php select options in different steps - php
VERY basic question: I currently have the following (array?) for a select box within a form but I need to change the jump/step in numbers that appear in the drop down. At the moment it is in multiples of 100, from 300-1500 but I need multiples of 100 up to 1000 and then 1250 and 1500.
for($i = 300; $i <= 1500; $i += 100)
Have you tried using the function range() ?
You can configure start, end, and step.
http://php.net/manual/en/function.range.php
Just add the options from 100 to 1000 with a for loop and then add the other two options afterwards.
<select>
<?php
for($i = 100; $i <= 1000; $i+=100)
{
echo "<option>" . $i . "</option>";
}
?>
<option>1250</option>
<option>1500</option>
</select>
$step = 100;
for($i = 300; $i <= 1500; $i += $step){
if ($i >= 1000) {$step = 250}...
or
$inc = 100;
foreach (range(300, 1500, $inc) as $number)
{
echo "<Option>".$number."</Option>";
if ($number >= 1000){$inc = 250}
}
but i am not 100% sure if the second one works
Related
Can a PHP for loop with number array start with NULL and also have 0?
I am working with some legacy PHP code so re-writing this isn't an option at this point but I have a dropdown for number of years and months of employment and currently they go from 0 - 11, 0 - 65. Can a PHP loop array numbers starting at NULL, which adds -Select- as the default forcing user to make a selection, but also have 0 as the starting number?
I've tried:
for ($i = NULL; $i <= 11; $i++) {
echo $i;
}
But 0 is no loner an option
This is what I have currently:
for ($i = 0; $i <= 11; $i++) {
echo $i;
}
I need it to display as:
-Select-
0
1
2
3
4
etc
Echo the text before echoing the numbers using a loop.
<?php
echo '-Select-';
for($i = 0; $i <= 11; ++$i) {
echo $i;
}
NULL++ would not increase the value of NULL, which is essentially nothing. Why not start at -1, and if ($i == -1) then echo select?
It would look like this:
for ($i = -1; $i <= 11; $i++) {
if ($i == -1) {
echo '-Select-';
} else {
echo $i;
}
}
PHP Dubble For Loop
I need to make a loop in php that says 1*9 = 9 2*9 = 18 and 1*8 = 8 and 1*7 = 7 etc in loop so the result will be this:
1*7 = 7
2*7 = 14
3*7 = 21
1*8 = 8
2*8 = 16
3*8 = 24
1*9 = 9
2*9 = 18
3*9 = 27
And I need to use a nested loop for that i tried some things out but i cant get it to work here is what is did. But for the most part i don't really understand what nested loops do and what there used for. hope you can help me thanks!
for ($tafel7 = 0; $tafel7 <= 9; $tafel7++) {
for($tafel7 = 0; $tafel7 <= 9; $tafel7++){
for($tafel7 = 0; $tafel7 <= 9; $tafel7++){
$antwoord9 = $tafel7 * 9;
echo "$tafel7 x 8 = $antwoord9 <br>";
$antwoord8 = $tafel7 * 8;
echo "$tafel7 x 8 = $antwoord8 <br>";
$antwoord7 = $tafel7 * 7;
echo "$tafel7 x 7 = $antwoord7 <br>";
};
};
};
Your question is not as clear but if you just want the exact result, here's the code:
for($i = 7; $i <= 9; $i++){
for($j = 1; $j <= 3; $j++){
$result = $i * $j;
echo $j." * ".$i." = ".$result."<br>";
}
}
If you need to print complete tables, then try this:
for($i = 1; $i <= 9; $i++){
for($j = 1; $j <= 10; $j++){
$result = $i * $j;
echo $i." * ".$j." = ".$result."<br>";
}
}
As far as the explanation is concerned, then listen:
A normal loop works by executing a block of statement(s) again and again until a condition is met.
So, nested loops are basically loops in another loop. This means that the block of code will be executed 'n' number of times by the inner loop, and then the inner loop will again be executed by the outer loop, again 'n' number of times.
'n' means whatever the number of times the conditions are met for the respective for loops.
Hope you got my explanation!
You need to loop over your two variables, (1) the left operand and (2) the right operand
for ($right = 7; $right <= 9; $right++) {
for ($left = 1; $left <= 3; $left++) {
$product = $left * $right;
echo "{$left} x {$right} = {$product} <br>";
}
}
In this example, I have to loops one with the times table you want eg 7, 8 and 9 "tafel" and the other one with how many times you want it to run for each times table "$hoeveel". This can easily be altered to more times tables or how many times you want to run, you also only need to export the values once as it cycles through the loops anyways. Hope it helps.
for ($tafel = 7; $tafel <= 9; $tafel++) {
for ($hoeveel = 1; $hoeveel <= 3; $hoeveel++) {
$antwoord = $tafel * $hoeveel;
echo "$tafel x $hoeveel = $antwoord <br>";
}
}
Adding or condition in for loop
How can I add or in for loop check?
for example:
for ($i=1; ($i <= $untilPoint) or ($i <= $points); $i++){
.. code ...
