PHP版本 - 較低版本的難度編碼

Question

我的客戶端使用的是PHP 5.2.0版，我有一些完美的代碼，但目前使用5.4.7。當我的代碼在他們的服務器上運行時，我收到一個解析錯誤。我在網上搜索了一種替代方法來編寫這段代碼，但我找不到任何東西。如果任何人有任何提示，這將是一個很大的幫助。

**$count = mysqli_query($con, "SELECT COUNT(*) FROM menuitem")->fetch_row()[0];** //I want to make sure the previous and next button are only displayed exactly to the number of items 
                        //in my database. I have to count those items, and turn that mysqli_query object into data. fetch_row[0] stores them 
                        //in an array, starting at 0, turning it into an integer I can use to count below. 
    $prev = $start - 3; 
    if ($prev >= 0) { 
     echo '<td><a href="?start=' . $prev . '">Previous Items</a></td>';// Set your $start - 3 since our offset is 3, and pass that to a variable. 
    }                  //if $start - 3 is still greater than 0, then display the previous button. Logically, that is page 2. 

    $next = $start + 3; 
    if ($next < $count) { 
     echo '<td><a href="?start=' . $next . '">Next Items</a></td>';// Since i used fetch_row[0] I now know how many rows I have. If $start + 3 is less than the $count query 
    }                 // than you will have a next button. If that number exceeds the number of rows I have, then no next button. 

echo "</tr>";  
echo "</table>"; 

在上面的代碼中，...-> fetch_row（）[0];是帶回錯誤的部分。 PHP 5.2.0不喜歡它。

Answer 1

在PHP 5.4之前，你需要could not dereference arrays returned from function calls。簡單地做：

$count = mysqli_query(...)->fetch_row(); 
$count = $count[0];