PHP: Check if 0?

Question

I am using a class which returns me the value of a particular row and cell of an excel spreadsheet. To build up an array of one column I am counting the rows and then looping through that number with a for() loop and then using the $array[] = $value to set the incrementing array object's value.

This works great if none of the values in a cell are 0. The class returns me a number 0 so it's nothing to do with the class, I think it's the way I am looping through the rows and then assigning them to the array... I want to carry through the 0 value because I am creating graphs with the data afterwards, here is the code I have.

// Get Rainfall
$rainfall = array();
for($i=1;$i<=$count;$i++)
{
    if($data->val($i,2) != 'Rainfall') // Check if not the column title
    {
        $rainfall[] = $data->val($i,2);
    }
}

For your info $data is the excel spreadsheet object and the method $data->val(row,col) is what returns me the value. In this case I am getting data from column 2.

Screenshot of spreadsheet

Try using var_dump() and print_r() on your data array. Those nifty functions are very helpful for debugging! — John Himmelman
– John Himmelman, Commented May 28, 2010 at 13:41
@John I have done so, which is why i am confused. I have 30 rows in the excel file and when i do a print_r on the array I only get the values that are not 0.. so it's as if it disregards 0 — tarnfeld
– tarnfeld, Commented May 28, 2010 at 13:44

Philippe · Accepted Answer · 2010-05-28 13:46:24Z

4

Did you try an array_push() ?

array_push($rainfall, $data->val($i,2));

answered May 28, 2010 at 13:46

Philippe

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1 Comment

BMBM Over a year ago

I am really surprised here. array_push() works but $array[] = $var; does not? Actually, array_push() for a single variable is nothing else then $array[] = $var; I thought?

Mark Byers · Accepted Answer · 2010-05-28 14:48:21Z

3

I would use a strict comparison with the not identical operator here instead of using the not equals operator:

if($data->val($i,2) !== 'Rainfall')

If $data->val($i,2) is an integer and you use == both sides will be cast to integers which would give you the result that all integers would work as you expect except for zero. Here's a summary of the difference between == and === when comparing the string "RainFall" with zero:

0 == "RainFall" : true
0 != "RainFall" : false
0 === "RainFall" : false
0 !== "RainFall" : true

edited May 28, 2010 at 14:48

answered May 28, 2010 at 13:42

Mark Byers

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1 Comment

Tommy Over a year ago

Doesn't PHP get goofy with 0s and comparisons, meaning it treats a 0 as false instead of an integer in certain situations? I think I saw that somewhere before...

Tommy · Accepted Answer · 2010-05-28 13:50:17Z

1

I think that the array is treating the 0 like false, which could explain it not going into the array. Would something like this work (if you are using integers)?

(int)($data->val($i,2));

or

(float)($data->val($i,2);)

answered May 28, 2010 at 13:50

Tommy

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Comments

Erik Töyrä Silfverswärd · Accepted Answer · 2010-05-28 14:05:43Z

The problem lies in the if statement. You're trying to compare a string with an integer, which according to the PHP documentation will typecast both to integers.

If you compare a number with a string or the comparison involves numerical strings, then each string is converted to a number and the comparison performed numerically. These rules also apply to the switch statement. The type conversion does not take place when the comparison is === or !== as this involves comparing the type as well as the value.

You can read more here http://php.net/manual/en/language.operators.comparison.php.

Update: The if statement won't work in the case of 0 because (int)"Rainfall" will by typecasted into 0 by PHP causing the statement to be if (0 != 0) { ... }.

If $i represents the row number, why don't you start from 2 instead of 1?

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