Hello,
I have question: simply how having what is on pic. A to make what is on pic. B by script ?
I guess it is not so easy so I can pay for making this.
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More explanation: I have railways as lines and railway stations as dots. I can easy make stations with a proper symbol (white rectangle), but it should to be aligned along the railway. And I have many stations like these.
Thank you,
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You might be able to replace the dots with type on a path. Wingdings "n" will give you a solid box. Color it white and add a black stroke, scale the box width as needed. Separate the stations with a tab.
This is a dirty mock-up which does it. (Dirty because it uses Divide Objects Below through executeMenuCommand to get the tangent of the curve, which—as said—is the crux of the problem.) Be warned, there are many variables which will cause failure. But the point is, it can be done.
// select dot
var items = app.activeDocument.pathItems;
var railway = items["railway"];
railway.locked = true;
app.executeMenuCommand("Find Appearance menu item");
var rects = [];
for (var i = selection.length - 1; i > -1; i--) {
var x = selection[i].position[0] + (selection[i].width/2);
var y = selection[i].position[1] - (selection[i].height/2);
selection[i].remove();
rects[i] = items.rectangle(y + 12.5, x - 5, 10, 25);
rects[i].name = "station" + i;
rects[i].strokeColor = app.activeDocument.swatches["Black"].color;
rects[i].fillColor = app.activeDocument.swatches["White"].color;
rects[i].locked = true;
}
for (var i = 0; i < rects.length; i++) {
rects[i].locked = false;
var params = [rects[i].top, rects[i].left, rects[i].width, rects[i].height];
var tempRailway = railway.duplicate();
tempRailway.name = "tempRailway";
tempRailway.move(rects[i], ElementPlacement.PLACEBEFORE);
tempRailway.locked = false;
tempRailway.selected = true;
app.executeMenuCommand("Knife Tool2");
app.activeDocument.selection = null;
if (items[1].top > items[0].top) {
items[1].moveBefore(items[0]);
}
items[0].translate(0, -1);
items[0].selected = true;
app.executeMenuCommand("Knife Tool2");
items[0 + 1].remove();
var T = items[0].height / items[0].width;
var x1 = items[0].pathPoints[0].anchor[0];
var y1 = items[0].pathPoints[0].anchor[1];
var x2 = items[0].pathPoints[2].anchor[0];
var y2 = items[0].pathPoints[2].anchor[1];
if ((x1 < x2) && (y1 > y2)) {
var degree = 90 - (Math.atan(T) * (180 / Math.PI));
} else if ((x1 < x2) && (y1 < y2)) {
var degree = -90 + (Math.atan(T) * (180 / Math.PI));
}
items[0].remove();
rects[i] = app.activeDocument.pathItems.rectangle(params[0], params[1], params[2], params[3]);
rects[i].rotate(degree);
rects[i].move(railway, ElementPlacement.PLACEAFTER);
rects[i].locked = true;
}
railway.move(rects[0], ElementPlacement.PLACEAFTER);
Type on a path, above stacked stroke graphic style.
Beff,
Here are comparative views of appearances based upon the Tangent way and the suggested Two step way, for a gently curved stretch as in your screenshot and for a more strongly curved one.
The stations in the middle, including the original one, are at the inflexion.
Like Kurt, I also considered station shapes following the line shape but abandoned it, possibly wrongly assuming your preferring rectangles as specified in your explanation as opposed to a variety os shapes from straight to (more or less strongly) curved and including inflexion.
It would also be conceivable to prefer an intermediate positioning building on both ways, which would be more complicated than the sum of both single ways.
I have kept forgetting one aspect, namely that the Tangent way requires an actual Anchor Point on the line path at each station (irreversibly changing the stretch(es) in question) whereas the Two step way can be used on the line path as it is.
Bonjour,
Jacob, je n'ai pas tout compris ce que tu veux dire (problème de traduction)
Cependant je viens d'éditer un script rapide qui fait le travail suivant:
Je n'ai pas sélectionné les points extrémitées du tracé afin de montrer le sens du tracé.
Le résultat es identique si j'inverse le sens du tracé.
L'élement placé est un symbole (station mise à l'échelle 60), il peut avoir une toute autre forme..
If this is what you are working with, you can simplify the curves to some extent (Path> Simpify) but you might introduce some errors to the exact location of the tracks. The size of the artboard is 500 cm, is this what you want?
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