I have an array of objects, any array in php. How do i skip the last element in the foreach iteration?
问题:
回答1:
Use a variable to track how many elements have been iterated so far and cut the loop when it reaches the end:
$count = count($array);
foreach ($array as $key => $val) {
if (--$count <= 0) {
break;
}
echo "$key = $val\n";
}
If you don't care about memory, you can iterate over a shortened copy of the array:
foreach (array_slice($array, 0, count($array) - 1) as $key => $val) {
echo "$key = $val\n";
}
回答2:
There's various ways to do this.
If your array is a sequentially zero-indexed array, you could do:
for( $i = 0, $ilen = count( $array ) - 1; $i < $ilen; $i++ )
{
$value = $array[ $i ];
/* do something with $value */
}
If your array is an associative array, or otherwise not sequentially zero-indexed, you could do:
$i = 0;
$ilen = count( $array );
foreach( $array as $key => $value )
{
if( ++$i == $ilen ) break;
/* do something with $value */
}
回答3:
If you don't want to delete the last array entry with pop, you could skip it like this
$array = array('v1','v2','v3',...)
$counter = 1;
foreach($array as $value)
{
//do your thing in loop
if($counter == count($array)) continue; // this will skip to next iteration if last element encountered.
$counter++;
}
回答4:
What you are trying to do will defeat the purpose of foreach loop. It is meant to loop through the entire array and make our job easy.
for ex: You can get the array size using COUNT function in php and then can use for loop and set the limit to arraysize-2, so the last array will be omitted
回答5:
$count = count($array);
$i=0;
foreach ($arr as &$value)
{
$i++;
if($i==($count-1))
{
echo 'skip';
}
else
{
echo $value;
}
}