Here's the issue:
I retrieve the following data string from my database:
$row->exceptions = '1,2,3';
After explode I need the below code to check each one of the exploded pieces
$exceptions = explode(",", $row->exceptions);
//result is
//[0] => 1
//[1] => 2
//[2] => 3
for ($i = 0; $i <= $row->frequency; $i++) {
if ($exceptions[] == $i) {
continue;
} else {
//do something else
}
}
How can I make $exceptions[] loop through all keys from the exploded array so it evaluates if ==$i?
Thanks for helping.
It should suffice to substitute:
if($exceptions[] == $i)
with:
if(in_array($i,$exceptions))
By the way, it eliminates the need for a nested loop.
Ah, should be straightforward, no?
$exceptions = explode(",", $row->exceptions);
for ($i = 0; $i <= $row->frequency; $i++) {
foreach($exceptions as $j){
if($j == $i){
// do something
break;
}
}
}
I think I understand what you are asking. Here's how you would test within that loop whether the key equals $i.
for ($i = 0; $i <= $row->frequency; $i++)
{
foreach ($exceptions as $key => $value)
{
if ($key == $i)
{
continue;
}
}
}
Related
Here is my code, and it is not removing $arr[5] element so that I am trying to remove strings starting with # from my array
this is code
<?php
$arr = [
'#EXTM3U',
'#EXTINF:177,Paul Dateh & Oren Yoel - Be More',
'Be More.mp3',
'#EXTINF:291,Christopher Toy - Just Because',
'Just Because.mp3',
'#EXTINF:238,Magnetic North - Drift Away',
'Drift Away.mp3'
];
for ($i = 0; $i <= count($arr); $i++) {
if ($arr[$i]{0} == '#') {
echo $arr[$i] . "\n";
unset($arr[$i]);
}
}
print_r($arr);
?>
Reason:- You are counting array length inside the loop and every time when any value got unset() from the array, length of array decreased and value of count($array) changed (simply decreased)
So logically your 5th and 6th element never goes through if condition (they never get traversed by loop because of decreasing length of the array )
Solution 1:- Put count outside and it will work properly:-
$count = count($arr);
//loop start from 0 so use < only otherwise, sometime you will get an undefined index error
for ($i = 0; $i < $count; $i++) {
if ($arr[$i]{0} == '#') {
//echo $arr[$i] . "\n";
unset($arr[$i]);
}
}
print_r($arr);
Output:-https://eval.in/996494
Solution 2:- That's why i prefer foreach() over for() loop
foreach($arr as $key=> $ar){
if ($ar[0] == '#') {
unset($arr[$key]);
}
}
print_r($arr);
Output:-https://eval.in/996502
more spacific :
for ($i = 0; $i < count($arr); $i++) {
if (strpos($arr[$i], '#') !== false) {
echo "<br/>";
} else {
echo $arr[$i]."<br/>";
}
}
Try to use additional array to push right values. You calc count($arr); each iteration and when you do count($arr); your array gets smaller and count($arr); returns smaller values, so last elements won't be comparing, try to use variable to calc count before loop make changes:
<?php
//...
$start_count = count($arr);
for ($i = 0; $i <= $start_count; $i++) {
if ($arr[$i]{0} == '#') {
echo $arr[$i] . "\n";
unset($arr[$i]);
}
}
Or remove bad element with a help of additional array, put good elements in new array and don't delete them from input array:
<?php
$arr = [
'#EXTM3U',
'#EXTINF:177,Paul Dateh & Oren Yoel - Be More',
'Be More.mp3',
'#EXTINF:291,Christopher Toy - Just Because',
'Just Because.mp3',
'#EXTINF:238,Magnetic North - Drift Away',
'Drift Away.mp3'
];
$cleared_from_mess_array = array();
for ($i = 0; $i <= count($arr); $i++) {
if ($arr[$i]{0} != '#')
{
array_push($cleared_from_mess_array,$arr[$i]);
}
}
print_r($cleared_from_mess_array);
exit;
I have a foreach loop. There has uncountable value. I want to print last 3 values.
foreach($arr as $key=>$value) {
//print last 3 value.
