Biztalk XLANG transform() output Biztalk 编排内部循环中的相同随机值

Question

I wrote a map to generate an HL7 message header (MSH). For the MSH.10 segment, by definition should be unique so I put the following in my map.

    public string MessageControlId()
    {
       //return System.DateTime.Now.ToString("yyyyMMddHHmmssffff");


       string firstPart = System.DateTime.Now.ToString("yyyyMMdd");
       string middlePart = new Random().Next( 1000, 9999 ).ToString();
       string lastPart = System.DateTime.Now.ToString("ffff");
        
       return firstPart + middlePart + lastPart;
    }

Then in my orchestration I call the header map multiple time in a loop. My goal is to generate multiple HL7 messages, each with its own message header and a unique MSH.10 value.

The code below is based on Microsoft Biztalk XLANG syntax which invokes the map to transform and create the message header via the transform() statement.

tMapType = System.Type.GetType(msgBre.HeaderMapName);

transform (msgHeader) = tMapType(msgBilling);

However, when I tested this out I see multiple HL7 message generated but many of them have duplicate value in term of their MSH.10 segment. I grouped them in different color below.

I expect separate value each time because in my code I generate a random number between 1000 and 9999. Plus I also generate the time value to the thousand of a second.

Do you know why this occur? My only theory is when I call the tranform() function, it does not really invoke the map to recreate the header each time...that seems wrong to me.

Answer 1

You are using Random object which is a pseudo-random number generator, so it returns the same sequence of numbers with a same seed. You are not giving a seed explicitly to the constructor, so it uses default seed which is based on a system clock. If you are creating Random objects in a tight loop with a default seed then next() function will return you the same number multiple times, which I think is what happens here.

You should either give unique seed explicitly or use the same Random object all the time (if it is possible).