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我被困在一个非常基本的 sql 查询脚本中。Mybe有人可以注意到我看不到的东西。我检查了从 mysqladmin 执行的 sql 查询是否正常：

 <?
 include ('gps_db_connect.php'); 
 $query = "SELECT * from gps WHERE server_time > '20130124'";
 echo $query;
 $result = mysqli_query($connection, $query) or die(' Error getting data'); 
 echo '    After query';
 while ($row = mysql_fetch_array($result, MYSQL_ASSOC))
  {   
   echo $row['server_time'];
  }

 ?>

这是屏幕显示/输出：

SELECT * from gps WHERE server_time > '20130124' After                      
query
PHP Error Message 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a9440109/public_html/test.php on line 7
4

1 回答 1

0

试试这个 

 $result = mysql_query($query, $connection) or die('Error');
于 2013-01-25T00:26:42.223 回答