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在PHP中显示MySQL数据库的最后两行_Php_Mysql - Fatal编程技术网

在PHP中显示MySQL数据库的最后两行

在PHP中显示MySQL数据库的最后两行,php,mysql,Php,Mysql,我试图只显示表StudentInfo最后两行的timein和timeout列。 代码如下: $query = mysql_query(" SELECT timein, timeout FROM studentInfo WHERE name = 'Student' ORDER BY time DESC LIMIT 2 ") or die("Error querying database ".mysql_error()); while($m

我试图只显示表StudentInfo最后两行的timein和timeout列。 代码如下:

$query = mysql_query("
    SELECT timein, timeout 
    FROM studentInfo
    WHERE name = 'Student' 
    ORDER BY time DESC
    LIMIT 2
    ")
    or die("Error querying database ".mysql_error());
while($minutes = mysql_fetch_array($query)){
  echo "timein: ".$minutes['timein']."<br />";
  echo "timeout: ".$minutes['timeout']."<br />";
$query=mysql\u查询(“
选择timein,timeout
来自学生信息
其中name='Student'
按时间顺序描述
限制2
")
或者死(“查询数据库时出错”.mysql_Error());
而($minutes=mysql\u fetch\u数组($query)){
回显“timein:”.$minutes['timein'].“
”; 回显“超时:.$minutes['timeout']。”
试试这个

$query = mysql_query("SELECT timein, timeout  FROM studentInfo WHERE name = 'Student'  ORDER BY id DESC LIMIT 2") or die(mysql_error());

while($minutes = mysql_fetch_assoc($query)){
  echo "timein: ".$minutes['timein']."<br />";
  echo "timeout: ".$minutes['timeout']."<br />";
}
$query=mysql\u query(“选择timein,timeout FROM studentInfo,其中name='Student'ORDER BY id DESC LIMIT 2”)或die(mysql\u error());
while($minutes=mysql\u fetch\u assoc($query)){
回显“timein:”.$minutes['timein'].“
”; 回显“超时:.$minutes['timeout']。”
; }
--


尝试使用mysql mysqli,因为它已被弃用

use
print\r($minutes)
相反。除了使用已弃用且易受攻击的
mysql
库之外,我看不出问题出在哪里?您获取所有行,然后从最后一行到第一行排序,然后从结果集中获取前两行。您欣赏表和数据库之间的差异吗?为什么不呢你告诉我们那个错误?!
echo mysql_error();
而不是
die