hello i'm beginner for programming I've got a homework. googled it but couldnt find anything...
i need to get the total value of numbers from 1 to 10. this need to be done in loop. but couldn't figure which loop should i use. if you can also give me an example code thats would be great.
This is a homework question, I'm not sure why people are just giving you an answer to copy-paste.
Achieving the sum of numbers 1..10 is pretty simple. You will need to initialise an empty int var before your loop, and for each iteration from 0 up to and including 10 you will add your int var to the current iteration.
For example:
sum = 0;
for num in range 1 to 10:
sum = sum + num;
<?php
$start = 0; // set the variable that will hold our total
for($i=1;$i<11;$i++){ // set a loop, read here: http://php.net/manual/en/control-structures.for.php for more info
$start += $i; // add $i to our start value
}
echo $start; // display our final value
I would use a for loop.
$total = 0;
for($i = 1; $i <= 10; $i++){
$total += $i;
}
Using the for loop:
<?php
$sum = 0;
for($i = 1; $i <= 10; $i++){
$sum += $i;
}
Using the foreach loop:
<?php
$sum = 0;
foreach(range(1,10) as $num){
$sum += $num;
}
echo $sum; // prints 55
And disregarding your assignment, here is an easier way:
echo array_sum(range(1,10));
Related
Recently, my exams got over. My last exam was based on PHP. I got the following question for my exam:
"Convert the following script using for loop without affecting the output:-"
<?php
//Convert into for loop
$x = 0;
$count = 10;
do
{
echo ($count. "<BR>");
$count = $count - 2;
$x = $x + $count;
}
while($count < 1)
echo ($x);
?>
Please help me as my computer sir is out of station and I am really puzzled by it.
Well, If I understand well, You have to use "for" loop, instead of "do...while", but the printed text must not change.
Try:
$count = 10;
$x = 0;
$firstRun = true;
for(; $count > 1 || $firstRun;) {
$firstRun = false;
echo ($count . "<BR>");
$count -= 2;
$x = $x + $count;
}
echo ($x);
By the way loop is unnecessary, because $count will be greater than 1 after the first loop, so the while will get false.
EDIT
$firstRun to avoid infiniteLoop
$count in loop
EDIT
Fixed code for new requirement
Removed unnecessary code
Hmmm I don't know if your teacher wanted to own you... but the do{} will execute only once since $count is never < 1.
The output of your teacher's code is:
10
8
I presume there was a mistake in the code and the while would be while($count > 1) which would make more sense (since it's weird to ask for a loop to output only 10 8) and would result in this output:
10
8
6
4
2
20
Then a good for() loop would have been :
$x = 0;
$count = 10;
for($i = $count; $i > 1; $i -= 2)
{
$count -= 2;
echo $i . "<br>";
$x += $count;
}
echo $x;
Which will output the same values.
If you can, ask your teacher for this, and comment the answer ^^ ahahah
I need to find the averege after using a loop counting ever 3rd to a 100. The loop part is easy enough, but I need to sum every value then divide the sum on the total of values.
for ($x = 3; $x < 100; $x+=3) {
echo $x.", ";
}
This is the loop I need to use. How to I sum the values this produces and how do I find how many values this loop produces?
I believe the intention here is to learn about loops, otherwise this stuff can be done without looping too.
For learning purpose, you can simply introduce two variables count and sum and compute them inside the loop. For count, you just increment it on each iteration. For sum, you add the current value of x into sum. After the loop you print both variables.
$count = 0;
$sum = 0;
for ($x = 3; $x < 100; $x+=3) {
echo $x.", ";
$count++;
$sum+=$x;
}
echo $sum;
echo $count;
add your elements into an array and then use array_sum to sum the array elements , then divide the sum by the count of your array
$arr = [];
for ($x = 3; $x < 100; $x+=3) {
// echo $x.", \n";
$arr[] = $x;
}
print_r(array_sum($arr) / count($arr));
// Output : 51
$i=0;
$tempx=0;
for ($x = 3; $x < 100; $x+=3) {
//total sum
$tempx = $tempx + $x;
//count of how many times the loop ran in this case 33 times
$i++;
}
//first $i was 0 so we add 1
$i=$i + 1;
//getting the average
$average=$tempx / $i;
echo $average;
//output
For the last answer i think we should not do:
//first $i was 0 so we add 1 $i=$i + 1;
Regards
I need to get the total value of this array of all the numbers above or equal to 0.
