Does increasing the length of wire increase the work (energy) needed to move electrons from the negative terminal to the positive one? Imagine a circuit which consists of a battery and a resistor with a certain wire length. Suppose that the battery supplies 9 Volts. I know that it is the electromotive force which pushes the electrons from the negative terminal to the positive terminal. Does 9 Volts mean that the energy needed to move one coulomb from the negative terminal to the positive terminal is 9 Joules? If so, then what happens if we have a longer wire? It seems that that more work is done to move the electron but the battery is still supplying 9 Volts so the work to move 1 coulomb is still 9 Joules. Shouldn't the work depend on distance?!
 A: Okay, if I have understood you correctly, your doubt is not whether the length of the wire affects the circuit, but if the work done is the same from one end to another, right?
Because you already know that ideal wires have zero resistance. That means that it is not relevant to have $1 \; \text{m}$ or $10 \; \text{m}$; if it is an ideal wire, the current is not affected by it, so the length doesn't matter. [unless yo care about wavelengths]
However, real wires have a tiny but non-zero resistance, so the longer they are, the more resistance you add to the circuit, so the current decreases (slightly)
This has already been addressed here, I think there's no doubt here. If so, add a comment.

Work done
So, if I understood you correctly, your question is something like: "why doesn't the length matter, if $W= \text{Force} \times \text{distance}$ and the battery applies a constant energy?" Am I right that it was your question?
Well, if so, the answer is an important conceptual idea: the electric field inside the battery is different than that of the rest of the circuit.
You know that the electrostatic force is conservative. That means that the work done by the electrostatic force does not depend on the path followed, the work done depends only on the initial and final positions.
When this happens, you can define an electrostatic potential (voltage).
Okay, but, with moving charges, we're not in electrostatics anymore. We talk about "electric" field, rather than electrostatic. This one is more general.
Can we still apply our knowledge of electrostatics? Yes, but only in points where $\vec{\nabla}\times\vec{E}=\vec{0}$
This happens in the rest of the circuit. That's why we can still use a electric potential, although the charges are moving.
But there is a region where $\vec{\nabla}\times\vec{E} \neq \vec{0}$. Otherwise there wouldn't be energy for the charges to move. That region is the battery.
Inside the battery, there is a different electric field, which is a non-conservative electric field, and that means that the work done by it does depend on the path.
Inside the battery, the work done rises the charges from $0\; \text{V}$ to $9\; \text{V}$, but following that specific path.
In sum, the conclusions are:

A battery is "something that takes one end at $0V$ and injects $9\; \text{V}$ at the other end". It's like a pump that takes $0 \;\text{V}$ and raises it to $9V$.
Inside the battery, the electric field behaves differently.
However, in the rest of the circuit, the electric field keeps being conservative, so that the voltage exists. And, what's more, the work done does not depend on the path.

In a less-technical way, a classical (non-rigorous) and easier analogy would be something like this:
The battery accelerates charges. There is force inside the battery.
After being released from the battery, the electrons do not suffer a force anymore. They are only slowed down in resistances and other elements of the circuit, but they are not accelerated forth.
If they are not accelerated, there is no force, there is no work, regardless of the path.
A: It doesn't matter how long the wire is if the voltage across the wire is 9 volts it means 9 Joules of work is done in moving 1 Coulomb of charge from one end of the wire to the other. But increasing the length of the wire increases its resistance and reduces the current for the same potential difference (voltage),  per Ohms law $V=IR$.  The work per unit charge in moving the charge along the length of wire does not change unless there is a change in potential difference between the ends of the wire. 
However, a reduction in current means less Coulombs per second going from one end of the wire to the other. That results in a reduction in the power delivered by the battery. The rate at which work is done (power, $P$) is $P=VI$, = Joules/Coulomb x Coulombs/s= Joules/s= watts.

but why the length doesn't matter, the resistance of the wire is very
  tiny so we can consider the wire has no resistance.now, doesn't moving
  the electron a longer distance mean doing more work on it? .I want to
  know why the length of the wire doesn't matter?

If the wire has zero resistance, then from Ohms law, $V=IR$, the voltage (potential difference) between the ends of the wire must also be zero. Since voltage is defined as the work per Coulomb of charge to move the charge between the two points, if the voltage between the two points is zero, no work would be done to move the charge between the two points of a zero resistance wire, regardless of how long it is.  
So you might ask, what happens to the 9 volts the battery produces if it is connected to a zero resistance wire? 
A real battery always has internal resistance.  Otherwise it could deliver an infinite amount of current to a wire of zero resistance, since $I=\frac{V}{0}=∞$. 
The circuits below shows a real battery with an emf of 9 volts and internal battery resistance of $R_b$ in series with the wire. The top circuit shows the wire having resistance R. The bottom shows the wire having zero resistance, as you asked about.  
The current in the bottom circuit is
$$I=\frac{9v}{R_b}$$
All of the batteries 9 volt emf is across its internal resistance $R_b$.  There is no voltage across the wire. Therefore no work is done moving charge in the wire connected between the positive and negative terminals of the battery regardless of the length of the wire. All the work the battery does is in moving the charge through its internal resistance $R_b$..
Bottom Line: Changing the length of the wire only matters if it changes the potential difference (voltage) between the ends of the wire.
Hope this helps.

