獲取數據，沒有得到正確的數據

我試圖獲取基於時間的前10項（與time前10個值項柱）。我有以下代碼。

<?php 
    include_once("connect.php"); 
    $sql = "SELECT * FROM scores order by time desc limit 10"; 
    $query = mysql_query($sql) or die("systemResult=Error"); 
    $counter = mysql_num_rows($query); 

    if($counter>0) 
    { 
     print("systemResult=Success"); 
     $array = mysql_fetch_array($query); 

     foreach($array as $data) 
     { 
      $athleteName = $data["athleteName"]; 
      $email = $data["email"]; 
      $time = $data["time"]; 
      $timeStamp = $data["timeStamp"]; 
      $country = $data["country"]; 

      print "&athleteName=" . $athleteName; 
      print "&email=" . $email; 
      print "&time=".$time; 
      print "&timeStamp=".$timeStamp; 
      print "&country=".$country; 
     } 
    } 
    else 
    { 
     print("systemResult=Error"); 
    } 
?>

我得到的輸出是

systemResult=Success&athleteName=7&email=7&time=7&timeStamp=7&country=7&athleteName=7&email=7&time=7&timeStamp=7&country=7&athleteName=4&email=4&time=4&timeStamp=4&country=4&athleteName=4&email=4&time=4&timeStamp=4&country=4&athleteName=G&email=G&time=G&timeStamp=G&country=G&athleteName=G&email=G&time=G&timeStamp=G&country=G&athleteName=n&email=n&time=n&timeStamp=n&country=n&athleteName=n&email=n&time=n&timeStamp=n&country=n&athleteName=2&email=2&time=2&timeStamp=2&country=2&athleteName=2&email=2&time=2&timeStamp=2&country=2&athleteName=I&email=I&time=I&timeStamp=I&country=I&athleteName=I&email=I&time=I&timeStamp=I&country=I

可以看出，我得到的輸出是不是有什麼是在數據庫中的表。我越來越奇怪的價值觀。我究竟做錯了什麼？

來源

2013-03-10 Joe Slater

你並不需要使用每個在你的情況，如果是這樣，只是打印$的數據，儘量去除foreach循環，如果你想獲得的所有記錄，然後，使用時：

while($data = mysql_fetch_array($query)) 
    { 
     $athleteName = $data["athleteName"]; 
     $email = $data["email"]; 
     $time = $data["time"]; 
     $timeStamp = $data["timeStamp"]; 
     $country = $data["country"]; 

     print "&athleteName=" . $athleteName; 
     print "&email=" . $email; 
     print "&time=".$time; 
     print "&timeStamp=".$timeStamp; 
     print "&country=".$country; 
    }

來源

2013-03-10 11:06:17

感謝，此代碼的工作。 – 2013-03-10 11:12:30

嘗試

while($data = mysql_fetch_array($query)) { 
    $athleteName = $data["athleteName"]; 
    //...

來源

2013-03-10 11:07:21 Ziumin

獲取數據，沒有得到正確的數據

回答

相關問題