How do I check for the presence of an array?

Here's the code as it exists currently:

 $statements = Statements::getQueuedSmsStatements($qty);
echo count($statements);
//if ($statements !==false){ //this was originally "=," rather than "==".

What you're looking at is a snippet of some code that I'm trying to troubleshoot. I'm "smelling" something foul with the third line down in that "if($statements!=false) looked problematic in that typically, with an IF clause, you always use two "=" signs.

Having said that, now I'm second guessing myself in that I'm testing for the presence of an array. And the matter become further convoluted when I run a code in the database that is populating the database with what the SELECT statement is going to be looking for that will result in an array with some values in it, yet...

I keep getting an "empty array" result with the code I have above.

Bottom line: I need to test for the presence of data in the Array. I've tried "count" as well as what you see above and I'm still not getting anything other than a "false" result.

What am I missing?
brucegustPHP DeveloperAsked:
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Julian HansenConnect With a Mentor Commented:

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To expand on Marco's explanation

With code values like 1 TRUE "FRED" are called truthy values as they will evaluate to true
Values 0 NULL FALSE empty string, empty array are termed falsy values as they evaluate to false

What if you want to test whether a value is actually the value true or false.

That is where the === and !== come in

The statement in question is testing to see if the value returned by

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is the value FALSE rather than an empty array or similar

Consider this code
$x = array();
if ($x) {
   echo "It are";
else {
   echo "It are NOT!";

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The output from this is the "It are NOT!" - because $x is an empty array and evaluates to false.

The if statement is therefore distinguishing between the return of an empty array (which may be a valid success condition) and the value false which indicates a problem. If != had been used then the empty array would be misinterpreted as a false value whereas with !== the test is expressly for false
Marco GasiFreelancerCommented:
You can use isset() and is_array() functions:

if (isset($statements) && is_array($statements)){
        if (count($statements) > 0){
                //do stuf here

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Marco GasiConnect With a Mentor FreelancerCommented:
typically, with an IF clause, you always use two "=" signs
Why? You can check if a condition is true (==) of if it is false (!=) depending on your logic. Sometimes you use equal operators and ometimes you use identical operator (=== and !==) depending on the return value of the function, but it is not typical check for true with an if clause...
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brucegustPHP DeveloperAuthor Commented:
Here's my dilemma:

I can run the SELECT in MSSQL Studio and get two rows, which is what I'm planning on.

I'll then go and run the syntax as I've got it written and get a "false" result. Hence, my hesitation, as far as whether or not I was using a proper IF clause.

My thinking is that I need to check for, not just the presence of an array, but whether or not that array has data in it. That being the case, from what I can gather from your collective input, is that I want something like:


I've got values in the array
I may have a successful condition, but there's nothing in the array

Julian HansenCommented:
Yes - if you run that code I posted earlier you will see the result
At the risk of belabouring the point - here is another sample (working link below)
$data = isset($_POST['data']) ? $_POST['data'] : array();
if ($data) {
   echo "You sent me something, I am touched";
   echo "<pre>" . print_r($_POST, true) . "</pre>";
else {
   echo "Please send me something";
<form method="post">
   Data <input name="data" type="text" /> <input type="submit" />

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Working sample here
brucegustPHP DeveloperAuthor Commented:
Got it!
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