Hi!
I am kind of new to FME so this must be a simple issue to solve for anyone in here.
I have a xls-file with zip codes, 5 characters together. I want to add a blank space between the third and the fourth character. I just can´t find a simple way to do it with my limited knowledge...
Thanks,
Tobias
+50You could use an AttributeSplitter with format string 3s2s, this creates a list with 2 elements, let's call them _list{0} and _list{1}. The first one holds the first three characters, the second one the last two.
Then an AttributeCreator that creates a new attribute which is the value of _list{0}, then a space and then the value of _list{1}
+39Or just use the following formula in an attributecreator
@Left(@Value(zip),3) @Right(@Value(zip),2)
+44Or you could use an AttributeManager (or AttributeCreator) create a new attribute, open the text editor dialog, and enter:
@Left(@Value(zip),3) @Right(@Value(zip),2)Obviously there is a space between the two of those functions.
+44Or just use the following formula in an attributecreator
@Left(@Value(zip),3) @Right(@Value(zip),2)
+18Another way is using the AttributeCreator with the expression for the new value:
@Substring(@Value(test),0,3) @Substring(@Value(test),3,2)
This assumes the old attribute to be "test".
You can also use a StringReplacer, then you only need one transformer:
Replacement text has a space between \\1 and \\2
+50You can also use a StringReplacer, then you only need one transformer:
Replacement text has a space between \\1 and \\2
If the replacement text contains & or \0, then it is replaced in the substitution with the portion of string that matched the regular expression.
If replacement text contains \#, where # is a digit between 1 and 9, then it is replaced in the substitution with the portion of string that matched the n-th parenthesized subexpression of the regular expression.
The first paragraph is unreliable. If it was true, this setting for this case should work as expected.
In addition, this example within the help doc may not be correct. I think the Replacement Text should be --\0-\0, if the first paragraph was correct.
Source String: Bobby
Text to Find: b*y
Replacement Text: --\\0-\\0
Use Regular Expression: yes
Case Sensitive: yes
Result: Bo--bby-bbyLooks like all the ways I tried could have worked if I just had done it right!
Thanks to all of you!
I agree, the documentation needs to be reviewd on this subject. In FME 2016.1 it says:
If the replacement text contains & or \0, then it is replaced in the
substitution with the portion of string that matched the regular
expression.
I agree, the documentation needs to be reviewd on this subject. In FME 2016.1 it says:
If the replacement text contains & or \0, then it is replaced in the
substitution with the portion of string that matched the regular
expression.
Could possibly be a regression due to the homogenization efforts of the regex engine, @daleatsafe
+50
I agree, the documentation needs to be reviewd on this subject. In FME 2016.1 it says:
If the replacement text contains & or \0, then it is replaced in the
substitution with the portion of string that matched the regular
expression.
+44If the replacement text contains & or \0, then it is replaced in the substitution with the portion of string that matched the regular expression.
If replacement text contains \#, where # is a digit between 1 and 9, then it is replaced in the substitution with the portion of string that matched the n-th parenthesized subexpression of the regular expression.
The first paragraph is unreliable. If it was true, this setting for this case should work as expected.
In addition, this example within the help doc may not be correct. I think the Replacement Text should be --\0-\0, if the first paragraph was correct.
Source String: Bobby
Text to Find: b*y
Replacement Text: --\\0-\\0
Use Regular Expression: yes
Case Sensitive: yes
Result: Bo--bby-bby
+44You can also use a StringReplacer, then you only need one transformer:
Replacement text has a space between \\1 and \\2
1) I found some info that said:
You can use \\0 as a placeholder for the whole regex match, and \\1 through \\9 for the text matched by one of the first nine capturing groupsSo I think we need to update the doc to be more clear that the parts list starts at 1, and that 0 is the whole of the match.
2) The doc is also wrong in saying you need to use \\\\0 - I guess at some point you needed the first \\ as an escape character. Not any more.
3) It (\\0) no longer works in 2016.
I'll file all of these issues with our devs and let you know when they are fixed.
+44
1) I found some info that said:
You can use \\0 as a placeholder for the whole regex match, and \\1 through \\9 for the text matched by one of the first nine capturing groupsSo I think we need to update the doc to be more clear that the parts list starts at 1, and that 0 is the whole of the match.
2) The doc is also wrong in saying you need to use \\\\0 - I guess at some point you needed the first \\ as an escape character. Not any more.
3) It (\\0) no longer works in 2016.
I'll file all of these issues with our devs and let you know when they are fixed.
Or you could use an AttributeManager (or AttributeCreator) create a new attribute, open the text editor dialog, and enter:
@Left(@Value(zip),3) @Right(@Value(zip),2)Obviously there is a space between the two of those functions.
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