I think I have found the solution. The following three operations In xt = Table[{NSolve[x^2 - .001 i == 0 && x > 0]}, {i, 8}];

In[25]:= { {xt1 = xt /. {{Rule[a, b]}} :> b;}, {[Placeholder]} }

Out[25]= {{Null}, {[Placeholder]}}

In[26]:= xt1

Out[26]= {{0.0316228}, {0.0447214}, {0.0547723}, {0.0632456}, \ {0.0707107}, {0.0774597}, {0.083666}, {0.0894427}}

generates a 8 d vector, and by using Transpose, In xxt = Transpose[xt1]

Out[33]= {{0.0316228, 0.0447214, 0.0547723, 0.0632456, 0.0707107, 0.0774597, 0.083666, 0.0894427}} Taking the first element, In[34]:= xxt[[1]]

Out[34]= {0.0316228, 0.0447214, 0.0547723, 0.0632456, 0.0707107, \ 0.0774597, 0.083666, 0.0894427} this is the result I wanted. Thanks to you, because you, ivar johannesen,, oslomet, inspired me to look for transpose. This is the greatness of forum discussions, generate inspirations. Hong-Yee Chiu

Sorry, I ment Table[xt1[[n,2]],{n,8}]

xt is a list of elements containing two components: the x-value of the intended plot and - in some nested form - the y-value (the result of NSolve). I extracted this value simply using a replacement rule, which could as well be written like so:

xt1 = xt /. {{a_ -> b_}} :> b

which is probably more similar to what you are seeing in xt.

General hint: Use FullForm in combination with the excellent documentation, e.g.:

{{a_ -> b_}} :> b // FullForm