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请问一个PHP操作MYSQL的有关问题

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Release: 2016-06-13 10:45:26
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请教一个PHP操作MYSQL的问题

PHP code
<!--Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/-->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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