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社区首页 >问答首页 >在php中执行sql查询不起作用,但查询在phpmyadmin中工作

在php中执行sql查询不起作用,但查询在phpmyadmin中工作
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Stack Overflow用户
提问于 2015-09-12 06:48:20
回答 1查看 190关注 0票数 2

我有我的数据库sql查询文件,我想通过下面的代码在php中运行它

代码语言:javascript
复制
$query = file_get_contents('./sqlquery.txt');
print $query;
$conn->query($query);

但是它返回了这个错误

代码语言:javascript
复制
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'CREATE TABLE IF NOT EXISTS `ads` ( `id` int(11) NOT NULL AUTO_INCREMENT, `ur' at line 2 

我把打印输出复制到phpmyadmin,一切都很好,这里有什么问题吗?

我的sql查询如下

代码语言:javascript
复制
DROP TABLE IF EXISTS `ads`;
CREATE TABLE IF NOT EXISTS `ads` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `url` varchar(300) NOT NULL,
  `path` varchar(200) NOT NULL,
  `width` int(11) NOT NULL,
  `height` int(11) NOT NULL,
  `priority` int(11) NOT NULL,
  `adsalter` varchar(200) CHARACTER SET utf8 COLLATE utf8_persian_ci NOT NULL,
  `adstitle` varchar(200) CHARACTER SET utf8 COLLATE utf8_persian_ci NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=12 ;



DROP TABLE IF EXISTS `comment`;
CREATE TABLE IF NOT EXISTS `comment` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(120) CHARACTER SET utf8 COLLATE utf8_persian_ci NOT NULL,
  `comment` text CHARACTER SET utf8 COLLATE utf8_persian_ci NOT NULL,
  `contentid` int(11) NOT NULL,
  `parentid` int(11) NOT NULL DEFAULT '0',
  `date` varchar(250) NOT NULL,
  `haschild` int(1) NOT NULL DEFAULT '0',
  `visible` int(1) NOT NULL DEFAULT '0',
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=8 ;



DROP TABLE IF EXISTS `files`;
CREATE TABLE IF NOT EXISTS `files` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(200) NOT NULL,
  `size` int(14) NOT NULL,
  `type` varchar(50) NOT NULL,
  `newsid` int(11) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=11 ;



DROP TABLE IF EXISTS `frgpss`;
CREATE TABLE IF NOT EXISTS `frgpss` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `email` varchar(100) CHARACTER SET utf8 COLLATE utf8_persian_ci NOT NULL,
  `ip` varchar(31) CHARACTER SET utf8 COLLATE utf8_polish_ci NOT NULL,
  `token` varchar(27) CHARACTER SET utf8 COLLATE utf8_persian_ci NOT NULL,
  `date` int(15) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;



DROP TABLE IF EXISTS `news`;
CREATE TABLE IF NOT EXISTS `news` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `titrimage` varchar(100) NOT NULL,
  `titr` varchar(100) CHARACTER SET utf8 COLLATE utf8_persian_ci NOT NULL,
  `titralter` varchar(160) CHARACTER SET utf8 COLLATE utf8_persian_ci NOT NULL,
  `newsshurt` varchar(200) CHARACTER SET utf8 COLLATE utf8_persian_ci NOT NULL,
  `text` text CHARACTER SET utf8 COLLATE utf8_persian_ci NOT NULL,
  `keywords` text CHARACTER SET utf8 COLLATE utf8_persian_ci NOT NULL,
  `description` text CHARACTER SET utf8 COLLATE utf8_persian_ci NOT NULL,
  `author` varchar(200) CHARACTER SET utf8 COLLATE utf8_persian_ci NOT NULL,
  `branch` int(11) NOT NULL,
  `date` varchar(160) CHARACTER SET utf8 COLLATE utf8_persian_ci NOT NULL,
  `visible` int(1) NOT NULL DEFAULT '0',
  `visited` int(11) NOT NULL DEFAULT '0',
  `titrtitle` varchar(200) CHARACTER SET utf8 COLLATE utf8_persian_ci NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=11 ;


DROP TABLE IF EXISTS `signup`;
CREATE TABLE IF NOT EXISTS `signup` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(50) NOT NULL,
  `family` varchar(50) NOT NULL,
  `email` varchar(80) NOT NULL,
  `gender` varchar(50) NOT NULL,
  `username` varchar(80) NOT NULL,
  `picture` varchar(80) NOT NULL,
  `password` varchar(60) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;


DROP TABLE IF EXISTS `users`;
CREATE TABLE IF NOT EXISTS `users` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(60) NOT NULL,
  `lastname` varchar(60) NOT NULL,
  `email` varchar(100) NOT NULL,
  `password` varchar(40) NOT NULL,
  `userregistereddate` varchar(60) NOT NULL,
  `key` varchar(60) NOT NULL,
  `type` varchar(20) NOT NULL,
  `userphoto` varchar(50) NOT NULL,
  `usergroup` varchar(300) NOT NULL,
  `ipaddress` text NOT NULL,
  `telnumber` varchar(14) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=4 ;
EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2015-09-12 07:50:30

您在这里执行多个查询,它不会简单地工作,我非常肯定,在query函数($conn->query($query);)中,您使用的是mysql_query()mysqli_query(),但是您必须使用mysqli_multi_query()。看看这里的docx

代码语言:javascript
复制
$con = mysqli_connect($host, $user, $pass, $db) OR die(mysqli_error($con));
$query = file_get_contents('./sqlquery.txt');
mysqli_multi_query($con, $query);
mysqli_close($con);

希望这能帮到你

备注:假设所有查询在PHPMyAdmin中执行时都运行良好。

票数 2
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/32536069

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