


Warning</b>: mysql_free_result(): supplied argument is not a valid MySQL result
这是什么错?
<?php header('content-type:application/json;charset=utf8'); $host = '127.0.0.1:3307'; $root = 'root'; $pwd = 'apeg1996'; $con = mysql_connect($host,$root,$pwd); if($con == false){ echo "连接数据库失败!"; }else{ echo "连接数据库成功!"; } $sql = "select * from user"; function execute_data($sql){ $result = mysql_query($sql); mysql_free_result($result); mysql_close($conn); return $result; } echo execute_data($sql);?>
已显示连接成功,但是报这个错。。
连接数据库成功!<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 /><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 />
回复讨论(解决方案)
$result = mysql_query($sql);
改为
$result = @mysql_query($sql) or die(mysql_error());
$result = mysql_query($sql);
改为
$result = @mysql_query($sql) or die(mysql_error());
网页上显示的是No database selected, 这个是没有找到这个数据库吗?
但是我用的是Navicat 8 for mysql 这个视图工具。
我在这里面创建的数据库,并简历了表。 我应该怎么去查询这个数据库里的表呢?
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
版主回答的完美
$result = mysql_query($sql);
改为
$result = @mysql_query($sql) or die(mysql_error());
网页上显示的是No database selected, 这个是没有找到这个数据库吗?
但是我用的是Navicat 8 for mysql 这个视图工具。
我在这里面创建的数据库,并简历了表。 我应该怎么去查询这个数据库里的表呢?
你在mysql_query之前没有mysql_select_db,mysql不知道你需要在哪个db中执行查询操作。
在创建连接后,加一句
@mysql_select_db(' 这里填写数据库名',$conn) or die(mysql_error());
<?php header('content-type:application/json;charset=utf8'); $host = '127.0.0.1:3307'; $root = 'root'; $pwd = 'apeg1996'; $con = mysql_connect($host,$root,$pwd); if($con == false){ echo "连接数据库失败!"; }else{ echo "连接数据库成功!"; } @mysql_select_db('这里填写数据库名',$conn) or die(mysql_error()); // 加入这句 $sql = "select * from user"; function execute_data($sql, $conn){ // 加入$conn 参数 $result = mysql_query($sql); mysql_free_result($result); mysql_close($conn); return $result; } echo execute_data($sql, $conn);?>

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