in general
As others have pointed out, a voltage source and a resistor can be used. But... with the following rule:
Start with a lot of voltage and throw almost all of it away!If you know the voltage compliance range over which a certain worst case current regulation must exist, then:
$$\%I_{_\text{SRC}}=\%V_{_\text{COMPLIANCE}}\cdot\frac1{\frac{V_{_\text{SRC}}}{V_{_\text{COMPLIANCE}}}-1}$$
illustration making and clarifying the above point
Suppose you need \$I_{_\text{SRC}}=100\:\mu\text{A}\$ and regulation to no more than 1%, or \$\%I_{_\text{SRC}}=\frac{\Delta\,I_{_\text{SRC}}=1\:\mu\text{A}}{I_{_\text{SRC}}=100\:\mu\text{A}}=0.01\$.
Suppose the voltage compliance range is \$0\:\text{V}\dots 10\:\text{V}\$ so \$\%V_{_\text{COMPLIANCE}}=\frac{\Delta\, V_{_\text{COMPLIANCE}}=5\:\text{V}}{V_{_\text{COMPLIANCE}}=5\:\text{V}}=1.00\$.
Then this means that the required \$V_{_\text{SRC}}=5\:\text{V}\cdot\left(1+\frac1{\frac{0.01}{1.00}}\right)=505\:\text{V}\$!
simulate this circuit– Schematic created using CircuitLab
If you short out the load, then \$I=\frac{505\:\text{V}}{5\:\text{M}\Omega}=101\:\mu\text{A}\$. If the load requires \$10\:\text{V}\$ (the maximum given in the earlier specification above), then \$I=\frac{505\:\text{V}-10\:\text{V}}{5\:\text{M}\Omega}=99\:\mu\text{A}\$.
Note that this meets the specification that I've written above.
summary
The above shows the theory and applies it to one extreme circumstance demanding only 1% variation in order to make a point.
So. Yes. You can have a current source without transistors and amplifiers and only using a voltage source and a resistor to get there.
But... there are caveats.