I just want to ask if there's a function in excel/phpexcel that can SUM a range of
cells but skip to add every one cell.
Ex. Range is A1:G1 and it will only sum A1+C1+E1+G1 skipping every 1 cell.
Or how can I do that? Please take note that range is dynamic. It could be a1:g1 or further or less.
Build the formula dynamically in your PHP script:
$startCell = 'A';
$endCell = 'H';
$row = 1;
$formula = '=' . $startCell . $row;
while ($startCell++ != $endCell && $startCell++ != $endCell) {
$formula .= '+' . $startCell . $row;
}
EDIT
Note that I've also added a pure Excel formula answer to your previous question
Related
I am struggling to find a way to increment a specific pattern required. For each new member, they are given a unique ID such as ABC000001. Each new members ID should increment by one. So the second member's ID would be ABC000002. I am using PHP and MySQL to keep track of each member but I have not been able to come up with a way to properly increment using the string format above.
What is the best way to approach this?
As #axiac mentions this is probably not a good idea but it's pretty easy to manage.
$memberid = 'ABC000001';
list($mem_prefix,$mem_num) = sscanf($memberid,"%[A-Za-z]%[0-9]");
echo $mem_prefix . str_pad($mem_num + 1,6,'0',STR_PAD_LEFT);
Split your current member number into the alpha and numeric parts then put them back together bumping the number when you do it. I use this as a function and pass the previous ID and what I get back is the next ID in the sequence.
You can extract only digits using regex to increment and using str_pad for create a prefix :
$memberid = 'ABC000001';
$num = preg_replace('/\D/', '',$memberid);
echo sprintf('ABC%s', str_pad($num + 1, "6", "0", STR_PAD_LEFT));
Possible answer without regex.
Runs through each character and checks if it is a number or not.
Then uses sprintf() to make sure leading 0s are still there.
$str = "ABC000001";
$number = "";
$prefix = "";
$strArray = str_split($str);
foreach ($strArray as $char) {
if (is_numeric($char)) {
$number .= $char;
} else {
$prefix .= $char;
}
}
$length = strlen($number);
$number = sprintf('%0' . $length . 'd', $number + 1);
echo $prefix . $number;
This works for this instance but would not work if the prefix had numbers in it.
just use the PHP increment operator ++
as long as you've sufficient leading zeros in the pattern, then PHP will correctly increment the numeric component
This code:
<?php
$name = 'ABC0000098';
print ++$name . PHP_EOL;
print ++$name . PHP_EOL;
print ++$name . PHP_EOL;
Outputs:
ABC0000099
ABC0000100
ABC0000101
Read more in the PHP docs:
https://www.php.net/manual/en/language.operators.increment.php
aha - just saw there is already a link to a similar example in the comment posted by "axiac" above
(Working on a search algorithm) I want to iterate over possible matches with two bits set in a 16bit word. Seems like a silly problem with a currently overly-complex solution.
Iteration should return (decimal) 3,5,6,9,10,12,17...
What's the proper word for the problem? Bit-mask-looping?
Any clever function for this?
Current code - now updated:
(As it stands, i guess there's no easier way around this.)
<?php
function biterate($numBits=8, $setBits=2, $maxval=null) {
//init
if(is_null($maxval)) $maxval = (pow(2,$setBits)-1) * pow(2,$numBits - $setBits);
$err = 0;
header('content-type:text/plain');
echo '-- ' . $setBits . ' of ' . $numBits . " --\r\n";
$result = str_pad('', $numBits - $setBits, '0') . str_pad('', $setBits, '1');
do {
$err++;
if($err > 200) die('bad code');
//echo and calc next val.
echo $result . ' : ' . bindec($result) . "\r\n";
//count set bits and search for '01' to be replaced with '10'. From LSB.
$bitDivend = '';
$hit = false;
for($i=$numBits;$i>0;$i--) {
if(substr($result,$i-2,2) == '01') {
$hit = true;
//do the replacement and replace the lower part with bitDivend.