This is the array
$aReeks = array(23,245,1,2,12,-10,46,6,66,9999,-55,348,56,6,66,983);
This is the code i have so far, but it only shows the highest number of the array and doesnt count the values up and shows the total.
$totaal = 0;
for($y=0; $y < count($aReeks); $y++)
{
if($totaal < $aReeks[$y] && $aReeks[$y] > 0)
$totaal = $aReeks[$y];
}
I have to do it with a for loop.
Here's a quick way:
$total = array_sum(array_filter($aReeks, function($n) { return $n > 0; }));
Filter the array for values greater than 0
Sum that array
Oh I now see the "I have to do it with a for loop.", so this won't be accepted for your homework I imagine.
$aReeks = array(23,245,1,2,12,-10,46,6,66,9999,-55,348,56,6,66,983);
$total=0;
for($i =0 ; $i< count($aReeks) ; $i++)
{
if($aReeks[$i]>=0)
{
$total+= $aReeks[$i];
}
}
echo $total ;
?>
Output
11859
You are making 2 major mistakes one is in if condition which is $totaal < $aReeks[$y] you don't need this check at all. Secondly rather than summing up the value of each item to the total of all previous items... you are simply assigning the value to the $totaal variable inside the loop.
$aReeks = array(23,245,1,2,12,-10,46,6,66,9999,-55,348,56,6,66,983);
$totaal = 0;
for($y=0; $y < count($aReeks); $y++)
{
if($aReeks[$y] > 0)
$totaal = $totaal + $aReeks[$y];
}
I have the PHP code below:
<?php
$length = $_GET["length"];
$maxValue = $_GET["maxValue"];
$distribution = array();
for($j = 0; $j < $maxValue; $j++) {
$distribution[j] = 5;
}
$x = 0;
$x++;
for($j = 0; $j < $maxValue; $j++) {
echo $distribution[j] , " ";
}
echo $x;
?>
$x starts as 0 and is incremented by 1. However, just below $x is incremented, I am also incrementing the first element of the "distribution" array - $distribution[0]. And it's not working. It worked fine when I was initializing the elements (set them to 5).
Any ideas on why it might now be working? I am probably referencing the array element wrong. But this seems inconsistent.
When you say $distribution[j] -> php doesn't understand the j as a variable - but rather as an undefined constant
It looks like you are trying to say $distribution[$j] - which is partially -why your increments aren't working - -
The other reason would be that you aren't ever calling $distribution[$j]++ --- so there is no incrementation happening...
I am having some problems with PHP.
I used while to sum a number's digits always that it has more than two digits, some how, it gets into an infinity loop.
e.g: 56 = 5 + 6 = 11 = 1+1= 2.
Here is the code:
$somaP = 0;
$numPer = (string)$numPer; //$numPer = number calculated previously
while (strlen($numPer) > 1){
for ($j = 0; $j < strlen($numPer); $j++){
$somaP = $somaP + (int)($numPer[$j]);
}
$numPer = (string) $somaP;
}
Can anyone help me? Guess it is a simple mistake, but I couldn't fix it.
You need to reset the value of $somaP in your while loop.
Currently it continues to increase its value every time through the loop.
Try this:
$numPer = (string)$numPer; //$numPer = number calculated previously
while (strlen($numPer) > 1){
$somaP = 0;
for ($j = 0; $j < strlen($numPer); $j++){
$somaP = $somaP + (int)($numPer[$j]);
}
$numPer = (string) $somaP;
}
Take a look at this line:
$numPer = (string) $somaP;
It seems that the length of $somaP is never lesser (or equal) than 1. So the length of $numPer is never lesser (or equal) than 1.
What are you trying to do?
It's unclear to me.
This for example would add every number in a string together?
E.g "1234" = 1+2+3+4 = 10
$total = 0;
for($i = 0; i < strlen($string); $i++){
$total += $string[$i];
}
echo $total;
This looks cleaner I would say:
$numPer = 56;
while ($numPer > 9){
$numPer = array_sum(str_split($numPer));
}
echo $numPer;
PHP handles all string <> number conversions for you, so no need to do (string) on a number unless really needed.