我认为这可能是一个死锁问题,但我只读取表数据。或者一些我看不见的很简单的东西。
我的$msqli obj可以用于db访问。
但我对这个错误有点困惑..。我被告知clients表没有被LOCK TABLES锁定
欢迎任何建议。谢谢,这是我的PHP代码
$mysqli = $GLOBALS['mysqli'];
$mysqli->query("LOCK TABLES clients READ, invoices READ, estimates READ");
$mysqli->query("SET @inv='Invoice'");
$mysqli->query("SET @est='Estimate'");
$sql = " (SELECT @inv, name AS client, invoices.id, invoices.ref AS ref, invoices.addTs AS ts \n"
. "FROM `clients` , `invoices` \n"
. "WHERE invoices.user_id = " . $_SESSION['user_id'] . " \n"
. "AND clients.id = invoices.client_id)\n"
. "UNION ALL\n"
. "(SELECT @est, name as client, estimates.id, estimates.ref AS ref, estimates.addTs AS ts \n"
. "FROM `clients` ,`estimates` \n"
. "WHERE estimates.user_id = " . $_SESSION['user_id'] . " \n"
. "AND clients.id = estimates.client_id) \n"
. "ORDER BY ts DESC LIMIT 5";
if ($result = $mysqli->query($sql) or die(mysqli_error($mysqli))) {
//Do stuff here
}当我在$sql中运行这个PhpMyAdmin查询时,它是成功的.谢谢你的帮助..。
********************SOLVED (?) ******************
固定(或似乎是):
来自http://bugs.mysql.com/bug.php?id=6588的报价
“不能在一个查询中多次使用锁定的表--为此使用别名。”
因此,查询被更新为
$mysqli->query("LOCK TABLES clients_t READ,clients READ ,
invoices READ, estimates READ");
$mysqli->query("SET @inv='Invoice'");
$mysqli->query("SET @est='Estimate'");
$sql = " " .
"(SELECT @inv, name AS client, invoices.id,
invoices.ref AS ref, invoices.addTs AS ts "
. "FROM `clients` AS clients_t , `invoices`\n"
. "WHERE invoices.user_id = " . $_SESSION['user_id'] . " \n"
. "AND clients_t.id = invoices.client_id)\n"
. "UNION ALL \n"
. "(SELECT @est , name AS client, estimates.id,
estimates.ref AS ref, estimates.addTs AS ts\n"
. "FROM `clients` ,`estimates` \n"
. "WHERE estimates.user_id = " . $_SESSION['user_id'] . " \n"
. "AND clients.id = estimates.client_id)\n"
. "ORDER BY ts DESC LIMIT 5";我还需要锁定别名表clients_t。无论如何,对于未来的用户,我希望这会有所帮助。
发布于 2013-07-20 02:04:19
虽然有点挑剔,但明智的做法是对$_SESSION‘’user_id‘输入进行加号或mysql_real_escape_string,以避免注入问题。
https://stackoverflow.com/questions/17650455
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