php Code Igniter 在插入数据时捕获错误(Active Record)
声明:本页面是StackOverFlow热门问题的中英对照翻译,遵循CC BY-SA 4.0协议,如果您需要使用它,必须同样遵循CC BY-SA许可,注明原文地址和作者信息,同时你必须将它归于原作者(不是我):StackOverFlow
原文地址: http://stackoverflow.com/questions/8202547/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Code Igniter catch errors on inserting data (Active Record)
提问by lodhb
I am inserting data into my database using the following code:
我使用以下代码将数据插入到我的数据库中:
$this->db->set('event_id', $event_id);
...
$this->db->set('creator_id', $creator_id);
$this->db->insert('event');
How can I make sure the process was successful and show the user a confirmation message, else an error message?
如何确保该过程成功并向用户显示确认消息,否则显示错误消息?
回答by Damien Pirsy
Uhm, IIRC the method returns true if succesful, so you could go with a simple
嗯,如果成功,IIRC 方法返回 true,所以你可以用一个简单的
if($this->db->insert('event'))
{
echo 'Row succesfully inserted!';
}
Otherwise, you might always count the affected rows:
否则,您可能总是计算受影响的行数:
//....
$this->db->insert('event');
if($this->db->affected_rows() > 0)
{
echo 'row succesfully inserted';
}
UPDATEafter comment:
评论后更新:
I believe your three insert_batch() are on three different tables, so a check should be down on each one. Anyway, what's puzzling me is the reason behin this check..Query don't fail randomly: possibily they should neverfail and when they do you'll know through errors (logged or dislayed).
我相信您的三个 insert_batch() 位于三个不同的表上,因此应该对每个表进行检查。总之,什么是我百思不得其解的原因behin这check..Query不要随意失败:possibily他们应该永远不会失败,当他们这样做,你就会知道通过错误(记录或dislayed)。
回答by TV-C-15
"If you need to get the last error that has occurred, the error() method will return an array containing its code and message." https://www.codeigniter.com/userguide3/database/queries.html
“如果您需要获取发生的最后一个错误,error() 方法将返回一个包含其代码和消息的数组。” https://www.codeigniter.com/userguide3/database/queries.html
$insert = $this->db->insert('event');
if(!$insert){
$error = $this->db->error(); // Has keys 'code' and 'message'
echo "<BR><BR>";
print_r($error);
echo "<BR><BR>";
}
echo "Error contact the database admin.";
die();
}else{
echo "SUCCESSFULLY INSERTED ITEM";
return TRUE;
}