Php validation code

[starbuck69400]

Hello every body,

I try to use a php validation code on a display element to get a value from a field in function of the value i've select in my databasejoin element (cnx_cnx___Adliv):

$adliv== '{cnx_cnx___Adliv}';
$db =&JFactory::getDBO();
$db->setQuery("SELECT `Nom` FROM `cnx_client` where $adliv = '1029092'");
$fieldA = $db->loadResult();
return $fieldA;

Cou'ld you tell me what i'am wrong?

Thanks for help.

 

[starbuck69400]

I had try this code but without more success:

$adliv = JRequest::getVar('cnx_cnx___Adliv');$db=$JFactory::getDBO();$db= ('SELECT nom FROM cnx_client where adliv= $adliv';return $db->loadResult();

Somebody can redirect me in the right direction?

Thank's for help

 

[troester]

where $adliv = '1029092'

What do you want to achieve/compare?
The syntax is WHERE table-column=some-value.
Is adliv a column of cnx_client and you want to select one record with adliv= content of your cnx_cnx___Adliv element?

$adliv== '{cnx_cnx___Adliv}'; 
$db =&JFactory::getDBO();
$q="SELECT `Nom` FROM `cnx_client` where adliv = '$adliv'";
//var_dump($q);exit; //uncomment this line for debugging your query
$db->setQuery($q); 
$fieldA = $db->loadResult(); 
return $fieldA;

 

[starbuck69400]

Is adliv a column of cnx_client and you want to select one record with adliv= content of your cnx_cnx___Adliv element?

Yes it's exactly that i want but with your code and by uncomment the debugging line i obtain this:

string(47) "SELECT `Nom` FROM `cnx_client` where adliv = ''"

It seems that $adliv== '{cnx_cnx___Adliv}'; don't return any value.
I don't understand why.

 

[troester]

php validation

I've overlooked this; in php validation you can't use placeholders. See the tooltip for how to access element data (try JRequest::getvar('cnx_cnx___Adliv'))

 

[starbuck69400]

Ok thank's for your help.
It's a little bit clearly for me but With this code:

$adliv=JRequest::getvar('cnx_cnx___Adliv'); 
$db =&JFactory::getDBO();
$q="SELECT nom FROM cnx_client where adliv = '$adliv'";
var_dump($q);exit; //uncomment this line for debugging your query
$db->setQuery($q); 
$fieldA = $db->loadResult(); 
return $fieldA;

I obtain this error:

string(51) "SELECT nom FROM cnx_client where adliv = 'Array'"

Thank's for help

 

[starbuck69400]

Sorry for my insistance but i don't find any solution to my problem.

I believed at first that I had to use (' cnx_cnx ___ Adliv_raw ') to obtain the value and not an array but my variable $adliv seems to return an array and my sql select can't know which value to get.

To express my need here is a plan:

In a mysql table "cnx_cnx" i have

Adliv Name
1034121 Toto
1034122 Hubert

In my form i've (for the moment) two ?l?ments.
The first is a database join on cnx_cnx_adliv and i would like that the second is a display ?l?ment which show the name value of table cnx_cnx in fonction of the raw data in the databasejoin ?l?ment.

So in fact:
if in databasejoin ?l?ment i've 1034121 i expect to have in my display ?l?ment TOTO.

Is there a way to obtain this result?

Thank's again for you help.

 

[troester]

For the code:
you can also use
var_dump($adliv);
to see what it is (so maybe $adliv[0] will contain what you need).

But I'm not sure what you are trying to do, it doesn't sound like a validation.

In a dbjoin element you can select the value (Adliv), this is what is stored in the database.
And you can select the label (Name) for what is displayed in the form/list.

If you want to fill a second element depending on the selection of the first one you can use the cascading dropdown element.
Maybe you'll have to update from GitHub to get it
http://fabrikar.com/forums/index.ph...s/#Installing_and_Enabling_Additional_Plugins

 

[starbuck69400]

Thank's troester
You're right! With cascading dropdown it work's. Not exactly how i would but it work.
Sometime the answer is front of us.

many thank's again.