[RESOLVED] Indexing

[genista]

Hi guys,

I have a join form connecting to a MySQL database. This form is a copy of another join form I have on the site for customers (this new one is for suppliers).

On the suppliers form I have added some extra columns, and yet when I try and submit to them I get an error - "undefined index" on all of the new columns only, the old columns post fine. Most of the new columns I want to be checkboxes and I have them set in MySQL as "char3" and then I have a few longtext.

I have gone over the PHP code and it is fine, any ideas on why I am having this error?

[tsinka]

Please post the relevant PHP code.

Regards,
Thomas

[genista]

Ok, here is the php function new user which gets passed to a config include for processing:

newUser($_POST["username"], $_POST["password"], $_POST["first_name"], $_POST["last_name"], $_POST["address_line1"], $_POST["address_line2"], $_POST["town"], $_POST["county"], $_POST["postcode"], $_POST["daytime_phone"], $_POST["mobile_phone"], $_POST["email_address"], [B]$_POST["test1"], $_POST["test2"], $_POST["test3"],[/B] 

The ones in bold are the new columns that are getting the error, in the database they look like so:

Name: test1, Type: Varchar, Length: 25, Null: Not null

In the html for the form it looks like so:

<tr><td>Test1:</td><td><input type="checkbox" name="test1" value="" >
</td></tr>

Does this help?

[tsinka]

Ok,

do you always get that warning or only for those post variables whose checkboxes haven't been checked ?

the post variables of checkboxes that are not checked will not be submitted and therefore not exist in the $_POST array.

You need to check if the post variables are set at all with the isset function. I'd suggest not to use the $_POST array directly in function calls.

Something like (simplified):

$strUsername = isset($_POST['username']) ? $_POST['username'] : "";
$strPassword = isset($_POST['password']) ? $_POST['password'] : "";
// do the same with all the other post variables
// then do some error checking
// finally insert the data
newUser($strUsername, $strPassword,.....);

Thomas

[Houdini]

I reformatted your code and removed the and which only would work with a quote not within code. This is not a function.

newUser($_POST["username"], $_POST["password"], $_POST["first_name"], $_POST["last_name"], 
$_POST["address_line1"], $_POST["address_line2"], $_POST["town"], $_POST["county"], 
$_POST["postcode"], $_POST["daytime_phone"], $_POST["mobile_phone"], $_POST["email_address"], 
$_POST["test1"], $_POST["test2"], $_POST["test3"],

a function looks like

<?php
function myFunk($test="Success")
{
//do something here
echo "myFunction prints this to the screen!<br />";
echo $test;
//end of doing something
}
$crud="Its a real mess";
myFunk($crud);
?>
}

This is of course a simplistic function and has absolutely no value at all, but what you have is not only not a function I am uncertain as to what you would call it.

Might I ask what it is you are trying to do, and if you need a function then what are all those $_POST variables about?

[genista]

You are right the error only comes up if the checkboxes are unchecked, however no records are posted to the database.

As for the otherpart you mentioned aboutt isset I am not sure I understand, my post code gets sent to the config file, which calls a functions file. Here it gets processed as the 'newuser' function.

[genista]

Hope that answers your post too, Houdini.

[tsinka]

Using isset will avoid those warnings about undefined indexes. The code I posted only tries to access a post value if it is set at all. The warning about undefined indexes is shown because in your code you try to access a post variable regardless if it is defined or not.

Thoms

[genista]

Ok, so in order for me to understand this how could it be applied to the following code?:

Join form:

newUser($_POST["username"], $_POST["password"], $_POST["first_name"], $_POST["last_name"], $_POST["address_line1"], $_POST["address_line2"], $_POST["town"], $_POST["county"], $_POST["postcode"], $_POST["daytime_phone"], $_POST["mobile_phone"], $_POST["email_address"], [B]$_POST["test1"], $_POST["test2"], $_POST["test3"],[/B]

Which then gets sent to a functions file (through a config file) and the function in this file looks like so:

function newUser ($username, $password, $first_name, $last_name, $address_line1, $address_line2, $town, $county, $postcode, $daytime_phone, $mobile_phone, $email_address, $test1, $test2 etc) {

global $link;

$query="INSERT INTO suppliers (username, password, first_name, last_name, address_line1, address_line2, town, county, postcode, daytime_phone, mobile_phone, email_address, test1, test2)
VALUES ('$username', '$password', '$first_name', '$last_name', '$address_line1', '$address_line2', '$town', '$county', '$postcode', '$daytime_phone', '$mobile_phone', '$email_address', '$test1', '$test2')";

$result=mysql_query($query, $link) or die("Died inserting login info into db.  Error returned if any: ".mysql_error());

return true; 

Thanks,

G

[tsinka]

My second post contains an example of how you can do it.

First assign the post values to variables and then use those variables with your function call.

