Double slit: why don't the edges of the slit constitute a measuring device? The usual explanation is that any interaction with a 'classical' system will collapse the wave function and destroy the interference pattern.  So, a recorder which is left on but never observed, will still collapse the wave function.
But the metal (or whatever) plate which contains the slits is also a 'classical' object and most of the wave/photon gets blocked by the plate.  Why does that blocking interaction not count as measurement?  The fact that the blockage happens all around and doesn't provide which-way information is a higher level semantic concept.  At the low level, the photon interacted with a classical object, so why no collapse?
 A: Those photons that are absorbed by the first screen can be considered to be measured by it, but they do not contribute to the interference pattern on the second screen. The photons that pass through the slits and do reach the second screen do not interact with the first screen and the first screen does not record their passing. Therefore those photons are not measured by it.
A similar picture also describes the alternative histories of a single photon. The amplitude for a photon to reach a particular point on the second screen is the sum of the amplitudes of paths that reach that point. In those histories the photon does not get absorbed by the first screen. The paths that lead to absorption at the first screen contribute to the amplitude for the photon not reaching the second screen at all.
If a detector is placed at one of the slits to record which way the particle went (I think this only makes sense for massive particles like electrons), then paths through different slits lead to different final states: either just a mark at a point on the final screen or a mark on the final screen and a recording on the detector at the slit. Amplitudes leading to different final states are summed separately. The probability of observing a particle at some point on the screen is the sum of the probabilities, not amplitudes, of different final sites that have an electron at that position. This makes the interference pattern disappear.
A: This is an extended answer to the comment by @BenitoCiaro on the answer by @DanielMahler (which is correct and complete).
The photon does not interact with itself. The wave functions are superposed, that has nothing whatsoever to do with interaction!
If the photon does not pass through any slit, i.e. not contribute to the interference pattern anyway, it will (if you prefer that formulation) collapse due to "being measured" by the plate (i.e. interacting with the plate and getting absorbed).
By tracing out the environmental states (i.e. micro-states of the plates) you will get a density matrix of the form 
$$\rho = p_\text{abs} \left|0\right>\left<0 \right| + (1 - p_\text{abs}) \frac 1 2 \big( \left|l\right> + \left|r\right>\big)\big(\left<l\right| + \left< r \right|\big)$$
for the photon, where $p_\text{abs}$ is the probability the photon is absorbed by the plates, and $\left|0\right>$, $\left|l\right>$, $\left|r\right>$ are the states where there is no photon and where the photon passed through the left resp. right slit. That is, you will have a classical probabilities you'll get one of two "collapsed" options: In the one case the system collapses into a state where there is no longer a photon (it got absorbed, if you will measured, by the wall) and in the other state there is a coherent superposition of the photon that went through the left, resp. the right slit.
In other words: Measurement is just a special interaction where you entangle your classical measurement apparatus with your quantum system. This results in classical amplitudes once you trace out the micro-states of your measurement device/environment (which are macroscopically indistinguishable). There is no magic in measurement per se which causes the system to enter a position eigenstate.
(If you like I can try to provide a more complete model calculation which demonstrates that the density matrix I give above is indeed correct).
A: In a double slit experiment a measurement can be made by either blocking a slit or watching the experiment with light. Either way you will interfere with the photon in question and eliminate or change the pattern.
 Its interesting that you mention the edges. What if you were to hypothetically follow one single photon? When it gets to the wall that has the slits, the photon will either go through one of the slits or it will impact the wall. If it impacts the wall it will either be absorbed or it will be absorbed and re emitted as a new photon in a random direction but not through a slit. If the photon goes through a slit it will continue and become part of the pattern on the far detection screen.
 There should be a third option and if a photon happened to impact on one of the four edges of the two slits it would again be absorbed. This time when a new photon is re emitted in a random direction it will sometimes travel in the direction of the detection screen. Photons emitted from any of these four edges will form a single edge diffraction patterns on the far screen. Single edge patterns are different than slit patterns. When you combine and overlap these four single edge patterns you will get a double slit pattern. Even if you send one photon at a time. 