$result = substr($result, 0, $i-2) . '10';
$result .= str_pad('',$numBits - $i - strlen($bitDivend),'0');
$result .= $bitDivend;
//exit loop
$i = 0;
}
if($result[$i-1] == '1') $bitDivend .= '1';
}
} while($hit && bindec($result) <= $maxval);
}
biterate(8,2);
biterate(8,7);
biterate();
If you just want all the 16 bit ints with 2 bits set, the following code should do it:
<?php
for($i=1;$i<16;$i++)
{
for($j=0;$j<$i;$j++)
{
echo (1<<$i)|(1<<$j) , "\r\n";
}
}
?>
If you look at the bit patterns of the numbers you can see how it works:
11 3
101 5
110 6
1001 9
1010 10
1100 12
10001 17
10010 18
10100 20
11000 24
etc. You just move the most significant bit one place to the left (another power of 2) for each iteration of the outer loop, and inside the inner loop you iterate from the least significant bit (1) to 1 place to the right of the most significant bit.
If you wanted to generalise this to support an arbitrary number of bits and places, you could extend the above algorithm using recursion:
<?php
function biterate_recursive($numBits=8, $setBits=2, $initialValue=0, $maxval=null) {
for($i=$setBits-1;$i<$numBits;$i++)
{
if(!is_null($maxval) && ($initialValue|(1<<$i)) > $maxval)
break;
if($setBits==1)
echo $initialValue|(1<<$i) , "\r\n";
else
biterate_recursive($i, $setBits-1, $initialValue|(1<<$i), $maxval);
}
}
biterate_recursive(16, 2);
?>
You can also think of the problem as just choosing combinations i.e. C(16,2) choosing 2 numbers a,b from the set 0-15, and then calculating (1<<a)|(1<<b). However you have to be careful about your choice of combination algorithm if you want to get the numbers in order.
I have written a little script to automate some calculations for me. It is pretty simple.
1*3=3
2*3=6
3*3=9 and so on. When the answer(product) is more than one digit long I want it to add the digits of the answer.
3*4=12 || 1+2=3.
I want it to automatically add the answer if it is more than one digit no matter how many times adding the answer arrives larger than a single digit.
as in when you reach 13*3=39 || 3+9=12 || 1+2=3
Currently my code does not loop, and i cannot figure out how to make it loop here.
http://screencast.com/t/MdpQ9XEz
Also, it is not adding up more than 2 digit answers see 34*3=102. any help here to allow it infinity addition would be great.
As each line is produced the answers get larger, so it should add as many digits there are in the answer.
here is my code:
$i = 1; //Start with one
function count_digit($number) {
return strlen((string) $number);
};
while($i < 500){
$product = $i * 3; //Multiply set
$number_of_digits = count_digit($product); //calls function to count digits of $sum
if($number_of_digits > 1){ //if $sum has more than one digit add the digits together
$addproduct = strval($product);//seperates the digits of $sum
$ii = 0;
while($ii <= $number_of_digits -1){
$io = $ii + 1;
if($io < $number_of_digits){
$one = intval($addproduct[$ii]);
$two = intval($addproduct[$io]);
$sum = $one + $two;
print($i . ' * 3 = ' .$product. ' || ' .$one. ' + ' .$two. ' = ' .$sum. '<br>');
};$ii++;
};
}else{
Print ($i . ' * 3 = ' .$product.'<br>'); //Print Set
};
$i++; //add one
}
For a start, you might replace the $i=1, while ..., and $i++ into one statement: for ($i=0; $i<500; $i++) {.... - see http://nz1.php.net/manual/en/control-structures.for.php for more information about for loops. It comes out to effectively the same code, but keeps the three parts (initialisation, test, increment counter) together; it makes it clear that they are related, and quicker to understand.
Next I would replace conversion of the number to a string, seeing if the length of the string is greater than one, etc.... with:
while ($product>10) {
$out=0;
while ($product) {
$out+=$product%10;
$product=(integer)($product/10);
}
$product=$out;
}
Note that at no time am I treating a number as a string - I avoid that wherever possible.
In case you are not familiar with it, $something+=$somethingelse is the same as $something=$sometime+$somethingelse, just a shorthand. $out is the running total, and each time around the inner loop (while $product), $out is increased by the rightmost digit (the ones column): $out+=$product%10 - see http://nz2.php.net/operators.arithmetic for more info, then $product is divided by 10 and converted to an integer, so 12 becomes 1.2, then 1, dropping the right-most digit.
Each time the inner loop is complete (all the digits of $product are used up, and $product is 0), the result (in $out) is copied to $product, and then you get back to the outer loop (to see if the sum of all these digits is now less than 10).
