LabVIEW
What does the Search 1-D array vi return with more than one match?
The "Search 1D Array" function will only return the index of the first instance of the search item.
If you want to find all matching indexes, place the search function in a loop and pass the "Index of Element" terminal through a shift register to the next iteration of the loop and connect it to the "Start Index" terminal, but increment it by 1. This will search the rest of the array.
Take all the indexes of mathing elements and build into an array and pass that out of the loop.
The attached example shows this.
Ed
If you want to find all matching indexes, place the search function in a loop and pass the "Index of Element" terminal through a shift register to the next iteration of the loop and connect it to the "Start Index" terminal, but increment it by 1. This will search the rest of the array.
Take all the indexes of mathing elements and build into an array and pass that out of the loop.
The attached example shows this.
Ed
Using the Abort button to stop your VI is like using a tree to stop your car. It works, but there may be consequences.
Ed,
I finally got to a PC with LV 7.1. I'm using LV 5.1 and I don't believe I can use the "stop if" option in the while loop. I tried the "continue looping" option with >= 0 as the condition. It works except I also get -1 as the last match. Do you know how I could execution of the last iteration or would I just have to ignore the -1 entry in the matrix?
Thank,
Jeff
Well the -1 is required as it is the termination criteria for the loop no way to skip that as you don't know beforehand that there won't be any more instance. So what I usually do in such cases is wiring out the loop iteration value and use it as input to the Resize Array function to effectively remove the last element. The new Delete Array node in LabVIEW 6 and higher could also do the same as by default it removes the last element in an array.
Rolf Kalbermatter
