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Overpass API: for statement

[out:csv("name";false)];

area[name="Pfungen"];

way(area)[highway][name];

for (t["name"]) (
   make x name=_.val;
   out; 
);

[out:csv(
   'id', 'key', 'val';
    false;
   '|'
)];

node[name='Freienstein'] -> .nd;

for.nd -> .id (id())  {
  for.id -> .ky (keys()) {    
    for.ky -> .vl(t[ky.val]) {
      make x
          id  = id.val, 
          key = ky.val,
          val = vl.val;
      out;
    }
  }
};

for is GROUP BY in disguise

The for statement behaves like the SQL GROUP BY clause where the expression in the parantheses (here: t['name']) is the value on which the other elements are grouped (and/or aggregated). 
[ out : csv(
   name,
   total_length,
   cnt_ways,
   min_length,
   max_length,
   avg_length,
   all_lengths
)];

{{geocodeArea: Pfungen }};

way[highway](area);

for (t['name']) {

   make Beispiel
     name         = _.val,
     total_length = sum(length()),                // Aggregate all lengths
     cnt_ways     = count(ways)  ,
     min_length   = min(length()),
     max_length   = max(length()),
     avg_length   = sum(length()) / count(ways),  // There is no function avg()
     all_lengths  = set(length())                 // Concatenate the individual lengths with semicolons.
   ;

   out;
}
Of course, we can use a constant expression as group by expression so that we can, for example, calculate the total length of multiple ways: 
[out:csv(
     length,
     cnt,
     name
)];

(
   way(228531783);  // Length = 60.520
   way(228533515);  // Length = 38.373
) -> .ways;


for.ways (1) { // Group by constant value -> get aggregate for all elements
  make x
    cnt    = count(ways),
    length = sum(length());
  out;
}

See also

The foreach statement.

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