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 $data->val(row,col) is what returns me the value. In this case I am getting data from column 2.
Screenshot of spreadsheet
Did you try an array_push() ?
array_push($rainfall, $data->val($i,2));
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
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);)
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?
Related
Why below code is printing the "here", it should be "there"
$a = "171E-10314";
if($a == '0')
{
echo "here";
}
else
{
echo "there";
}
PHP automatically parses anything that's a number or an integer inside a string to an integer. "171E-10314" is another way of telling PHP to calculate 171 * 10 ^ -10314, which equates to (almost) 0. So when you do $a == '0', the '0' will equate to 0 according to PHP, and it returns true.
If you want to compare strings, use the strcmp function or use the strict comparator ===.
when you use the == comparison, PHP tries to cast to different data types to check for a match like that
in your case:
'0' becomes 0
"171E-10314" is still mathematically almost 0 but I think PHP just rounds it to 0 due to the resolution of float.
check this link for a visual representation:
http://www.wolframalpha.com/input/?i=171E-10314
As per the answers given I tried to convert the string into numerical:
$a = "171E-10314" + 0;
print $a;
And I got output as 0.
Thats why it is printing here.
Use === /*(this is for the 30 character)*/
I was trying to understand the in_array behavior at the next scenario:
$arr = array(2 => 'Bye', 52, 77, 3 => 'Hey');
var_dump(in_array(0, $arr));
The returned value of the in_array() is boolean true. As you can see there is no value equal to 0, so if can some one please help me understand why does the function return true?
This is a known issue, per the comments in the documentation. Consider the following examples:
in_array(0, array(42)); // FALSE
in_array(0, array('42')); // FALSE
in_array(0, array('Foo')); // TRUE
To avoid this, provide the third paramter, true, placing the comparison in strict mode which will not only compare values, but types as well:
var_dump(in_array(0, $arr, true));
Other work-arounds exist that don't necessitate every check being placed in strict-mode:
in_array($value, $my_array, empty($value) && $value !== '0');
But Why?
The reason behind all of this is likely string-to-number conversions. If we attempt to get a number from "Bye", we are given 0, which is the value we're asking to look-up.
echo intval("Bye"); // 0
To confirm this, we can use array_search to find the key that is associated with the matching value:
$arr = array(2 => 'Bye', 52, 77, 3 => 'Hey');
echo array_search(0, $arr);
In this, the returned key is 2, meaning 0 is being found in the conversion of Bye to an integer.
Try adding a third parameter true (strict mode) to your in_array call.
This is a result of the loose comparison and type juggling.
Loose comparison means that PHP is using == not === when comparing elements. == does not compare that the two variable types are equal, only their value, while === will ensure that they match in type and value (e.g. compare 0 == FALSE and 0 === FALSE).
So, basically, your in_array function is checking:
0 == 'Bye'
0 == 'Hey'
0 == 77
Note that the 52 will get lost due to the way you created your array.
So, note that if you do:
print (0 == 'Bye');
You will get 1. Apparently, PHP is type juggling the 'Bye' to 0, which is the same thing that will happen when you cast a string to an int, e.g. (int) 'string' will equal 0. Specific reference from the String conversion to numbers doc (Emphasis added):
The value is given by the initial portion of the string. If the string
starts with valid numeric data, this will be the value used.
Otherwise, the value will be 0 (zero). Valid numeric data is an
optional sign, followed by one or more digits (optionally containing a
decimal point), followed by an optional exponent. The exponent is an
'e' or 'E' followed by one or more digits.
Apparently, the integer type takes precedence over the string type (i.e. it could just as easily be doing the comparison by casting the int 0 to a string, which would then return False). This is specified in the comparison operators doc:
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.
Thanks for an interesting question that led me to do some research and learn something new!
in_array is supposed to be used on indexed arrays ([0], [1], [2] etc), not with a dictionary as you have defined (key-value store).
If you want to check if your array $arr includes '0', try using the PHP function array_key_exists instead - http://php.net/manual/en/function.array-key-exists.php.
var_dump(array_key_exists(0, $arr));
Type compare (third parameter) needs more system resources and more time.
