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I have a table feedback having columns ans1,ans2,ans3 ...., each column has values from 0 to 2. I want to count all columns where value is 1
In frontend I have displayed 0 as Poor, 1 as Good and 2 as Excellent. I want to count total number of Poor Feedback, Good Feedback and Excellent Feedback. for example if the table have two 0s in ans1, three 0s in ans2 so total should be five 0s.
One option, using summation of boolean expressions:
SELECT
SUM((ans1 = 0) + (ans2 = 0) + ... + (ansN = 0)) AS poor_count,
SUM((ans1 = 1) + (ans2 = 1) + ... + (ansN = 1)) AS good_count,
SUM((ans1 = 2) + (ans2 = 2) + ... + (ansN = 2)) AS excellent_count
FROM feedback;
maybe double looping can solve it,
$feedback = $this->db->get('feedback'); //get your data table
$poor = 0; $good = 0; excel=0;
foreach($feedback as $data1){
foreach($data1 as $data2){
if($data2 == 0){
$poor++;
}
if($data2 == 1){
$good++;
}
if($data2 ==2){
$excel++;
}
}
Related
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I want to have an php check for easy passwords patterns like
123456
123456789
abcde
98764321
101010
202020
Is it possible to have a check like this that doesn't rely in maintaining an array of pre-defined strings ?
Just iterate to each letter and check if the next one is same, one less or one more as current. So you can create a score of uniqueness.
For last case you can check if there are duplicate characters and remove from score, too.
$passwords = ['123456', '123456789', 'abcde', '98764321', '101010', '202020', 'xhdgszu'];
foreach ($passwords as $password) {
$score = $length = strlen($password);
$chars = str_split($password);
for ($i = 0; $i < $length - 1; $i++) {
$currentChar = $chars[$i];
$nextChar = $chars[$i + 1];
if ($currentChar === $nextChar
|| ord($currentChar) - 1 === ord($nextChar)
|| ord($currentChar) + 1 === ord($nextChar)
) {
$score--;
continue;
}
}
$unique = array_unique($chars);
$score -= $length - count($unique);
echo "{$password}: {$score} / {$length} ", ($score < $length) ? 'Bad' : 'Good', PHP_EOL;
}
123456: 1 / 6 Bad
123456789: 1 / 9 Bad
abcde: 1 / 5 Bad
98764321: 2 / 8 Bad
101010: -3 / 6 Bad
202020: 2 / 6 Bad
xhdgszu: 7 / 7 Good
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When using increment on variables in Javascript, lint says that x += 1 is preferred over x ++.
But what about in PHP?
Is there any difference between the += and ++ or does it just not really matter?
Well whatever you might say about conventions, try running the following...
$i = 1; $s = 's';
$i++; $s++;
echo $i.'<br>'.$s.'<br>';
$i = 1; $s = 's';
$i += 1; $s += 1;
echo $i.'<br>'.$s.'<br>';
the output is somewhat unexpected...
2
t
2
1
so I would say it could matter very much which is chosen!
x += 1 is rather equivalent to ++x.
All those expressions (x += 1, x++ and ++x) increment the value of num by one, but the value of x++ is the value x had before it got incremented.
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This is my solution to Euler Project Problem 14
<?php
$count = 0 ;
$max = 0;
for($n = 2 ; $n < 1000000 ; $n++){
while ($n > 1)
{
if ($n % 2 == 0 )
{
$n = $n/2;
}
else
{
$n = 3*$n + 1 ;
}
$count += 1;
if($count > $max )
{
$max = $count;
$final = $n;
}
}
}
echo $final;
>?
It took so long to run.I looked some other solutions and they were very similar to my code logically,but they were running way too faster than mine.
My question is,what is it that makes my code inefficient? What am I missing here?
Thanks ^^
Your approach is straight forward and unpolished.
You can use dynamic programming in order to improve the runtime (complexity approach).
Let's say you want to compute for 5. You will have :
5 -> 16 -> 8 -> 4 -> 2 -> 1.
But if you do that you will also have computed the value for 8 and 16.
The idea is to store the value that you have already computed in order to save work when you will need them later.
Working on the complexity of an algorithm will be the key of some problems, so better get used to it early.
An other reason why it is slow, is the choice of the language. For example try with C it will run a lot faster.
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I have a simple form and a submit button. After pressing the submit button, get the values (they will be numbers) and calculate the sum. And again calculate the sum of individual digits stored in sum.
A better explanation: 1+2+3+4=10 where 1,2,3 and 4 are user inputs from the form. And the sum 10 has to be split-ted and summed again as 1+0=1. And this would be my final result. So far I did the task where it gives me the first sum. I don't know what to do to display the second result which I want to be the finale.
<?php
$a='';
$sum='';
$total='';
if (!isset($_POST['submit'])) {
echo " ";
}
else {
$a = array($_POST['textfield'],$_POST['textfield1'],$_POST['textfield2'],$_POST['textfield3'],$_POST['textfield4'],$_POST['textfield5'],$_POST['textfield6'],$_POST['textfield7'],$_POST['textfield8'],$_POST['textfield9'],$_POST['textfield10'],$_POST['textfield11'],);
for ($i=0; $i<=12; $i++) {
$sum= array_sum($a);
$total= ;
}
}
echo "sbora e: $total ";
?>
$total = array_sum(str_split($sum));
Using Artefacto's method.
On another note,
$a = array($_POST['textfield'],$_POST['textfield1'],$_POST['textfield2'],$_POST['textfield3'],$_POST['textfield4'],$_POST['textfield5'],$_POST['textfield6'],$_POST['textfield7'],$_POST['textfield8'],$_POST['textfield9'],$_POST['textfield10'],$_POST['textfield11'],);
can be written as,
$a = array();
for ($i = 0; $i <= 11; $i++){
if (isset($_POST["textfield_$i"]))
array_push($a, $_POST["textfield_$i"]);
}
if the names of the fields are:
textfield_0, textfield_1, textfield_2...
So you want to build the cross sum of your first result:
$result2 = 0;
//cast integer to string
$strTotal = (string) $total;
//loop over "chars" and add them to your second result
for($i=0; $i < strlen($strTotal); $i++)
{
$result2 += $strTotal{$i};
}
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I need to divide an integer value into 3 parts using php
for eg: I have a value 9 the answer is like 3 + 3 + 3
if i have value 10 the ans like 3 + 3 + 4 or something like that
You can use a modulus function.
<?php
$yourInt=10;
$remainder=$yourInt % 3;
$third=floor($yourInt/3);
$lastBit=$third+$remainder;
echo "The numbers are $third + $third + $lastBit.";
?>
Output:
The numbers are 3 + 3 + 4.
If you have a value that is >= 3 you can do the following:
$number = 10;
$number1and2 = floor($number/3);
$number3 = $number - (2*$number1and2);
$num = 11;
$d = [$num/3,$num/3,$num/3];
$round = array_map('round', $d);
list($first, $second, $third) = $round;
$round = (is_float($num/3)) ? $num-(round($num/3)*3) : 0;
echo $first.' '.$second.' '.($third += (is_float($num/3)) ? $num-(round($num/3)*3) : 0);