我写一条自定义sql语句，表名用了@#代替，好像执行的时候报错？

迅睿框架更新 2020-12-07 19:23:36

tvv

请问php文件，我写一条自定义sql语句，表名用了@#代替，好像执行的时候报错？@#只能用到模板文件中吗？

sql语句没有问题，我在adminphp下测试过！

\Phpcmf\Service::M()->db->query("UPDATE @#1_news as a,@#1_wgone as b SET b.updatetime ='2020-12-1' where a.title=b.title");

方案

迅睿框架创始人 #1 · 2020-12-07 14:49:42

满意答案

表名@#只能用在模板的sql标签里面，你在php里面需要写上前缀函数
tvv #2 · 2020-12-07 14:52:10

前缀函数?是完整的数据库表名吗？
黄老邪 #3 · 2020-12-07 14:55:48

表前缀是：.$this->prefix.
黄老邪 #4 · 2020-12-07 14:56:35

SELECT * FROM `".$this->prefix."member_menu` WHERE
迅睿框架技术-实习 #5 · 2020-12-07 14:58:35

\Phpcmf\Service::M()->db->prefix
tvv #6 · 2020-12-07 15:21:55

\Phpcmf\Service::M()->db->query("UPDATE '".$this->prefix."1_news' as a,'".$this->prefix."1_wgone' as b SET b.updatetime ='2020-12-1' where a.id=b.cid");报错，哪里错了呢？CRITICAL - 2020-12-07 15:11:43 --> Using $this when not in object context#SQL：SELECT *FROM `dr_cron`WHERE `id` = 50#URL：http://172.16.12.19/admin.php?c=cron&m=do_add&id=50#AGENT：Mozilla/5.0 (Windows NT 6.1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/55.0.2883.87 UBrowser/6.2.4098.3 Safari/537.36#REFERER：http://172.16.12.19/admin.php?c=cron&m=index
迅睿框架技术-实习 #7 · 2020-12-07 15:36:14

明显错了
黄老邪 #8 · 2020-12-07 15:41:56

.updatetime ='2020-12-1 错了，这里应该是10位的整数，来是格式化后的时间
黄老邪 #9 · 2020-12-07 15:42:15

不是
迅睿框架技术-实习 #10 · 2020-12-07 15:42:37

$this->prefix这个写错了，改成五楼的写法
tvv #11 · 2020-12-07 15:44:52

\Phpcmf\Service::M()->db->query("UPDATE '".$this->prefix."1_news' as a,'".$this->prefix."1_wgone' as b SET b.updatetime =unix_timestamp(NOW()) where a.id=b.cid");我改时间格式了，还是报错！
黄老邪 #12 · 2020-12-07 15:47:53

我自己的是这样写的
易学习 #13 · 2020-12-07 16:01:57

同意十楼的观点
tvv #14 · 2020-12-07 16:05:56

\Phpcmf\Service::M()->db->prefix(); $sql="UPDATE '".$this->prefix."1_news' as a,'".$this->prefix."1_news_form_tet' as b SET b.qy =unix_timestamp(NOW()) where a.id=b.cid"; \Phpcmf\Service::M()->db->query($sql);CRITICAL - 2020-12-07 15:57:21 --> Call to undefined method CodeIgniter\Database\MySQLi\Connection::prefix()
南哥 #15 · 2020-12-07 16:29:55

$sql="UPDATE '".\Phpcmf\Service::M()->db->prefix."1_news' as a,'".$this->prefix."1_news_form_tet' as b SET b.qy =unix_timestamp(NOW()) where a.id=b.cid";十楼的方法啊
tvv #16 · 2020-12-07 16:49:30

$sql="UPDATE '".\Phpcmf\Service::M()->db->prefix."1_news' as a,'".$this->prefix."1_news_form_tet' as b SET b.qy =unix_timestamp(NOW()) where a.id=b.cid"; $rt = \Phpcmf\Service::M()->db->query($sql);if ($rt) { // 成功} else { //失败 echo \Phpcmf\Service::M()->db->error();}还是这个错误提示CRITICAL - 2020-12-07 16:40:50 --> Using $this when not in object context
夏侯渊 #17 · 2020-12-07 16:54:04

同意十楼的观点
tvv #18 · 2020-12-07 16:58:21

$sql="UPDATE '".\Phpcmf\Service::M()->db->prefix."1_news' as a,'".\Phpcmf\Service::M()->db->prefix."1_news_form_tet' as b SET b.qy =unix_timestamp(NOW()) where a.id=b.cid"; $rt = \Phpcmf\Service::M()->db->query($sql);if ($rt) { // 成功} else { //失败 echo \Phpcmf\Service::M()->db->error();} ERROR - 2020-12-07 16:48:37 --> 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 ''1_news' as a,'1_news_form_tet' as b SET b.qy =unix_timestamp(NOW()) where a.id=' at line 1#SQL：UPDATE '1_news' as a,'1_news_form_tet' as b SET b.qy =unix_timestamp(NOW()) where a.id=b.cid
易学习 #19 · 2020-12-07 16:59:38

撒比了，sql语法谱写错误了

先去phpmyadmin工具里面运行一下吧
tvv #20 · 2020-12-07 17:06:14

UPDATE test.dr_1_news as a,test.dr_1_news_form_tet as b SET b.qy =unix_timestamp(NOW()) where a.id=b.cid数据库里操作没问题啊！
步兵过河 #21 · 2020-12-07 17:07:40

UPDATE test.dr_1_news as a,test.dr_1_news_form_tet as b SET b.qy =unix_timestamp(NOW()) where a.id=b.cidUPDATE '1_news' as a,'1_news_form_tet' as b SET b.qy =unix_timestamp(NOW()) where a.id=b.cid你对比啊，少了什么，长度都不一样了，而且还多了‘号
tvv #22 · 2020-12-07 17:20:13

$sql="UPDATE test.dr_1_news as a,test.dr_1_news_form_tet as b SET b.qy =unix_timestamp(NOW()) where a.id=b.cid"; $rt = \Phpcmf\Service::M()->db->query($sql);if ($rt) { // 成功} else { //失败 echo \Phpcmf\Service::M()->db->error();}我用原数据库名称放进去都报错，提示执行失败？为啥啊？数据库里是成功的！

易学习 #23 · 2020-12-07 17:27:02

$sql="UPDATE dr_1_news as a,dr_1_news_form_tet as b SET b.qy =unix_timestamp(NOW()) where a.id=b.cid";
$rt = \Phpcmf\Service::M()->db->query($sql);
if ($rt) {
   // 成功
} else {
    //失败
    echo \Phpcmf\Service::M()->db->error();
}

tvv #24 · 2020-12-07 19:23:36

迅睿框架创始人：谢谢了，我解决了！

迅睿系统漏洞

我写一条自定义sql语句，表名用了@#代替，好像执行的时候报错？

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