Simply do that:
$x=0;
$result=in_array($x.'',array('ABC','BAC','12c','54'));
var_dump($result);
The array resAlloc contains 10 columns and 5 rows. All entries are equal to 0. So, I expect the following IF statement be TRUE, but it's false for some reason... Why?
if ($resAlloc[$i][$j] != 'ts' && $resAlloc[$i][$j] != 't' && $resAlloc[$i][$j] != 'st') {
$count++;
}
!= evaluates 0 as false. Use !== which is more strict.
IIRC anything equaling 0 with != will return FALSE.
The problem is that you're comparing a string and an integer, and PHP is "helpfully" casting the string to an integer -- the integer zero. 0!='ts' evaluates as false, because the comparison it ends up doing after conversion is 0!=0. You can prevent this by explicitly treating the contents of your array as a string:
strval($resAlloc[$i][$j]) != 'ts'
This will do the comparison '0'!='ts', which correctly evaluates to true. If you pass strval() a string it returns it unchanged, so this should be safe to use regardless of what's in your array.
Alternately, as Samy Dindane said, you can just use !== which won't do any type conversion.
I had an if statement similar to the following in my code and it took me forever to figure out what the problem was.
$a = 0;
if($a == 'something')
{
//this was being output when I didn't want it to be
}
Using
$a = '0';
fixed it, but I don't really know what's going on here.
One's a string, one's an integer. PHP will translate between the two as needed, unless you're using the 'strict' operators:
(0 == '0') // true
(0 === '0') // false (types don't match).
In your case, you'r comparing an integer 0 to a string 'something'. PHP will convert the string 'something' to an integer. If there's no digits in there at all, it'll conver to an integer 0, which makes your comparison true.
Just a guess, but I assume it's trying to cast the string to an integer.
intval('something') I expect will return 0.
You were comparing a numeric value ($a = 0;) to a string. In this case the string is casted to number, and PHP cast strings to 0 if there is no number in the beginning, so is true.
In the other case however you campared two strings, which are different, so it is false.
Here is some code I have: (p just echos plus adds a newline)
foreach ($vanSteps as $k => $reqInfo)
{
p($k);
if ('van' == $k) { p('The key is the van, continue'); continue; }//continue if we reached the part of the array where van is key
//do stuff
}
and I'm getting this output:
0
The key is the van, continue
1
2
3
van
The key is the van, continue
Why does the if statement return true when the key is 0? This foreach loop handles logic that applies when the key == 0 (and any other key except if the key is 'van') and this messes up the logic because it's return true when key is 0.
Any help?
Thank you.
Use === for this comparison. When PHP compares string and integer it first casts string to integer value and then does comparison.
See Comparison Operators in manual.
In PHP 'van' == 0 is true. This is because when using == to compare a string and a number, the string is converted to a number (as described in the second link below); this makes the comparison internally become 0 == 0 which is of course true.
The suggested alternative for your needs, would be to use a strict equality comparison using ===.
See Comparison Operators and String conversion to numbers
In PHP, when you compare 2 types, it has to convert them to the same type. In your case, you compare string with int.
Internally this gets converted to
if((int)'van'==0)....
and then
if((int)'van'==1)....
(int)'any possible string' will be 0:) So you either have to manually convert the both values to the same type, or use === as a comparison operator, instead of the loose =.
An exception from this rule(as pointed out in the comments) would be if the string start with a number, or can be interpreted as a number in any way(1, 0002, -1 etc). In this case, the string would be interpreted as a number, diregarding the end of the non-numeric end-of-string
Take a look at http://php.net/manual/en/types.comparisons.php for more details.
This works fine:
$array = array(0=>"a",1=>"b","van"=>"booya!");
function p($v){ echo "{$v}<br />"; }
foreach ($array as $k => $reqInfo)
{
p($k);
if ('van' === $k) { p('The key is the van, continue'); continue; }//continue if we reached the part of the array where van is key
//do stuff
}
Output:
0
1
van
The key is the van, continue
Note the ===.
Read the Comparison with Various Types table
When one of the operand is number, the other operand is converted to number too. Since 'van' is non-numeric sting, it's converted to 0. You should use === operator in the case, which also checks the variable type
That's becuase 'van' == 0 (true).
Instead, you should use 'van' === 0 (false).
In short, use === instead of ==.
Its interpreting the 'van' as a boolean value (false) which 0 is equal to.
To check for exact matches in type and value in PHP you must use === instead of ==