Also important is exactly where in each loop you print what. The first part, with the multiplication, is immediately in the '500' loop; the '||' happens once for each time '$product' is reduced to '$out'; the adding of the digits happens inside the innermost loop, with '+' either BEFORE each digit except the first, or AFTER each digit except the last. Since 'all except the last' is easier to calculate ($product>=10, watch those edge cases!), I chose that one.
Also note that, since I'm adding the digits from the right to the left, rather than left-to-right, the addition is reversed. I do not know if that will be an issue. Always consider if reversing the order will save you a lot of work. Obviously, if they NEED to be done in a particular order, then that might be a problem.
The result is:
for ($i=0; $i<500; $i++) {
$product = $i * 3; //Multiply set
print ($i . ' * 3 = ' .$product);
while ($product>10) {
print ' || ';
$out=0;
while ($product>0) {
print ($product%10);
if ($product>=10) {
print " + ";
}
$out+=$product%10;
$product=(integer)($product/10);
}
$product=$out;
print(' = ' .$product);
}
print "<br>";
};
I've decided to add this as a separate answer, because it uses a different approach to answer your question: the least amount of changes to fix the problem, rather than rewriting the code in an easy-to-read way.
There are two problems with your code, which require two separate fixes.
Problem 1: Your code does not add 3 digit (or more) numbers correctly.
If you look carefully at your code, you will notice that you are adding pairs of numbers ($one and $two); each time you add a pair of numbers, you print it out. This means that for the number 102, you are adding 1 and 0, printing out the results, then adding 0 and 2 and printing it out. Look carefully at your results, and you will see that 102 appears twice!
One way to fix this is to use $three and $four (4 digits is enough for 500*3). This is not good programming, because if you later need 5 digits, you will need to and $five, and so forth.
The way to program this is to start with a $sum of zero, then go through each digit, and add it to the sum, thusly:
$ii = 0;
$sum = 0;
while($ii <= $number_of_digits - 1){
$one = intval($addproduct[$ii]);
$sum = $sum + $one;
$ii++;
}
print($i . ' * 3 = ' .$product. ' || ' . $sum . '<br>');
};
The inner if disappears, of course, since it relates to $io, which we are no longer using.
Of course, now you lose your ability to print the individual digits as you are adding them up. If you want those in, and since we don't know in advance how many there are, you would have to keep a record of them. A string would be suitable for that. Here is how I would do it:
$ii = 0;
$sum = 0;
$maths = "";
while($ii <= $number_of_digits-1){
$one = intval($addproduct[$ii]);
if ($maths=="") {
$maths=$one;
} else {
$maths=$maths . " + " .$one;
}
$sum = $sum + $one;
$ii++;
}
print($i . ' * 3 = ' .$product. ' || ' . $maths . " = " . $sum . '<br>');
};
Note the lines with $math in them; there's one at the top, setting it to an empty string. Then, once we've decided on what number we're adding, we add this to $maths. The first time round is a special case, since we don't want "+" before our first digit, so we make an exception - if $maths is empty, we set it to $one; later times around the loop, it's not empty, so we add " + " . $one;
Problem 2: if the result of the addition is more than 1 digit long, go around again. And again, as many times as required.
At this point I would suggest a little refactoring: I would break the print statement into several parts. First the multiplication - that can easily be moved to the top, immediately after the actual multiplication. Printing the <br> can go at the end, immediately before the $i++. This also means you no longer need the else clause - it's already done. This only leaves the ' || ' . $maths . " = " . $sum. This is going to have to happen every time you go through the adding.
Now, replace your if($number_of_digits... with a while($number_of_digits...; this means it goes through as many times as is needed, if it's zero or 100. This does mean that you're going to have to update $product at the end of your loop, with $product = $sum; - put this immediately after the print statement. You will also need to update $number_of_digits; copy the line from the top to the end of the loop.
