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Post History
In response to a Q on Web Applications [WA] the asker self-answered with: =CONCATENATE("@ £/€ rate: ",text(J21,"0.00000")) The issue was that, with "in a cell J21 I have a set value 1.10000":. ...
#3: Post edited
by
Alexei
·
2020-11-01T06:58:15Z (about 4 years ago)
#2: Post edited
by
pnuts
·
2020-10-17T16:18:00Z (about 4 years ago)
In response to a [Q](https://webapps.stackexchange.com/questions/147879/concatenation-truncating-numberic-value) on Web Applications the asker self-answered with:- =CONCATENATE("@ £/€ rate: ",text(J21,"0.00000"))
- The issue was that, with "*in a cell J21 I have a set value 1.10000*":.
- =CONCATENATE("@ £/€ rate: ",(J21))
- was returning `@ £/€ rate: 1.1` rather than `@ £/€ rate: 1.10000`.
- Is there a shorter and/or clearer formula to the same effect?
- In response to a [Q](https://webapps.stackexchange.com/questions/147879/concatenation-truncating-numberic-value) on Web Applications [WA] the asker self-answered with:
- =CONCATENATE("@ £/€ rate: ",text(J21,"0.00000"))
- The issue was that, with "*in a cell J21 I have a set value 1.10000*":.
- =CONCATENATE("@ £/€ rate: ",(J21))
- was returning `@ £/€ rate: 1.1` rather than `@ £/€ rate: 1.10000`.
- Is there a shorter and/or clearer formula to the same effect?
#1: Initial revision
by
pnuts
·
2020-10-17T16:10:05Z (about 4 years ago)
Loss of precision when concatenating an exchange rate with a string
In response to a [Q](https://webapps.stackexchange.com/questions/147879/concatenation-truncating-numberic-value) on Web Applications the asker self-answered with:
=CONCATENATE("@ £/€ rate: ",text(J21,"0.00000"))
The issue was that, with "*in a cell J21 I have a set value 1.10000*":.
=CONCATENATE("@ £/€ rate: ",(J21))
was returning `@ £/€ rate: 1.1` rather than `@ £/€ rate: 1.10000`.
Is there a shorter and/or clearer formula to the same effect?