扫描QR码以添加记录或如果记录存在则将+1添加到[count] [英] Scan QR code to either add a record or if record exists add +1 to [count]
问题描述
我正在研究试剂库存数据库。
目前我有2张桌子
1）罗氏与[UPC] [计数]和一些计算基于[UPC]的字段
2）MaterialNum根据罗氏表的计算数据提供试剂名称。
我有一个表格我可以搜索UPC，如果它存在填充表格，我可以修改[计数]并从那里保存。
如果通过搜索框找不到记录我可以添加一个新的记录并添加它，并从那里调整[Count]。
这实在是效率低下。我想要做的是扫描QR，检查记录，如果没有找到添加它。并且一旦添加或发现在那里，[Count]就会添加+1。
I am working on an reagent inventory db.
Currently I have 2 tables
1) Roche with [UPC][Count] and some calculated fields based off of [UPC]
2) MaterialNum which provides the name of the reagent based on the calculated data from Roche Table.
I have a form that I can search for the UPC and if it exists populates the form and I can modify the [Count] and save from there.
If the record is not found through the search box I can add a new record and add it, and adjust the [Count] from there.
This really is inefficient. What I want to be able to do is scan the QR, that check for record, if not found add it. And once added or found to be there a +1 is added to the [Count].
推荐答案
Abhean，
我真的很困惑。你说你当前的过程效率很低，但是你提出的解决方案执行的功能完全相同。
具体来说，你在这个帖子中要求的是什么？
Abhean,
I am really confused. You say that your current process is really inefficient, but your proposed solution performs the same exact functions.
Exactly what, specifically, are you asking for in this thread?
它会做同样的功能。麻烦就是这个过程。如果在表中找不到，则用户不希望必须扫描两次项目。一次用于搜索，然后再次用于添加记录。
为了能够扫描一次，通过查找记录的逻辑运行它，然后添加，如果不存在而不是trowing如果存在错误，则在所选记录的[count]中添加+1。对于用户来说，这将更加用户友好和更快。
It would do the same function. The trouble is the process. User will not like having to scan an item twice if it is not found in the table. Once for the search and then again for the add record.
To be able to scan once, run it through the logic of looking for the record then adding if not there and not trowing an error if it is there, then adding a +1 in the [count] of the selected record. This would be more user friendly and faster for the user.
然后，我仍然感到困惑。如果你有你想要的代码，你为什么需要扫描两次？
但是，我仍然不知道你在这里问我们到底是什么。
您需要帮助我们并询问 具体的 问题。
Then, I am still confused. If you have the code that you are looking for, why do you need to scan it twice?
But again, I still don''t know exactly what you are asking from us here.
You need to help us out and ask a specific question.
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