Add to array if set

Question

Is there a better/shorter way to do this?

Each variable is a MySQL table value that is called. I have a main table and an override table, so call the first table, extract the resulting array, then call the override table and extract those results to override the first extract.

if (isset($Price1)){
    $AllPrices[] = $Price1;}
if (isset($Price2)){
    $AllPrices[] = $Price2;}
if (isset($Price3)){
    $AllPrices[] = $Price3;}    
if (isset($Price4)){
    $AllPrices[] = $Price4;}
if (isset($SPrice1)){
    $AllPrices[] = $SPrice1;}
if (isset($SPrice2)){
    $AllPrices[] = $SPrice2;}
if (isset($SPrice3)){
    $AllPrices[] = $SPrice3;}

"All variables are pulling from MySQL" what does this mean? That is, how are these variables being set specifically? — erisco
– erisco, Commented Mar 1, 2013 at 21:23
Not really, it looks like you need to check each individual variable, which you're doing. Unless those vars were originally in an array or something... — Adam Plocher
– Adam Plocher, Commented Mar 1, 2013 at 21:23
can you show us the full code (the part that pulls from the DB and loops for example)? — Green Black
– Green Black, Commented Mar 1, 2013 at 21:23
There is a way using variable variables, but I don't recommend it. Best would be to originate all those vars like $Price1 as array keys rather than global vars. — Michael Berkowski
– Michael Berkowski, Commented Mar 1, 2013 at 21:26
Each variable is a table value that is called. I have a main table and an override table, so call the first able, extract, the values, then call the override and extract those values to override. Then I run the code above. — EricImprint
– EricImprint, Commented Mar 1, 2013 at 21:34

Thomas Kelley · Accepted Answer · 2013-03-01 21:31:53Z

4

I've done things like this in the past, but I wouldn't recommend it:

$variables = array("Price1", "Price2", "Price3", "Price4", "SPrice1", "SPrice2", "SPrice3");

$AllPrices = array();

foreach ($variables as $variable)
{
    if (isset($$variable))
        $AllPrices[] = $$variable;
}

See here for more information on how the $$ syntax works in PHP.

But I do agree with the comments to your post... You really should take another look at how you're getting this data. This solution is not ideal in the least.

edited Mar 1, 2013 at 21:31

answered Mar 1, 2013 at 21:26

Thomas Kelley

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2 Comments

Green Black Over a year ago

I think it is a neat solution for the information you got from the question :-)

EricImprint Over a year ago

Doesn't work for what I am doing (see my last comment), but nice solution.

user1646111 · Accepted Answer · 2013-03-01 21:37:53Z

0

You can also do like this:

for($i = 1; $i < 4; $i++){
    $current = ${'AllPrices'.$i};
    isset($current) && array_push($AllPrices, $current);
}

As also posted in above answer, using array of variable names is more efficient.

answered Mar 1, 2013 at 21:37

user1646111

Comments

Roy · Accepted Answer · 2013-03-01 21:40:46Z

0

This structure is not recommended.

Instead of using $price1, $price2, ... use an array - $price[1], $price[2], ... (Do the same for $SPrice).

Then you could use a simple array_merge:

$AllPrices = array_merge($price, $SPrice);

answered Mar 1, 2013 at 21:40

Roy

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