


Warning</b> mysql_free_result(): supplied argument is not a valid MySQL result,该怎么解决
Warning: mysql_free_result(): supplied argument is not a valid MySQL result
这是什么错?
<br /><?php<br /> header('content-type:application/json;charset=utf8');<br /> $host = '127.0.0.1:3307';<br /> $root = 'root';<br /> $pwd = 'apeg1996';<br /><br /> $con = mysql_connect($host,$root,$pwd);<br /> if($con == false){<br /> echo "连接数据库失败!"; <br /> }else{ <br /> echo "连接数据库成功!";<br /> }<br /><br /> $sql = "select * from user";<br /> function execute_data($sql){<br /> $result = mysql_query($sql);<br /> mysql_free_result($result);<br /> mysql_close($conn);<br /> return $result;<br /> }<br /> echo execute_data($sql);<br />?><br />
已显示连接成功,但是报这个错。。
<br />连接数据库成功!<br /><br /><b>Warning</b>: mysql_free_result(): supplied argument is not a valid MySQL result resource in <b>F:\environment\WAMP\wamp5\wamp\wamp\www\test\sql.php</b> on line <b>17</b><br /><br /><br /><br /><b>Warning</b>: mysql_close(): supplied argument is not a valid MySQL-Link resource in <b>F:\environment\WAMP\wamp5\wamp\wamp\www\test\sql.php</b> on line <b>18</b><br /><br />
------解决思路----------------------
你只连接了数据库系统,但没有选择待操作的数据库
mysql_select_db('库名');
否则怎么知道你在查询那个库中的 user 表呢?
另外,你的 mysql_close($conn) 在函数中,而 $conn 并未传到函数中。
所以会有第2条错误
------解决思路----------------------

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