}
Exactly what you typed is valid PHP. For example, this program:
<?
$untilPoint = 3;
$points = 5;
for ($i=1; ($i <= $untilPoint) or ($i <= $points); $i++){
echo("$i\n");
}
?>
prints this:
1
2
3
4
5
(tested in PHP 5.2.17, run from the Bash prompt).
That said, I wonder if maybe you really need and rather than or? If $i is an index into an array of points, where $points is the number of points and $untilPoint is an arbitrary cutoff, then you want and, not or.
This is how it should be done, probably it runs slightly faster.
You can separate with , see http://php.net/manual/en/control-structures.for.php for more information
<?
$untilPoint = 3;
$points = 5;
for ($i=1; $i <= $untilPoint, $i <= $points; $i++){
echo($i , "\n");
}
?>
how to find the sum of all the multiples of 3 or 5 below 1000 in php, issue?
i have an small issue with the way this problem is resolved.
some would say: println((0 /: ((0 until 1000).filter(x => x % 3 == 0 || x % 5 == 0))) (_+_)) is the solution witch adds to 233168
my way was to do:
$maxnumber = 1000;
for ($i = 3; $i < $maxnumber; $i += 3)
{
$t += $i;
echo $i.',';
}
echo '<br>';
for ($j = 5; $j < $maxnumber; $j += 5)
{
$d += $j;
echo $j.',';
}
echo '<br>';
echo $t;
echo '<br>';
echo $d;
echo '<br>';
echo $t+$d;
this will give me :
3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99,102,105,108,111,114,117,120,123,126,129,132,135,138,141,144,147,150,153,156,159,162,165,168,171,174,177,180,183,186,189,192,195,198,201,204,207,210,213,216,219,222,225,228,231,234,237,240,243,246,249,252,255,258,261,264,267,270,273,276,279,282,285,288,291,294,297,300,303,306,309,312,315,318,321,324,327,330,333,336,339,342,345,348,351,354,357,360,363,366,369,372,375,378,381,384,387,390,393,396,399,402,405,408,411,414,417,420,423,426,429,432,435,438,441,444,447,450,453,456,459,462,465,468,471,474,477,480,483,486,489,492,495,498,501,504,507,510,513,516,519,522,525,528,531,534,537,540,543,546,549,552,555,558,561,564,567,570,573,576,579,582,585,588,591,594,597,600,603,606,609,612,615,618,621,624,627,630,633,636,639,642,645,648,651,654,657,660,663,666,669,672,675,678,681,684,687,690,693,696,699,702,705,708,711,714,717,720,723,726,729,732,735,738,741,744,747,750,753,756,759,762,765,768,771,774,777,780,783,786,789,792,795,798,801,804,807,810,813,816,819,822,825,828,831,834,837,840,843,846,849,852,855,858,861,864,867,870,873,876,879,882,885,888,891,894,897,900,903,906,909,912,915,918,921,924,927,930,933,936,939,942,945,948,951,954,957,960,963,966,969,972,975,978,981,984,987,990,993,996,999
5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100,105,110,115,120,125,130,135,140,145,150,155,160,165,170,175,180,185,190,195,200,205,210,215,220,225,230,235,240,245,250,255,260,265,270,275,280,285,290,295,300,305,310,315,320,325,330,335,340,345,350,355,360,365,370,375,380,385,390,395,400,405,410,415,420,425,430,435,440,445,450,455,460,465,470,475,480,485,490,495,500,505,510,515,520,525,530,535,540,545,550,555,560,565,570,575,580,585,590,595,600,605,610,615,620,625,630,635,640,645,650,655,660,665,670,675,680,685,690,695,700,705,710,715,720,725,730,735,740,745,750,755,760,765,770,775,780,785,790,795,800,805,810,815,820,825,830,835,840,845,850,855,860,865,870,875,880,885,890,895,900,905,910,915,920,925,930,935,940,945,950,955,960,965,970,975,980,985,990,995
$t - 166833
$d - 99500
and total:
266333
why am i wrong?
Some numbers are multiples of both 3 and 5. (Your algorithm adds these numbers to the total twice.)
Because 6 * 5 == 30 and 10 * 3 == 30, you're adding the some numbers up twice.
$sum = 0;
$i = 0;
foreach(range(0, 999) as $i) {
if($i % 3 == 0 || $i % 5 == 0) $sum += $i;
}
Because you double-count numbers that are multiple of both 3 and 5, i.e. multiples of 15.
You can account for this naively by subtracting all multiples of 15.
for ($j = 15; $j < $maxnumber; $j += 15)
{
$e += $j;
echo $j.',';
}
$total = $total - $d;
In your case, if it is 15, you will add the number twice.
Try this:
$t = 0;
$d = 0;
for ($i = 0; $i <= $maxnumber; $i++){
if ($i % 3 == 0)
$t+= $i;
else if ($i % 5 == 0)
$d += $i;
}
echo $t.'<br>'.$d;
I think that in your code, if a number is a multiple of 3 and 5, it is added twice. Take 15 for example. It's in your list of multiples of 3 and in the list of multiples of 5. Is this the behaviour you want?
