请问一个PHP操作MYSQL的有关问题
Release: 2016-06-13 13:16:42
Original
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请教一个PHP操作MYSQL的问题
PHP code
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function writeLeft()
{
$retArray = array();
[color=#FF0000]$sql = "select id,name,strr from jh_lanmu";[/color]
$result = mysql_query($sql,$conn);
while($myrow = mysql_fetch_array($result,MYSQL_BOTH))
{
$retArray[] = $myrow;
}
mysql_free_result($result);
foreach($retArray as $tmp)
{
echo '
Copy after login
'.$tmp[0].''.chr(13);
}
}
红色的SQL语句在Navicat中查询是没有报错的,能够查出结果,但是PHP页面上始终报错:
mysql_query(): supplied argument is not a valid MySQL-Link resource in F:\PHPWebSite\jinghong\config.php on line 11
其他的SQL操作都能正常运行。
数据库代码如下:
DROP TABLE IF EXISTS `jh_lanmu`;
CREATE TABLE `jh_lanmu` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(20) DEFAULT NULL,
`strr` varchar(50) DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=gbk;
-- ----------------------------
-- Records of jh_lanmu
-- ----------------------------
INSERT INTO `jh_lanmu` VALUES ('1', '产品配料', 'cppl.php');
INSERT INTO `jh_lanmu` VALUES ('2', '产品系列', 'cpxl.php');
------解决方案--------------------
function writeLeft()
{
global $conn;
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