If you've been following along, you should have:
$i = 1; //Start with one
function count_digit($number) {
return strlen((string) $number);
};
while($i < 500){
$product = $i * 3; //Multiply set
print($i . ' * 3 = ' .$product); // Print basic multiplication
$number_of_digits = count_digit($product); //calls function to count digits of $sum
while ($number_of_digits > 1){ //if $sum has more than one digit add the digits together
$addproduct = strval($product);//seperates the digits of $sum
$ii = 0;
$sum = 0;
$maths = "";
while($ii <= $number_of_digits-1){
$one = intval($addproduct[$ii]);
if ($maths=="") {
$maths=$one;
} else {
$maths=$maths . " + " .$one;
}
$sum = $sum + $one;
$ii++;
}
print(" || " . $maths . " = " . $sum ); // print one lot of adding
$product=$sum;
$number_of_digits = count_digit($product); //calls function to count digits of $sum
};
Print ("<br>"); //go to next line
$i++; //add one
}
I'm using PHP Excel to create an Excel using a template Excel file. The problem is I have a datagrid and which I styled the header and first row in template. Here how it looks like:
The top leftmost coordinate is C49.
If I have 100 rows, I need to copy style of first row and paste it 100 times.
Here is my code
$cstart = 2;
$rstart = 49;
$count = 1;
$input = $worksheet->getStyle(num2char($cstart) . $rstart);
foreach ($b_data['rawData'] as $value) {
$worksheet->setCellValueByColumnAndRow($cstart, $rstart, $count++)
->setCellValueByColumnAndRow($cstart + 1, $rstart, $value['key'])
->setCellValueByColumnAndRow($cstart + 5, $rstart, $value['value']);
$interval = num2char($cstart) . $rstart . ':' . num2char($cstart+5) . $rstart;
$worksheet->duplicateStyle($input, $interval);
$rstart++;
}
function num2char($num) {
$numeric = $num % 26;
$letter = chr(65 + $numeric);
$num2 = intval($num / 26);
if ($num2 > 0) {
return num2char($num2 - 1) . $letter;
} else {
return $letter;
}
}
However, I had the following:
but what I expected is:
Is it bug or am I doing something wrong?
As Mark pointed out in the comments; a merged cell is structural, not style. So copying the style will not automatically copy the merged cells.
There is a feature request to be able to duplicate entire rows including merged cells
One way to deal with this is to check if the cells are part of a merge using the isInMergeRange() function like this:
$workbook = new PHPExcel; // prepare the workbook
$sheet = $workbook->getActiveSheet(); // get the sheet
$sheet->mergeCells('A1:E1'); // merge some cells for tesing
$cell = $sheet->getCell('A1'); // get a cell to check if it is merged
$cell->isInMergeRange() // check if cell is merged
^This returns a Boolean value indicating if it is part of a merged cell.
Another function that may interest you is the isMergeRangeValueCell() function:
$cell->isMergeRangeValueCell()
^This returns a Boolean value indicating it is part of a merged cell and the cell contains the value for the merged range; it returns false in all other situations.
I have system in PHP in which I have to insert a Number which has to like
PO_ACC_00001,PO_ACC_00002,PO_ACC_00003.PO_ACC_00004 and so on
this will be inserted in Database for further reference also "PO and ACC" are dynamic prefix they could different as per requirement
Now my main concern is how can is increment the series 00001 and mantain the 5 digit series in the number?
>> $a = "PO_ACC_00001";
>> echo ++$a;
'PO_ACC_00002'
You can get the number from the string with a simple regex, then you have a simple integer.
After incrementing the number, you can easily format it with something like
$cucc=sprintf('PO_ACC_%05d', $number);
Create a helper function and a bit or error checking.
/**
* Takes in parameter of format PO_ACC_XXXXX (where XXXXX is a 5
* digit integer) and increment it by one
* #param string $po
* #return string
*/
function increment($po)
{
if (strlen($po) != 12 || substr($po, 0, 7) != 'PO_ACC_')
return 'Incorrect format error: ' . $po;
$num = substr($po, -5);
// strip leading zero
$num = ltrim($num,'0');
if (!is_numeric($num))
return 'Incorrect format error. Last 5 digits need to be an integer: ' . $po;
return ++$po;
}
echo increment('PO_ACC_00999');
Sprintf is very useful in situations like this, so I'd recommend reading more about it in the documentation.
<?php
$num_of_ids = 10000; //Number of "ids" to generate.
$i = 0; //Loop counter.
$n = 0; //"id" number piece.
$l = "PO_ACC_"; //"id" letter piece.
while ($i <= $num_of_ids) {
$id = $l . sprintf("%05d", $n); //Create "id". Sprintf pads the number to make it 4 digits.
echo $id . "<br>"; //Print out the id.
$i++; $n++; //Letters can be incremented the same as numbers.
}
?>