PHP: Dynamically find how many variables aren't empty

Question

I have a list of 14 variables that may or may not be empty. Let's say for example we have many individual variables that holds a select ingredient for a recipe. The number of ingredients in a recipe will differ by recipe, obviously. Is there a way, using PHP, that we can see how many of the 14 variables are not empty (i.e. contain an ingredient)?

Example (using 7 ingredients):

    $ing_1 = "carrots"
    $ing_2 = "apples"
    $ing_3 = "sugar"
    $ing_4 = "celery"
    $ing_5 = ""
    $ing_6 = ""
    $ing_7 = ""

How would I get the awnser 4?

Further Info:

I am using mysqli_fetch_array to load to variables.

if you do this as an array - $ing[1]/$ing[2]/etc - instead of your var structure, then stackoverflow.com/questions/4422889/… or stackoverflow.com/questions/19696281/… or stackoverflow.com/questions/7672562/… could help. — Sean
– Sean, Commented May 12, 2015 at 5:22
Disclaimer: This is not PHP, but an alternative with (my)SQL: As you say, that you use mysqli_fetch_array I think you got a database in the back. You could select an addional field (count numbers of ingredients) and then just "read" this field where the number of ingredients is saved. — bish
– bish, Commented May 12, 2015 at 5:25
if you stay with your current var structure, then you could do a loop - $count=0; for($i=1;$i<=7;$i++){ if(!empty(${"ing_$i"})) $count++; } echo $count; — Sean
– Sean, Commented May 12, 2015 at 5:37

Sean · Accepted Answer · 2015-05-12 21:06:28Z

0

given your current variable structure, you could loop over your variable like-

$count=0; 
for($i=1;$i<=7;$i++){ 
    if(!empty(${"ing_$i"})) $count++; 
} 
echo $count;

with the $i<=7 could be increased to $i<=14 if you have 14 variable/ingredients.

answered May 12, 2015 at 21:06

Sean

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