But you should add some error checking code (valid username, valid password and so on)

Thomas

[genista]

Hi thats great, I have done the following with the code:

newuser($strUsername = isset($_POST['username']) ? $_POST['username'] : "", 
		$strPassword = isset($_POST['password']) ? $_POST['password'] : "", 
		$strfirst_name = isset($_POST['first_name']) ? $_POST'first_name'] : "",
		$strlast_name = isset($_POST['last_name']) ? $_POST'last_name'] : "", 
		$straddress_line1 = isset($_POST['address_line1']) ? $_POST['address_line1'] : "", 
		$straddress_line2 = isset($_POST['address_line2']) ? $_POST['address_line2'] : "", 
		$strtown = isset($_POST['town']) ? $_POST['town'] : "",  

		$strcounty = isset($_POST['county']) ? $_POST['county'] : "",
		$strpostcode = isset($_POST['postcode']) ? $_POST['postcode'] : "",
		$strdaytime_phone = isset($_POST['daytime_phone']) ? $_POST['test1'] : "");

I am now without my errors, but the checkbox results are not being posted to the database, any ideas on this?

[tsinka]

That's not exactly what I was writing about. I never tried it that way and I am a little bit surprised that it works at all.

Look at the last argument to the function:
isset($POST['daytime_phone']) ? $POST['test1'] : ""

Do you want to save the daytime phone number if checkbox test1 has been checked ? Or vice versa ?

Please post the script (as attachment). The way you call the function can be improved.

Thomas

[genista]

Ok, that was a typo, sorry about that, you will find below the details of the join form as it stands now and then below that the function as defined in the functions form:

Join form:

 if(empty($messages)) {
        // registration ok, get user id and update db with new info:
        newuser($strUsername = isset($_POST['username']) ? $_POST['username'] : "", 
		$strPassword = isset($_POST['password']) ? $_POST['password'] : "", 
		$strfirst_name = isset($_POST['first_name']) ? $_POST['first_name'] : "",
		$strlast_name = isset($_POST['last_name']) ? $_POST['last_name'] : "", 
		$straddress_line1 = isset($_POST['address_line1']) ? $_POST['address_line1'] : "", 
		$straddress_line2 = isset($_POST['address_line2']) ? $_POST['address_line2'] : "", 
		$strtown = isset($_POST['town']) ? $_POST['town'] : "",  

		$strcounty = isset($_POST['county']) ? $_POST['county'] : "",
		$strpostcode = isset($_POST['postcode']) ? $_POST['postcode'] : "",
		$strdaytime_phone = isset($_POST['daytime_phone']) ? $_POST['daytime_phone'] : "",
		$strmobile_phone = isset($_POST['mobile_phone']) ? $_POST['mobile_phone'] : "",
		$stremail_address = isset($_POST['email_address']) ? $_POST['email_address'] : "",
		$strtest1 = isset($_POST['test1']) ? $_POST['test1'] : "",
		$strtraining = isset($_POST['test2']) ? $_POST['test2'] : "",

	$strinsurancedetails = isset($_POST['insurancedetails']) ? $_POST['insurancedetails'] : "");

Functions include:

function newUser ($username, $password, $first_name, $last_name, $address_line1, $address_line2, $town, $county, $postcode, $daytime_phone, $mobile_phone, $email_address, $test1, $test2) {
/*
    Creating a New User Record in the DB:
    In this function we create a new user record in the db.

We first build a query and save it into the $query variable.
The query statement says:
line 75
    'Insert the value of $login and $password into the 'login'
    and 'password' columns in the 'users' table' (line 77)
    */

global $link;

$query="INSERT INTO suppliers (username, password, first_name, last_name, address_line1, address_line2, town, county, postcode, daytime_phone, mobile_phone, email_address, test1, test2)
VALUES ('$username', '$password', '$first_name', '$last_name', '$address_line1', '$address_line2', '$town', '$county', '$postcode', '$daytime_phone', '$mobile_phone', '$email_address', '$test1', '$test2')";
$result=mysql_query($query, $link) or die("Died inserting login info into db.  Error returned if any: ".mysql_error());

return true;
} // end func newUser($username, $pass)

Hope this helps...

g

[tsinka]

In one of your last post I saw this HTML code:

<input type="checkbox" name="test1" value="" >

The problem:

You'll get no value even if someone checks the box because the value attribute is empty. So to test if the checkbox stuff works, set the value.

What kind of info do you want to submit with the checkboxes ?

Thomas

[genista]

I would want the value to be true if ticked and false if unticked.

Thanks

g

[genista]

Ok so I put true in the values and it is all working now, I think we are done.

Now time for a new post on my search script....

[genista]

Hi all,

I know this Post is marked as Resolved, but I have a problem with the script posted here. I can sign up a new user and they get redirected to the members section, but on inspection nothing is getting posted to the database. I get no errors or any indication that this is not happening, any ideas?

[tsinka]

Ok,

please repost the current version of your script(s). Post the code as attachment (e.g. in a ZIP file) if it is too large.

Thomas

[genista]

Its ok I have solved it (starting to get the hang of this now...)