2012-09-03 37 views
2

我有兩個組合框,一個報告成績和一個設定誰的進球,我怎樣才能使它所以如果在後期[「提交」]如果帖子值等於其他帖子值以外的回顯「失敗」;?

如果$ _ POST [「Score1」]和$ _POST ['Score2'] 不等於 $ _POST ['homegoalscorer1']和$ _POST ['awaygoalscorer1'] 然後回顯「失敗」;

有點像;

if(isset($_POST['submit'])) 
{ 
$homescore = $_POST['Score1']; 
$awayscore = $_POST['Score2']; 

$homegoalscorer = $_POST['homegoalscorer1']; 
$awaygoalscorer = $_POST['awaygoalscorer1']; 

    if '$homescore' + '$awayscore' != $homegoalscorer + $awaygoalscorer { 
     echo "failed"; 
    } 
     else { 

      } 
    } 

任何想法?

+1

不看了看剩下的問題還沒有,但是'「$ homescore」 +「$ awayscore''應該是$ homescore + $ awayscore,失去了引號。 –

回答

3

變量上的單引號會將該變量介紹爲不帶執行的字符串。你也忘了加括號:

if(isset($_POST['submit'])) { 
    $homescore = (float)$_POST['Score1']; 
    $awayscore = (float)$_POST['Score2']; 

    $homegoalscorer = (float)$_POST['homegoalscorer1']; 
    $awaygoalscorer = (float)$_POST['awaygoalscorer1']; 

    if (($homescore+$awayscore) != ($homegoalscorer+$awaygoalscorer)) { 
     echo "failed"; 
    } else { 

    } 
} 
1

在你的if語句中使用一些括號來強制條件設置在正確的上下文中 - 你爲什麼要用單引號封裝你的變量?

if (($homescore + $awayscore) != ($homegoalscorer + $awaygoalscorer)) 
{ 
// Your code continues.... 
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