Could you check my logic on this, please?

The dealbreaker, at this point, is the presence of an array. At least, that's what I've been able to deduce thus far.

Here's the first part of the process:

protected function processStatementSql ($sql) {

        $statements = StatementImage::runQuery($sql);
        $ret = array();

        //Second and Third DB Calls
        //TXN TABLE STUFF
        // NOTE - I don't know there is a reason to process these as separate queries
        // and I think they can be reasonably combined with the query held in getSQL()
        // and processed through to get the same dataset more quickly.
        if ($statements) {
            foreach($statements as $st) {
                $accountid = $st['AccountID'];

                if (!$st['StatementRollUp'] || $st['templateid'] == '9') {
                    $st['newtxns'] = StatementImage::getNewTransactions($st, $st['showphysician'] == '1');
                } else {
                    $st['newtxns'] = array();
                }

Open in new window


Look at line 15. $statements is a SELECT statement that's just been fired. We're going to parse it out as an array and when we get to $st['StatementRollUp'] we pause and ask a question...

if (!$st['StatementRollUp'] || $st['templateid'] == '9') {

Which, in plain language is asking, if the StatementRollUp value is not defined OR the template id is 9, then proceed with the getNewTransactions method. At which point, you've defining an index within an array ($st['newtxns'])  as the values represented by the getNewTransactions method.

Otherwise...

You've got the $st['newtxns'] array, but there's nothing in it.

The very next line of the code is this:

foreach($st['newtxns'] as $txn){
                    $st['lastclaimdate'] = $txn['firstdos'] < $st['lastclaimdate'] || $st['lastclaimdate'] === null ? $txn['firstdos'] : $st['lastclaimdate'];
                }

Open in new window


So, now that $st['newtxns'] is $txn.

I know. It's a mess. But be that as it may, I've been hired to fix what's there and not write new code. There's six years worth of original programming and the layers of nonsense are eternal.

So, here comes that $txn array and it's being looked at in the context of another IF statement:

  if (isset($encounter['txns'])) {

From here, the PDF that's being generated is formatted according to the result of this IF statement. My thought is that if your $encounter['txns'] array is devoid of content, then the result of the above IF clause is FALSE.

Agreed?
Bruce GustPHP DeveloperAsked:
Who is Participating?

[Product update] Infrastructure Analysis Tool is now available with Business Accounts.Learn More

x
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

Julian HansenCommented:
Think of it like this isset will return true if $encounter['txns'] exists i.e. if $encounter is an array AND index 'txns' is defined.

If $encounter is not an array or does not have an index 'txns' set, then isset will return false
0
Bruce GustPHP DeveloperAuthor Commented:
So...

It doesn't have anything to do with content. If both the array and the index have been defined, then this:

 if (isset($encounter['txns'])) {

...is TRUE. Both the array and the index have been defined and what's IN the array is not what's being looked at as much as it's THAT it's an array.

Yes?
0
Ray PaseurCommented:
It doesn't have anything to do with content...

Not quite.  Variables may have been defined, then subsequently set to NULL or unset().  PHP isset() determines if a variable is set and is not NULL.  It returns TRUE if both cases are true for all of its arguments.

There really isn't any such thing as "devoid of content" unless you make an assignment of NULL to a variable.  And to fully understand what is happening to the PHP symbol table, you need to know the scope of the variables.

In the case of the $encounter array, it can be defined like this, but one of its member positions remains undefined:
<?php // demo/temp_brucegust.php
/**
 * https://www.experts-exchange.com/questions/28959883/Could-you-check-my-logic-on-this-please.html
 *
 * http://php.net/manual/en/function.var-dump.php
 * http://php.net/manual/en/function.error-reporting.php
 */

// WITH FULL ERROR REPORTING -- ALWAYS RECOMMENDED
error_reporting(E_ALL);

$encounter = [];
var_dump($encounter);
if (isset($encounter)) echo "encounter is defined";
if (isset($encounter['txns'])) echo "encounter.txns is defined";

Open in new window

Outputs:
array(0) { } encounter is defined

Open in new window

0

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
PHP

From novice to tech pro — start learning today.