I have 2 variables named $min and $sec.
If $min = 5 and $sec = 15 I want to print this "5m 15s", echo "$min m $sec s";
But that gives "5 m 15 s". How do I get rid of the blank space?
You can use braces to do this, same as many shells:
$min = 7;
$sec = 2;
echo "${min}m ${sec}s";
The output of that is:
7m 2s
The braces serve to delineate the variable name from the following text in situations where the "greedy" nature of variables would cause problems.
So, while $minm tries to give you the contents of the minm variable, ${min}m will give you the contents of the min variable followed by the literal m.
echo $min."m ".$sec."s"; is one way of doing it
u can try this
echo $min."m ".$sec."s ";
edit>
the output is
5m 15s
Related
First I thought this was a stupid question, and i should do some search and it would be easy to solve. But I am afraid I just ain't getting anywhere!
The thing i need to do is simple. I have a U$ value and i want to divide it by 12. Thats it.
Well, the thing is that this value is outputed by a function, and echoes ok, look:
<?php
$preconormal = wpsc_the_product_price(); // it echoes like 99.90
$precoja = str_replace (".", "", $preconormal);
echo $precoja; //echo ok -> 9990
$quantas = '12';
$parcela = $precoja/$quantas; // ok, so divide 9990 by 12, right?
echo $parcela; //no!!!!! it echoes 0 :(
?>
I really hope you can help me!
You are trying to divide strings, if you used numbers say
$quantas = 12;
$precoja = 9990;
What happens?
It should fix the division, in which case, prior to the mathmematics, convert your vars to integs by
$quantas = intval($quantas);
$precoja = intval($precoja);
//your manipulation here..l
Remove the quotes...
$quantas = '12';
to
$quantas = 12;
$precoja = floatval($preconormal)*100;
$preconormal = $precoja / 12;
I'd change your 5th line by removing the single quotes and/or 6th line with $parcela = (int)$precoja / (int)$quantas; because as soon as you use the function str_replace then $precoja becomes a string. Also having the single quotes earlier on = '12' it is also a string and that division returns 0.
There is a problem in Interview Street challange. Maybe the most easiest of all challenges. "Unfriendly Numbers", is the name and question goes like this.
There is one friendly number and N unfriendly numbers. We want to find how many numbers are there which exactly divide the friendly number, but does not divide any of the unfriendly numbers.
Input Format:
The first line of input contains two numbers N and K seperated by spaces. N is the number of unfriendly numbers, K is the friendly number.
The second line of input contains N space separated unfriendly numbers.
Output Format:
Output the answer in a single line.
I did a PHP programming like this:
<?php
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
$handle = fopen ("php://stdin","r");
$input = fgets($handle);
$num_unfriendly_number=substr($input,0,1);
$friendly_number=substr($input,2,1);
$input2=fgets($handle);
for($i=0;$i<=($num_unfriendly_number); $i=$i+2){
$unfriendly_numbers[$i]=substr($input2,$i,1);
}
//truncates additional input
//now getting divisiors of given friendly numbers
$check_num=1;
//one is always a divisor of any number
$divisior[0]=1;
$arrayindex=1;
for($check_num; $check_num<=$friendly_number; $check_num++){
$hold_var=$friendly_number%$check_num;
if($hold_var==0){
$divisor[$arrayindex]=$check_num;
$arrayindex++;
}
}
$index=0;
foreach($divisor as $test_div){
$output=true;
foreach($unfriendly_numbers as $test_unfrnd){
if($test_unfrnd%$test_div){
$output=false;
}
}
if ($output){
$outputarray[$index]=$test_div;
$index++; //edited afterwards after #Boris's suggestion but didn't work :(
}
}
$num_of_output=count($outputarray);
define('STDOUT',fopen("php://stout","r"));
fwrite(STDOUT,$num_of_output);
?>
The above programme worked fine for 2 testcases but did not applied for other tests. I did some research but did not found any errors. Any helps please. Thanks in advance.
Fist of all I would like to mention that I do not know php. However, I think this is simple enough I can try to help.
