This question made me think: suppose you have an irregular hexagon $ABCDEF$, and you know the edge lengths $AB,BC,CD,DE,EF,FA$ as well as the diagonal lengths $AD,BE,CF$. Then counting degrees of freedom suggests that up to isometric transformations there should be only finitely many hexagons with the given lengths. (That's $6\times2$ coordinates for the vertices, minus $3$ real degrees of freedom for translation and rotation, leaves $9$ degrees to be fixed by the $9$ given lengths.)
But am I right, or is there some hidden relation between the $9$ lengths so that one of them can be computed from all the others and therefore does not add any information? And if I am right, how many distinct solutions are there at most? Is the solution perhaps even unique (counting mirror images as equivalent)?
And how to construct these solutions using ruler and compass? A description showing how to construct $C$ given $A,B$ and all the lengths should be enough to complete the whole hexagon by repeating the same process for every subsequent edge.
The problem appears hard because no proper subset of vertices forms a configuration that would be rigid under the given length constraints. One can't simply construct one triangle after the other because no three of the given segments form a triangle.
This picture shows two hexagons $ABCDEF$ and $ABC'DEF'$, with equal sides and diagonals but not congruent between them. So the construction is in general not possible, unless you specify some constraint.
EDIT
In general, it is not difficult to construct two hexagons like that with GeoGebra (see diagram below). Start with a side $AB$, common to both hexagons, then add vertices $C$ and $C'$ with the only constraint that $BC=BC'$ (second sides of hexagons). Now add the fourth vertex $D$ of the first hexagon: vertex $D'$ is then fixed by the conditions $CD=C'D'$ and $AD=AD'$. The same goes for $E$ (chosen at will) and $E'$ (fixed by $DE=D'E'$ and $BE=BE'$).
Finally, one has to choose the last vertex $F$ of the first hexagon, but $F'$ has then to satisfy three conditions: $EF=E'F'$, $AF=AF'$ and $CF=C'F'$. That is, the three dashed circles in the diagram (represented with the same color as their centers) should meet at the same point $F'$. This condition, of course, is not in general satisfied for any choice of $F$. But "experimentally" it is not difficult to adjust $F$ so that $F'$ exists.
To turn that construction into an existence proof should then be tedious but possible.