以下部分为必填项.请给一个解决方案.
Required suggestion on below part.please any one give solution.
我们已经从850映射到FlatFile
We have mapping from 850 to FlatFile
X12/PO1Loop1/PO1/PO109,我需要映射到字段" VALUE ",该字段在Option记录下,是不受限制的.
X12/PO1Loop1/PO1/PO109 and I need to map to field VALUE which is under record Option which is unbounded.
将PO109拆分为以'.'分隔的子字符串,在第一个字符串之后进行forsub递减,并使用value = substring创建新的Option
Split PO109 into substrings delimited by '.', foreach subsring after the first, create new Option with value=substring
因此,在输入样本中,我们的值类似于147895632qwerqtyuui.789456123321456987
So in input sample we have value like 147895632qwerqtyuui.789456123321456987
类似地,该字段在POLoop1下重复出现.
Similarly the field repeats under POLoop1.
所以我需要基于(.)分割值,然后将值传递给选项记录(无界)下的value字段.
So I need to split value based on (.) then pass a value to value field under option record(unbounded).
我尝试使用下面的代码段
I tried using below code snippet
public string SplitValues(string strValue) { string[] arrValue = strValue.Split(".".ToCharArray()); foreach (string strDisplay in arrValue) { return strDisplay; } }但是它不起作用,而且我对String方法并不十分熟悉,并且不确定是否有简单的方法可以做到这一点.我有一个字符串,其中包含几个用."分隔的值.
But it doesn't works, and I am not really familiar with the String methods and I am not sure if there's an easy way to do this. I have a String which contains couple of values delimited with "." .
所以我需要基于delimiter(.)分隔值并将值传递给字段.
So I need to separate values based on delimiter(.) and pass value to field.
我该怎么做
推荐答案正如我提到的,不太清楚您的目标是什么,但我认为您想将具有某种分隔符的节点拆分为多个节点...如果是这样,请尝试以下操作: seroter.wordpress/2008/10/07/splitting-delimited-values-in-biztalk-maps/
As I mentioned, not too clear what is your objective, but I think you want to split a node that has some kind of delimiter into multiple nodes... if so, Try this: seroter.wordpress/2008/10/07/splitting-delimited-values-in-biztalk-maps/
他就是这么做的.给定一个以 a | b | c | d 为值的节点,输出多个节点,每个节点包含被 | 分割后的值,因此node1 = a,node2 = b,node3 = c,node4 = d.
He is doing exactly that. Given a node with a|b|c|d as value, output multiple nodes, each containing the value after splitted by |, so node1 = a, node2 = b, node3 = c, node4 = d.
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