find in structure a certain string

raw is a cell.

I would like to find the identification number in the example which is 13161509800019, but I only want the cell that contains 1-2-3, as the cell below it contains a trailing 4-5-6. Also how can I do that for any number of identification numbers containing 1-2-3 column 1 excluding first row 'SERIAL'?

Answer 1

[Azzi Abdelmalek]

C={'serial';'123456 1-2-3';'1452145 4-5-6';'745841 3-2-4';'85471245 1-2-3'}
C=C(2:end);
idx=~cellfun(@isempty,regexp(C,'1-2-3'))
D=C(idx)
out=regexp(D,'.+(?=1-2-3)','match')
out=str2double([out{:}])

[Leonardo Wayne]

Hi Azzi, thanks for the answer. I am stuck with another issue. If you take a look the images 1-4. The cell "get_motors_no_workbook2_NO_123" being created in image 1 has two index numbers when I click on the cell values and reach the image 4. How can that be, does this mean that this is a cell inside another cell?

Also, if I want "get_motors_no_workbook2_NO_123" to have only 6x1 cell and those individual rows in image 2 to be char values as shown in 4 not 1x1 cell as is shown in 3.

In other words: e.g. refer to image 2: I want row 1 which is '13161509800019' to be a char and not a 1x1 cell and so on. Image 1 has the code!

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[Leonardo Wayne]

never mind. I have solved it.