}
Try this:
$i = 0;
foreach($arr as $key=>$value) {
if (count($arr) - $i < 4)
print($value);
$i++;
}
Try this you can use like this
$i = 0;
$len = count($array);
foreach ($array as $item) {
if ($i == 0) {
// first
} else if ($i == $len - 3) {
// last
}
else if ($i == $len - 2) {
// last
}
else if ($i == $len - 1) {
// last
}
// …
$i++;
}
I am trying to find duplicated values/string in an array using for loop
<?php
$b=array('a','b','c','a','b');
$c=count($b);
$d=array();
for($i=0;$i<=($c-1);$i++)
{
for($j=1;$j<=($c-1);$j++)
{
if($b[$i]!=$b[$j])
{
$flag=1;
}
}
if($flag==1)
{
$d[$i]=$b[$i];
}
}
print_R($d);
?>
where is my mistake? I have used array $d to display non duplicate values.....
NOTE: I need to try this only with for loop - I know how to do it using array functions.
You should reverse your test, because there are almost always values, which are different from the one you're testing. And you must reset your $flag before the inner loop, otherwise it will always be true.
When you want to find unique values, you can just test against $d only. If the value is already in $d, skip it.
$c1 = count($b);
for ($i = 0; $i < $c1; $i++) {
$dup = 0;
$c2 = count($d);
for ($j = 0; $j < $c2; $j++) {
if ($b[$i] == $d[$j])
$dup = 1;
}
if (!$dup)
$d[] = $b[$i];
}
print_r($d);
If you want to find values, which don't have duplicates instead
for ($i = 0; $i < $c; $i++) {
$dup = 0;
for ($j = 0; $j < $c; $j++) {
if ($i != $j && $b[$i] == $b[$j])
$dup = 1;
}
if (!$dup)
$d[] = $b[$i];
}
function has_dupes($array){
$dupe = array();
foreach($array as $val){
if(++$dupe[$val] > 1)
return true;
}
return false;
}
could do something like this.. this would check for dupes, then u can print the uniques
Why are you making a simple task complex .. simply
$b = array('a','b','c','a','b');
var_dump(customCount($b));
Output
array (size=3)
'a' => int 2 //duplicate
'b' => int 2 //duplicate
'c' => int 1
Function Used
function customCount($array) {
$temp = array();
foreach ( $array as $v ) {
isset($temp[$v]) or $temp[$v] = 0;
$temp[$v] ++;
}
return $temp ;
}
I need some help, even though I think I'm checking for the length of the array and I should be breaking out of the loop, I still get warnings on my [else if ($value....] line. So either I'm missing something crucial or I've been staring at this code segment too long and its obvious. Any insight would be appreciated.
$count = count($filter); //Filter is an array
if ($count > 1 ){
//Compare values and generate a range to choose from
$i = 1;
foreach($filter as $value){
//Break the loop if at the end of the array
if ($i >= $count){
//throw new exception($i .' '.$count);
break;
}
//if the value is smaller then the next procceding value, because they are already in order of presidence,
//add it to our range of potentials.
else if($value < $filter[$i]->value){
array_push($range, key($filter));
}
$i++;
}
}else {
return false;
}
I suspect that there are gaps in your array. Try this:
$filter = array_values($filter); // this will remove any gaps in the array
$count = count($filter);
if ($count <= 1)
return false;
for ($i = 0; $i < $count; $i++)
{
if ($i != $count-1 && $filter[$i]->value < $filter[$i+1]->value)
array_push($range, key($filter));
}
Your array might have non-numeric keys. Then try this:
foreach($filter as $key=>$value)
{
// test for $filter[$key];
}
Or your $filter array doesn't hold objects, then you can't use the -> in
$filter[$key]->value
try this code.... no need of checking count..
$range = array();
$i = 1;
foreach($filter as $value)
{
if(isset($filter[$i]) && $value < $filter[$i]->value)
{
array_push($range, key($filter));
$i++;
}
else
{
break;
}
}
Is there a way of iterating through an array but performing an operation on every other element?
ie If I have an array with 13 elements how do I do something to only elements 2,4,6,8,10 and 12?
foreach($array as $val) {
if(($i++ % 2) == 0) {
...do stuff here...
}
}
for ($i=1; $i<sizeof($array); $i+=2) {
// do stuff to $array[$i]
}
You can integrate it into a foreach loop too:
$i = 0;
foreach ($array as $v) {
if ($i++ & 1) continue;
// do stuff to $v
}
Note: $i & 1 is equivalent to ($i % 2) == 1 (or just $i % 2).
to fix cletuses answer for more speed and fix typos:
for ($i = 1, $j = count($array); $i < $j; $i += 2) {
// code
}
Another variation on the answers already posted... Similar to Phil Carter's answer. If the array has a numeric index you can use that in the foreach instead of managing a separate counter variable:
foreach ($array as $i => $v) {
if (! ($i % 2)) {
// do stuff to $v
}