php數據庫問題

Question

我寫了這個PHP代碼，工作正常...然後我想要一個特定的代碼段作爲一個函數工作，但只要我這樣做，我沒有得到正確的結果...我很困惑出了什麼問題，可以有人請...幫助我非常感謝......

下面的代碼給我的錯誤...

$arr1=array(); 

$date = date("D"); 

$link = mysql_connect ('localhost', 'root', ''); 

$db = mysql_select_db ('dayevent', $link); 

function grabData($arr){ //works properly NOT as a function, but I want to make this code part act like a funciton 
    $i=0; 
    $sql = "SELECT event FROM events WHERE day = '$date'"; 
    $sel = mysql_query($sql); 

    echo $sel; //this prints Resource id #3 
    if (mysql_num_rows($sel) > 0) { // but doesn't go into if block 

      while($row = mysql_fetch_array($sel)) { 
       echo $row['event'] . '<br />'; 

       //storing DB query result in array 

       $arr[$i]=$row['event']; 

       $i=$i+1;  

      } 

      foreach($arr as $key => $value) { 
       echo $key . " " . $value . "<br />"; 

      } 

    } else echo 'Nothing returned!'; //prints this instead of the correct result 

} 

grabData($arr1); 

mysql_close(); 

Answer 1

移動這個自己的函數中：$date = date("D");。現在的情況是，$ date沒有被定義。如果您使用error_reporting(E_ALL)運行，您將馬上就會發現它。

Answer 2

測試下面的代碼：

EDITED

$arr1=array(); 
$date = date("D"); 

$link = mysql_connect ('localhost', 'root', ''); 
$db = mysql_select_db ('dayevent', $link); 

function grabData() 
{ 
    global $link,$date; 

    $arr = array(); 
    $i=0; 
    $sql = "SELECT event FROM events WHERE day = '$date'"; 
    $sel = mysql_query($sql,$link); 

    echo $sel; //this prints Resource id #3 
    if (mysql_num_rows($sel) > 0) 
    { 
     while($row = mysql_fetch_array($sel)) 
     { 
      echo $row['event'] . '<br />'; 
      $arr[$i]=$row['event']; 
      $i=$i+1;  
     } 
     foreach($arr as $key => $value) 
     { 
      echo $key . " " . $value . "<br>"; 
     } 
    } else echo 'Nothing returned!'; //prints this instead of the correct result 
    return $arr; 
} 

print_r(grabData()); 
mysql_close();