I am attempting to show data in rows of three like this (notice the number of items will not always be even):
abcd defg hijk
lmno pqrs tuvw
xyz1 2345 6789
1011 1213
I am struggling to get the logic right to do this (this is in a foreach() loop).
I know I have to have some if($i %3 == 0) logic in there.. But I'm a bit stuck.
Can anyone help me out?
$a = array('abcd','defg','hijk','lmno');
for ($i = 0; $i < count($a); $i++) {
if ($i && $i % 3 == 0)
echo '<br />';
echo $a[$i].' ';
}
It's better to use a for loop as:
// run $i for each index in the array.
for($i=0 ; $i<count($arr) ; $i++) {
// if $i is non-zero and is divisible by 3 print a line break.
if ($i && $i % 3 == 0) {
echo "<br />";
}
// print the element at index $i.
echo $arr[$i].' ';
}
Code in action
Pseudo-code since I don't know PHP (and you asked for the logic which tends to be the same across all procedural languages):
perline = 3
i = 0
foreach item in list:
if i > 0 and (i % perline) == 0:
print newline
if (i % perline) != 0:
print space
print item
i = i + 1
This will both output a line separator before elements 3, 6, 9 and so on (first element being 0) and place whatever desired spacing you want before the second and third elements on each line. You can just use a different value for perline to change the number output on each line.
Related
I'm somewhat new to PHP, been reading a few books and I've never seen a loop where it gets you all the even numbers(for example from 1 to 10), so I decided to try it myself:
for($i=0;$i<10 && $i % 2===0;$i++)
echo $i;
Tried with only double == as well.
And this,
$i=0;
do echo $i; while($i++<10 && $i % 2 ==0);
Can't seem to figure out how to use 2 conditions in the same statement.
Would appreciate the help!
Thanks.
Try to use this code
for( $i=0; $i<=10; $i++ )
{
if( $i%2 == 0 ){
echo $i;
}
}
The loop is breaking entirely when the second condition fails the first time. On the first iteration: 0 is less than 10, and it is even, so the loop iterates. On the second iteration: 1 is less than 10, but is odd, so the loop breaks.
Your code is the equivalent of this:
for($i=0; $i<10; $i++) {
if ($i % 2 !==0 ) {
break;
}
echo $i;
}
0
You can eliminate the second condition of your for loop to prevent the breakage and rely exclusive on a third expression to increment $i by two each iteration.
for($i=0; $i<10; $i = $i + 2) {
echo $i;
}
02468
The second statement in a for-loop is/are the condition(s) which gets checked every loop. so if it fails your loop stops. what you need will look somewhat like this:
for ($i = 0; $i < 10; $i++)
if ($i % 2 == 0)
echo $i;
So the loop will run over every number but only print out the even ones.
You don't need to loop.
Range can create a range with third parameter step 2.
$arr = range(0,20,2);
Echo implode(" ", $arr);
https://3v4l.org/S3JWV
you can use also regular loop and get the evens by formula:
for($i=0; $i<10 ;$i++) {
$j = $i * 2;
// do somthing with $j witch loop over 10 first evens...
}
I'm try to dynamically generate an HTML table containing data from a database. The output looks perfect, but upon closer inspection I realized it was omitting every fourth result. I know it has something to do with the structure of my while loop and the if/else statement within, but I'm not sure what it is exactly.
$i=0;
while ($person = $pull_person->fetch()){
if ($i <= 2){
echo "<td valign='top'>";
echo "<h3>" . $person['person_name'] . " - " . $person['person_id'] . "</h3>";
echo "<label style='background-image:url(" . $person['person_pic'] . ");'><input type='checkbox' name='person[]' value='" . $person['person_id'] . "''></label>";
echo "</td>";
$i++;
}
else{
echo "</tr>";
echo "<tr>";
$i=0;
}
}
It's gotta be something simple/obvious, but it's not registering with me. Thanks!
The loop is not hitting the fourth result because of the loop limiting logic.
$i=0;
while ($person = $pull_person->fetch()){
if ($i <= 2){
echo "<p>item: $i</p>";
$i++;
}
}
Iteration 1: $i = 0
Iteration 2: $i = 1
Iteration 3: $i = 2
Iteration 4: $i = 3
Iteration 4 is never hit because it checks and sees that $i must be less than or equal to 2. If you change this to be less than or equal to 3 it will work as you want.
if ($i <= 3)
Evidently... you only increment the variable $i when the condition is met:
$i=0;
while ($person = $pull_person->fetch())
{
if ($i <= 2)
{
//output
$i++;
}
else
{
//no output
$i=0;
}
}
So this happens:
iteration old $i new $i output
1 0 1 yes
2 1 2 yes
3 2 3 yes
4 3 0 no //condition not met
5 0 1 yes //loops...
...
What you observe here is that the code will skip the output of the iteration given by the number in the conditional plust 2. So, for example, if you use the condtion $i <= 3, the results are:
iteration old $i new $i output
1 0 1 yes
2 1 2 yes
3 2 3 yes
4 3 4 yes
5 4 0 no //condition not met
6 0 1 yes //loops...
...