One of the best approach to this solution (to achieve optimum time complexity), run an Arithmetic Progression series and find the number of terms in all series by using AP formula: T=a+(n-1)d, then find sum by : S=n/2[2*a+(n-1)d]
where : a=first term ,n=no. of term , d=common deference, T=nth term
The code solution below has been implemented to suit the question above - so the values 3 and 5 are hard-coded. However, the function can modified such that values are passed in as variable parameters.
function solution($number){
$val1 = 3;
$val2 = 5;
$common_term = $val1 * $val2;
$sum_of_terms1 = calculateSumofMulitples($val1,$number);
$sum_of_terms2 = calculateSumofMulitples($val2,$number);
$sum_of_cterms = calculateSumofMulitples($common_term,$number);
$final_result = $sum_of_terms1 + $sum_of_terms2 - $sum_of_cterms;
return $final_result;
}
function calculateSumofMulitples($val, $number)
{
//first, we begin by running an aithmetic prograssion for $val up to $number say N to get the no of terms [using T=a +(n-1)d]
$no_of_terms = (int) ($number / $val);
if($number % $val == 0) $no_of_terms = (int) ( ($number-1)/$val ); //since we are computing for a no of terms below N and not up to/exactly/up to N. if N divides val with no remainder, use no of terms = (N-1)/val
//second, we compute the sum of terms [using Sn = n/2[2a + (n-1)d]
$sum_of_terms = ($no_of_terms * 0.5) * ( (2*$val) + ($no_of_terms - 1) * $val );
// sum of multiples
return $sum_of_terms;
}
You can run a single loop checking whether the number is multiple of 3 OR 5:
for ($i = 0; $i < $maxnumber; $i++)
{
if($i%3 || $i%5){
$t += $i;
echo $i.',';}
}
I think the original code is not including numbers which are multiples of both 3 and 5 in the total: if the test for multiple of 3 matches, it takes that and goes on. If you total the multiples of 15 up to 1000, you get 33165, which is exactly the difference between your total, 266333, and the original total, 233168.
Here's my solution to the question:
<?php
$sum = 0;
$arr = [];
for($i = 1; $i < 1000; $i++){
if((int)$i % 3 === 0 || (int)$i % 5 === 0)
{
$sum += $i;
array_push($arr,$i);
}
}
echo $sum;
echo '<br>';
print_r($arr);//Displays the values meeting the criteria as an array of values
how to show hit counter then make it back to zero?
i want after i've been submit form it can show hit counter..
but i want after it reach "20" it can back to zero..bcoz the limit of submit is 20 times so it can't over the limit.
how do i make it works?I've been try to this code...
<?
$limit="20";
$Query_counter=mysql_query("SELECT model FROM inspec");
$Show_counter=mysql_fetch_array($Query_counter);
$show_counter = $show_counter["model"]+1;
if($show_counter > $limit[0]) {
$show_counter = 0;
}elseif ($show_counter > $limit[1]) {
$show_counter = 0;
}
$Query_update=mysql_query("UPDATE inspec SET model=$Show_counter");
$Show_counter=number_format($Show_counter);
$Show_counter=str_replace(",",".",$Show_counter);
echo "Hit:</br><strong>$show_counter</strong>";
?>
To make a counter go up to a certain value then loop back to zero, you can use the modulus operator, which in a lot of languages (including PHP and MySQL) is %
$x = 0;
$limit = 4;
for ($i = 0; $i < 10; ++$i) {
$x = ++$x % $limit;
echo $x;
}
// 1, 2, 3, 0, 1, 2, 3, 0, 1, 2
I hope that makes enough sense. I can't really figure out from the question what exactly you want... Perhaps something like this?
UPDATE `mytable` SET `mycounter` = (`mycounter` + 1) % {{the limit}}
The concept here is to test to see if the variable you're incrementing exceeds some acceptable range as a result of the next increment. Simple increment the variable, then test its value.
In your case, just add a test after you increment the counter:
$show_counter = $show_counter["model"]+1;
if($show_counter > $limit){
$show_counter = 0;
}
Be sure to define $limit to whatever number you want to cycle on.
If you want to do this for multiple thresholds you can add additional tests. Note that you could hard-code $limit to any number or any variable you want, it's just the thing you're testing against.
<?
$Query_counter=mysql_query("SELECT model FROM inspec");
$Show_counter=mysql_fetch_array($Query_counter);
$show_counter = $show_counter["model"]+1;
$x = 0;
$limit = 20;
for ($i = 0; $i < 30; ++$i) {
$x = ++$x % $limit;
echo $x;
}
$Query_update=mysql_query("UPDATE inspec SET model= (model + 1) % 20");
$Show_counter=number_format($Show_counter);
$Show_counter=str_replace(",",".",$Show_counter);
echo "Hit:</br><strong>$show_counter</strong>";
?>