I'm trying to get this piece of script to work, but it keeps dying
$currentdays = intval(date("t"));
echo $currentdays; //echoes 30 as we're in April
$i = 1;
while ($i <= $currentdays){
echo $day;
}
It keeps dying with no error. I feel like it's timing out, but it's certainly taking its time.
You need to increment $i. 1 will always be less than 30, creating an infinite loop.
$currentdays = intval(date("t"));
$i = 0;
while ($i++ < $currentdays){
echo $i; // outputs 1, 2, 3.. 30
}
$i never changes. Try this:
$currentdays = intval(date("t"));
echo $currentdays; //echoes 30 as we're in April
$i = 1;
while ($i++ < $currentdays){
echo $i;
}
You never increment $i after echoing $day. This will go into an infinite loop.
Related
what I am trying to achieve here is to be able to loop from 0 to 100 (100, 98, 96, 94 ...) but has to stop at 0. What is doing right now is it passes 0 and -2 -4 which crashes the server. What am I doing wrong?
for ($i = 100; $i <= 100; $i--){
echo $i--;
echo "<br>";
}
Maybe a little explanation would be useful.
The middle part of the for loop $i <= 100 is what makes it infinite. That expression is checked before each iteration of the loop, and the loop will continue as long as that expression evaluates to true.
Since you set $ito 100 in the first section of the loop, and you're doing nothing except making it smaller, it will always be <= 100, forever.
The loop will work fine just the way you have it written if you change the continuation condition.
for ($i = 100; $i >= 0; $i--){
echo $i--;
echo "<br>";
}
That way it will continue until $i is reduced to less than zero, then $i >= 0 will be false, and the loop will end.
The third argument in for loop is what will be executed at the end of the loop. So:
for ($i = 100; $i >= 0; $i -= 2){
echo "$i<br>";
}
Will do the trick
As you can read here https://secure.php.net/manual/en/control-structures.for.php
At the end of each iteration, expr3 is evaluated (executed).
Alternatively:
<?php
foreach(range(100, 0, -2) as $n) {
echo $n;
}
for ($i = 100; $i >= 0; $i-=1){
echo $i--;
echo "<br>";
}
I figured it out somehow. Has been studying JavaScript for a year; Loop is still confusing to me
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
This code will divide the score until it reaches the number 5.
The $rows[score] is equal to 6600 in the database.
<?php
$i = $rows[score]; //score is 6600 in the database
while ($i >= 5) {
echo $i = $i /2;
echo "<br>";
}
?>
This is what my browser outputs:
3300
1650
825
412.5
206.25
103.125
51.5625
25.78125
12.890625
6.4453125
3.22265625
I don't understand why the browser output the last 3.22 - how do I stop the loop from echo out the last one that is less than 5??
Nothing wrong here the last value you get is from 6.4453125 / 2 = 3.22265625 since 6.4453125 still greater than 5
because 6 is higher than 5? so it does one more loop making $i 3.2 where the loop stops
If ($i<5) it wont go into the loop but there is no way to know until you check. $i = 6.4453125 the last time it checks, so it goes into the loop and it divides it by 2, which makes it less than 5 so it doesn't go into the loop again and stops.
I found the way to answer my own question so 3,22 will not be viewed on the page.
<?php
$i = $rows[score]; //score is 6600 in the database
while ($i >= 5) {
$i = $i /2;
if($i >= 5) {
echo $i;
echo "<br>";
}
}
?>
Since you're dividing by two immediately before displaying the result, you want to stop your loop when $i >= (5*2) i.e. $i >= 10, not 5.
<?php
$i = $rows[score]; //score is 6600 in the database
while ($i >= 10) {
echo $i = $i /2;
echo "<br>";
}
?>
This gives:
3300
1650
825
412.5
206.25
103.125
51.5625
25.78125
12.890625
6.4453125
Ive managed to loop through a table and get the difference in days between 2 dates adjacent to each other in the table.
