also means fewer database queries. Another alternative would be to use hidden form objects in addition to your visible form objects. For example, <input type="hidden" name="db_street_address" value="123 Main Street">.

--Donnie

jamesh

DLB: Can you by any chance point me to where I can get more in-depth info on the hidden form objects and their uses? I have done some looking around, but I would prefer to be more proficient in using them.....

Thanks!

 --Jim

Anon

Forms have different types of objects (or elements): text, textarea, radio, select, HIDDEN, etc.

For more info, take a look at this tutorial.
http://hotwired.lycos.com/webmonkey/99/30/index4a_page4.html

Good luck.