If you have an o-chem book, your best bet is to flip to the benzene section, where you should be able to see a detailed description of e-withdrawing and e-donating properties.
Basically, there are 2 ways e donation/withdrawal can be made - through a resonance effect and through an inductive effect. Both of these could withdraw or donate, or it could be that a functional group donates e through resonance and withdraws them inductively - then it will depend on which effect is stronger.
Inductive effects occur due to sigma bond polarity - think C-Cl bond. Chlorine is highly electronegative and thus withdraws electrons through the sigma bond with carbon. It will pull the electrons away from the ring.
Resonance effects occur due to pi orbital overlap between the substituent atom and the ring. Basically, the substituent will be forming a double bond with the ring as one or more of the resonance structures. For example, a substituent that has oxygen bonded to the ring (OCH3, for example) can have the oxygen form a double bond, thus transferring some of the electron density from the oxygen to the ring. Hence the e-donating property of OCH3.
Hope this helps a bit...if you want, I can discuss this some more. However, I do have to say that while understanding some of the mechanics behind this helps, I thought it was fairly easy to memorize. Alkyl, alkoxy, hydroxy, oxy, phenyl, carboxy, and amine are ortho-para activators (electron donators), halogens are ortho-para deactivators (electron withdrawing b/c the resonance effect here is weaker than the inductive effect), and everything else is meta deactivating (electron-withdrawing).