Multiple entries have the same date, i have it now that when that date changes, it displays an image however i want it to display the image as many times as the difference in date
$sql = mysql_query("SELECT * FROM Films_Info")
or die(mysql_error());
$last_value = null;
while ($row = mysql_fetch_assoc($sql)) {
if (!is_null($last_value)) {
$a = new DateTime($row['FilmRelease']);
echo "<p>".$row['FilmName']."</p>";
$interval = $a->diff(new DateTime($last_value));
//echo $interval->format('%d days');
$i = 0;
}
$howManydays = $interval->days;
for ( $i; $howManydays; $i++) {
echo "<img src=\"day.jpg\" />";
$howManydays = 0;
}
$last_value = $row['FilmRelease'];
}
for ( $i = 0; $i < $howManydays; $i++) The second is a conditional statement telling when the loop should stop.
The first section in the for loop where it says $i = 0 initialized the variable $i to 0 and then tests the condition $i < $howManydays.
Let's say $howManydays equals 1. That means 0 < 1, so the loop will perform.
At the end of the loop, the third section is called ($i++), so $i is incremented and now equals 1. The second section is called again to test conditions $i < $howManydays which is asking if 1<1 which it's not, so the loop will exit.
So if $howManydays is greater than 0, the loop should happen the integer amount that is in $howManydays.
You will want to remove $howManydays = 0; within the for loop, if you don't want it to only fire once.
The for loop
for ( $i = 0; $i < $howManydays; $i++){
// ...
}
is somewhat equivalent to the while loop:
$i = 0;
while ( $i < $howManydays ){
// ...
$i++;
}
http://php.net/manual/en/control-structures.for.php for more information
In your code above, you should also check if interval was set. You should probably just do an if rather than a while, if only need one interval.
you could just do a simple division on the interval and print out the image that many times
$last_value_u = $last_value->format('U');
$a_u = $a->format('U');
$interval = $a_u - $last_value_u;
$image_count = intval($interval/86400);
for($i=0;$i<$image_count;$i++)
echo "<p><img src=\"day.jpg\" /></p>";
Update
An alternative option would be to loop through the interval:
for($i=$last_value_u;$i<$a_u;)
{
if(intval(($a_u - $i)/86400) == X)
{
// run special code for specific day
}
else
{
// run code for every other day
}
$i+=86400;
}
So I have a query, and though my query can possibly return 20 results, I only want it to show the first 9 results. There is a dumb reason I'm not just limiting the query results to 9, for this purpose I need to know how to stop the while function if $i reaches 9.
Code is
$i = 0;
while($array = mysql_fetch_array($queryresults) && $i <= '9')
{
echo $array['id'];
$i++;
}
How do I get it to stop putting out more echoes after 9th result? Thank you!
Putting the limiting condition first will also make sure you only fetch 9 rows, not and 10 and one discard because && is called lazy (if the first part is false, it wont even look at the next part).
$i = 0;
while($i < 9 && $array = mysql_fetch_array($queryresults)){
echo $array['id'];
$i += 1;
}
or if you want to be fancy:
$i = 9;
while($i-- > 0 && $array = mysql_fetch_array($queryresults)){
echo $array['id'];
}
But I find this error prone ..
$i = 0;
while($array = mysql_fetch_array($queryresults) && $i++ < 9)
echo $array['id'];
}
Notes:
No need to put 9 in quotes
You can increment-after-compare by doing $i++ in the while loop condition (although you don't have to, you could also put it after echo $array['id'] in the while loop body).
Careful! You want < 9 and not <= 9 since starting at zero, or you'll get 10 iterations of the loop.
Cheers
Your $i++ needs to be inside the while loop
You should also have $i <= 9 instead of $i <= '9' but it should still work as is
'9' needs to be 9
$i = 0;
while($array = mysql_fetch_array($queryresults) && $i < 9){
echo $array['id'];
$i += 1;
}