2013-03-05 23 views

回答

10
foreach (array_slice($categories, 0, 2) as $forum_category){ 
    //do something 
} 
1

<?php for($i=0;$i<2;$i++){you can do anything with $categories[$i]}?>

或者您可以在foreach語句之前添加參數$i=0和:

$i=0; 
<?php foreach ($categories as $forum_category) { $i++; if($i>2) return }?> 
5
<?php $i = 0; 
    foreach ($categories as $forum_category) { 
    if($i < 2) { 
     // display result 
    } 
    $i++; 
    } 
?> 
+0

在'if'子句裏面有一個'break'缺少,對嗎? – GeneSys 2013-03-05 12:55:25

+0

@GeneSys no。如果它會顯示結果。 – 2013-03-05 12:56:33

+1

@GeneSys你爲什麼要打破它?這意味着繼續foreach循環'直到最後,但只有做(如果條件裏面有什麼)如果$ i小於2 – George 2013-03-05 12:56:46

2

這是一種方法。

<?php 

    $count = 0; 
    foreach ($categories as $forum_category) { 
      if($count == 2) 
       break; 
      else { 
        // Your Code 
      } 
     $count++; 
    } 

?> 
0

只要保留一個櫃檯。 根據您所提供的有限的結構中,所有我可以建議是:

<?php 
    $categoryNumber = 0; 
    foreach ($categories as $forum_category) { 
     $categoryNumber++; 
     if($categoryNumber > 2){ 
      continue; 
     } 
     //... 
    } 
?> 
0
for ($i = 0; $i < 2; $i++) { 
    $categories[$i]; 
} 

foreach ($categories as $forum_category) { 
    if($i < 2) { 
     $categories[$i]; 
    } 
    $i++; 
} 

或只是

$categories[0]; 
$categories[1]; 
0
<?php $i = 0; 
    foreach ($categories as $forum_category) { 
    if($i <= 1) { 
     //code 
    } 
    $i++; 
    } 
?> 
+0

條件應該是'$ i <= 1' – vedarthk 2013-03-05 13:02:47

0

試試這個:

$i = 0; 
foreach ($categories as $forum_category) { 
    echo $forum_category; 
    if($i == 1) break; 
$i++; 
}