Several errors I see:
for($i=0;$i<=($num_unfriendly_number); $i=$i+2){
$unfriendly_numbers[$i]=substr($input2,$i,1);
}
Here you use substr($input2,$i,1);, this however assumes all your unfriendly numbers are digits, which might not always be the case. Better use the split function in php. Replace the whole while with the following:
$unfriendly_numbers = explode(" ", $input2);
After that:
$index=0;
foreach($divisor as $test_div){
$output=true;
foreach($unfriendly_numbers as $test_unfrnd){
if($test_unfrnd%$test_div){
$output=false;
}
}
if ($output){
$outputarray[$index]=$test_div;
}
}
Here you never increase the $index variable. Isn't this meaning that you will override the divisors one with other? USe the operator []=. It appends to an array in php:
if ($output){
$outputarray []= $test_div;
}
EDIT One more error I see is that you count on the friendly number to be a digit too. You can fix this too:
$friendly_number=substr($input,2,1);
->
$friendly_number=explode(" ", $input)[0];
I have the same problem I can't understand why this code can't finish in less than 16 seconds!
I would like to hear your tricks
a = raw_input()# this will read this line: 8 16
b = raw_input()# this will read this line: 2 5 7 4 3 8 3 18
al = a.split()
bl = b.split()
blint = []
fn = int(al[1])
fnlist = [fn]
half_fn = fn / 2 # only I go to half the number to save some time
k = 1
while k <= half_fn:
if fn % k == 0:
fnlist.append(k)
k += 1
plist = []
for j in bl:
blint.append(int(j)) # here I changed the bl list elements which are string to int
for i in fnlist:
for j in blint: #I have the int elements so I don't need every time bring the string and change it to int
if j % i == 0:
plist.append(i)
break
counter = len(fnlist) - len(plist)
print counter
To get time I do this
$day_timmer = time()
echo $day_timmer;
This is what printsout 1332533147
But when I add to it I get this
$day_timmer = time() * 1000;
echo $day_timmer;
This is what printsout 1.332533387E+12
No clue what I'm doing wrong.
So I get voted down for wondering why letters are in my numbers lol
That's called scientific notation, and it's all right. E+12 stands for 1012 and in your case, this is exactly the given number:
1.332533387E+12 = 1.332533387 * 10^12 = 1,332,533,387,000
To fix your problem use the following statement.
<?php
echo date("h:i a");
?>
Say I have 1.234 I want to get the .234
I tried echo 1.234%1 //I get 0
I am rusty, what is the correct way?
(The tags says PHP as this might be an issue only with PHP, but I really am looking for the general solution).
php's % modulo operator converts its arguments to integers. To get a floating-point modulus, use fmod:
echo fmod(1.234, 1);
You can remove the whole number from the number itself. in php its:
$num = 1.234;
echo $num - floor($num);
Subtract the integer portion of $x ((int)$x) from $x:
$x = 1.234;
$d = $x - (int)$x;
// $d now equals 0.234
echo $d;
Example
Just substract integer part 1.234 - (int)1.234
Try this:
echo 1.234 - intval(1.234);
How do I output a value as a number in php? I suspect I have a php value but it is outputting as text and not as a number.
Thanks
Here is the code - Updated for David from question below
<?php
if (preg_match('/\-(\d+)\.asp$/', $pagename1, $a))
{
$pageNumber = $a[1];}
else
{ // failed to match number from URL}
}
?>
If I call it in: This code it does not seem to work.
$maxRows_rs_datareviews = 10;
$pageNum_rs_datareviews = $pagename1; <<<<<------ This is where I want to use it.
if (isset($_GET['pageNum_rs_datareviews'])) {
$pageNum_rs_datareviews = $_GET['pageNum_rs_datareviews'];
}
If I make page name a static number like 3 the code works, if I use $pagename1 it does not, this gives me the idea $pagename1 is not seen as a number?
My stupidity!!!! - I used $pagename1 instead of pageNumber
What kind of number? An integer, decimal, float, something else?
Probably the easiest method is to use printf(), eg
printf('The number %d is an integer', $number);
printf('The number %0.2f has two decimal places', $number);
This might be blindingly obvious but it looks like you want to use
$pageNum_rs_datareviews = $pageNumber;
and not
$pageNum_rs_datareviews = $pagename1;
echo (int)$number; // integer 123
echo (float)$number; // float 123.45
would be the easiest
I prefer to use number_format:
echo number_format(56.30124355436,2).'%'; // 56.30%
echo number_format(56.30124355436,0).'%'; // 56%
$num = 5;
echo $num;
Any output is text, since it's output. It doesn't matter what the type of what you're outputting is, since the human eye will see it as text. It's how you actually treat is in the code is what matters.
Converting (casting) a string to a number is different. You can do stuff like:
$num = (int) $string;
$num = intval($string);
Googling php string to number should give you a beautiful array of choices.
Edit: To scrape a number from something, you can use preg_match('/\d+/', $string, $number). $number will now contain all numbers in $string.