If you want to insert something each n iterations, do as follows:
$n = 3; //number of items per row
$i = 0;
while ($person = $pull_person->fetch())
{
//output item
$i++;
if ($i == $n)
{
//something each $n iterations
$i=0;
}
}
The effect is the following (assuming $n = 3):
iteration old $i new $i new row
1 0 1 no
2 1 2 no
3 2 0 yes //condition is met, $i reset to 0
4 0 1 no
5 1 2 no
6 2 0 yes //condition is met, $i reset to 0
...
Note 1: every iteration outputs an item.
Note 2: you can adjust the initial value of i to have an offset.
Misread question but will leave the answer as it may provide use anyway.
Best way is to look at the array as a 'module' (mathematically) -- i.e. use Modular Arithmetic:
if ((i % 4) == 0)
That is to say, indices 0, 4, 8, 12, 16, ... will be targeted, only. And, actually, this is how most setInterval functions work under the hood.
Could someone help suggest in below php explode function, we are displaying script after 5th listing. How is it possible to display script exactly after 5th listing and 10th listing on a page which has more than 10 listings
We tried using
if ($i == 5 & $i== 10)
but it does not work
Below is original code - which displays script after 5th listing
<?php
$listings = explode("<hr/>", $list);
$numberOfListings = count($listings);
for($i = 0; $i < $numberOfListings; ++$i)
{
if ($i == 5)
{ ?>
<script> </script>
<?php }
echo $listings[$i] . "<hr/>";
}
?>
Edit
How is it like - if have to display a separate script on $i==9, could you advise.
Because $i starts at 0 (0 to 9 is 10, whilst 0 to 10 is 11). Try if ($i == 4 || $i== 9), with an or operator.
Also I would not use the && (the and operator), because it is unlikely $i will ever equal both 4 and 9. I'd suggest you read into Truth Tables (and maybe Propositional Calculus) because from seeing what you had tried originally, it would be helpful to understand how a truth table works.
(source: wlc.edu)
You can use the contine, continue is used within looping structures to skip the rest of the current loop iteration and continue execution at the condition evaluation and then the beginning of the next iteration.
$arr = range(0,9);
foreach($arr as $number) {
if($number < 5) {
continue;
}
print $number;
}
Ref: http://php.net/manual/en/control-structures.continue.php
Try using modulus operator
$listings = explode("<hr/>", $list);
$numberOfListings = count($listings);
for($i = 1; $i < $numberOfListings; ++$i)
{
if ($i%5 == 0)
{
echo "in";
?>
<script> </script>
<?php
}
echo $listings[$i-1] . "<hr/>";
}
Here we are looping from 1 and there for $i <= $numberOfListings
and while listing we will use $listings[$i-1]
DEMO CODE AT http://codepad.viper-7.com/lrTOgP
I have the following PHP code working 'successfully' to display URL's:
<?php
for ($i = 0; $i <= $json->domaincount; $i++) {
echo '<td class="domainList">'.$json->$i.'</td>';
}
?>
Every forth echo I want to also
echo </tr><tr>
to start a new line in the table.
Is there an easy way to know which is every forth count?
I have $i which increments from 0 up so when it gets to 3, 7, 11 etc I need to change table line.
thx
Try this:
if ( ( $i + 1 ) % 4 == 0 ) { echo '</tr><tr>'; }
This is using the modulus operator. It divides a number and returns the remainder, so 7 % 4 = 3 (because 4 fits in once, and three is left over) and 8 % 4 = 0 (because 4 fits in evenly and there are no left overs)
Look into modulo operands (% in PHP). So the check that i % 4 == 0 would give you every fourth row.
<?php
for ($i = 0; $i <= $json->domaincount;) {
echo '<td class="domainList">'.$json->$i.'</td>';
if (!(++$i & 3))
echo '</tr><tr>';
}
?>
!($i & 3) is just a quick way of writing $i % 4 === 0 without having to use modulus. You can only use that trick for modding by powers of 2.
I currently have:
$i = 1;
while {
echo $i;
$i++;
}
And it shows:
1
2
3
4 etc..
How would I make it display backwards?
For example
4
3
2
1 etc..
I basically want to do the exact same thing but flip it around.
$i = 10;
while($i>0) {
echo $i;
$i--;
}
Example - Print number through 0 to 5 with PHP For Loop
for($i=0; $i<=5; $i=$i+1)
{
echo $i." ";
}
In the above example, we set a counter variable $i to 0. In the second statement of our for loop, we set the condition value to our counter variable $i to 5, i.e. the loop will execute until $i reaches 5. In the third statement, we set $i to increment by 1.
The above code will output numbers through 0 to 5 as 0 1 2 3 4 5.
Note: The third increment statement can be set to increment by any number. In our above example, we can set $i to increment by 2, i.e., $i=$i+2. In this case the code will produce 0 2 4.
Example - Print number through 5 to 0 with PHP For Loop
What if we want to go backwards, that is, print number though 0 to 5 in reverse order? We simple initialize the counter variable $i to 5, set its condition to 0 and decrement $i by 1.
for($i=5; $i>=0; $i=$i-1)
{
echo $i." ";
}
The above code will output number from 5 to 0 as 5 4 3 2 1 0 looping backwards.
Good luck! :)
If you want to back-word sr number as per your count of rows in result then use this.
$num_rows = mysqli_num_rows($query);
$x = $num_rows;
$x--;
$i = 4;
while($i > 0) {
echo $i--;
}