从数据库中获取数据,而不是获取正确的数据

时间:2013-03-10 11:03:02

标签: php mysql database

我有一个看起来像这样的数据库 -

Image 1

我正在尝试根据时间(time列中包含前10个值的条目)获取前10个条目。我有以下代码。

    <?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");
    }
?>

我得到的输出是

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,我得到的输出不是数据库中的表。我变得越来越奇怪了。我做错了什么?

2 个答案:

答案 0 :(得分:2)

在你的情况下你不需要使用每一个,如果是这样,只需打印$ data,尝试删除foreach循环,如果你想获得所有记录,那么,使用while:

 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;
     }

答案 1 :(得分:2)

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