为什么Datomic在迭代时连续两次产生相同的临时ID？ [英] Why does Datomic yield the same temporary ID twice in a row when iterating?

问题描述

这将产生两个不同的id，这是伟大的：

 ＃db / id [：db.part / user] 
＃db / id [：db.part / user] 
  

（我试过很多想法到目前为止）将产生相同的ID两次，这不是我想要的：

  2（fn []＃db / id [：db.part / user]））
（for [n [1 2]]＃db / id [：db.part / user]）
  
 
 
 
全部产生类似
 （＃db / id [：db.part / user -1000774]＃db / id [：db.part / user -1000774]）
   
 
其中每个调用产生的数字是相同的。
 
 
 
我实际想要的是不产生数字的调用
 
 
 
任何想法？
 
 
 
只是要清楚，文档说，每次调用tempid都会生成一个唯一的临时ID。
 
 
 
 [在@maxthoursie注释之后编辑，重复会在任何情况下出现此问题。 
 
解决方案
使用
  datomic.api：as d]）
（重复2＃（d / tempid：db.part / user））
 ;; => （＃db / id [：db.part / user -1000118]＃db / id [：db.part / user -1000119]）
  
考虑＃...是读取器宏，意味着当读取表达式时它们的值将被解析，这自然只发生一次。仅当您编写文字事务数据（如模式）时，才使用＃...宏。使用datomic.api / tempid在运行时生成tempids。
 
This will produce two different ids, which is great:
#db/id[:db.part/user]
#db/id[:db.part/user]
but anything like the following (I tried a lot of ideas so far) will produce the same id twice, which is not what I want:
(repeatedly 2 (fn [] #db/id[:db.part/user]))
(for [n [1 2]] #db/id[:db.part/user])
All yield something like
(#db/id[:db.part/user -1000774] #db/id[:db.part/user -1000774])
where the number produced is the same for each call.

What I actually want is for the calls to NOT produce a number at all, so that I can just add the produced data via a transaction.

Any ideas?

Just to be clear, the documentation says, "Each call to tempid produces a unique temporary id."

[Edited after comment by @maxthoursie that repeat would be having this problem in any case.]
解决方案
Use
(require '[datomic.api :as d])
(repeatedly 2 #(d/tempid :db.part/user))
;; => (#db/id[:db.part/user -1000118] #db/id[:db.part/user -1000119])
Consider that #... are reader macros meaning that their value will be resolved when the expression is read which naturally happens only once. Use the #... macro only when you are writing literal transaction data (like a schema). Use datomic.api/tempid to generate tempids in runtime.

                        
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