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property display: make sure tags are as short as possible but unique

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mfranz 2008-05-14 15:40:00 +00:00
parent af38cc7c63
commit 356a8430a7
1 changed files with 44 additions and 5 deletions
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49
Nasal/screen.nas
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@ -194,18 +194,57 @@ var property_display = {
reset : func {
me.loopid += 1;
me.window.close();
me.nodes = [];
me.entries = [];
},
nameof : func(n) {
var name = n.getName();
if (var i = n.getIndex())
name ~= '[' ~ i ~ ']';
return name;
},
add : func(n) {
append(me.nodes, isa(n, props.Node) ? n : props.globals.getNode(n, 1));
if (!isa(n, props.Node))
n = props.globals.getNode(n, 1);
var path = n.getPath();
foreach (var e; me.entries) {
if (e[0].getPath() == path)
return;
e[1] = e[0];
e[2] = me.nameof(e[0]);
}
append(me.entries, [n, n, me.nameof(n)]);
# extend names to the left until they are unique
while (1) {
var uniq = {};
foreach (var e; me.entries) {
if (!contains(uniq, e[2]))
uniq[e[2]] = [];
append(uniq[e[2]], e);
}
var finished = 1;
foreach (var u; keys(uniq)) {
if (size(uniq[u]) == 1)
continue;
finished = 0;
foreach (var e; uniq[u]) {
e[1] = e[1].getParent();
if (e[1] != nil)
e[2] = me.nameof(e[1]) ~ '/' ~ e[2];
}
}
if (finished)
break;
}
me._loop_(me.loopid += 1);
},
update : func {
me.window.lines = [];
foreach (var n; me.nodes) {
if ((val = n.getValue()) == nil)
foreach (var e; me.entries) {
if ((var val = e[0].getValue()) == nil)
val = "nil";
append(me.window.lines, [n.getName() ~ " = " ~ sanitize(val, 1),
append(me.window.lines, [e[2] ~ " = " ~ sanitize(val, 1),
me.color[0], me.color[1], me.color[2], me.color[3]]);
}
me.window.show();