У меня многомерный массив в javascript. Я хочу сделать так, чтобы он показывал первую запись для каждого уникального "имени". Так, например, это:
[{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Aellen , Nick","col1":"Director","col2":"test","col3":"test","col4":"96.88%"},
{"name":"Armstrong , Peter","col1":"Director","col2":"test","col3":"test","col4":"95.15%"},
{"name":"Ashmore , Abigail","col1":"Vice President","col2":"test","col3":"test","col4":"99.80%"},
{"name":"Avent , Christopher","col1":"Vice President","col2":"test","col3":"test","col4":"90.11%"},
{"name":"Baker , Ian Howard","col1":"Director","col2":"test","col3":"test","col4":"67.20%"},
{"name":"Baker , Ian Howard","col1":"Director","col2":"test","col3":"test","col4":"67.20%"},
Включилось бы в это:
[{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Aellen , Nick","col1":"Director","col2":"test","col3":"test","col4":"96.88%"},
{"name":"Armstrong , Peter","col1":"Director","col2":"test","col3":"test","col4":"95.15%"},
{"name":"Ashmore , Abigail","col1":"Vice President","col2":"test","col3":"test","col4":"99.80%"},
{"name":"Avent , Christopher","col1":"Vice President","col2":"test","col3":"test","col4":"90.11%"},
{"name":"Baker , Ian Howard","col1":"Director","col2":"test","col3":"test","col4":"67.20%"},
Первоначально это не было многовариантным, и я мог бы использовать базовую функцию, но этот метод не будет работать для этого. Есть идеи? Довольно отчаянный! Благодарю.
Что-то вроде этого создаст новый массив с единственными уникальными записями
var obj = {},
new_arr = [];
arr.forEach(function(itm) {
if ( ! (itm.name in obj) ) obj[itm.name] = itm;
});
for (var key in obj) {
new_arr.push(obj[key]);
}
new_arr.push(obj)
? Вместо того, for in
цикле.
Это должно сделать трюк:
var foos = [{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Aellen , Nick","col1":"Director","col2":"test","col3":"test","col4":"96.88%"},
{"name":"Armstrong , Peter","col1":"Director","col2":"test","col3":"test","col4":"95.15%"},
{"name":"Ashmore , Abigail","col1":"Vice President","col2":"test","col3":"test","col4":"99.80%"},
{"name":"Avent , Christopher","col1":"Vice President","col2":"test","col3":"test","col4":"90.11%"},
{"name":"Baker , Ian Howard","col1":"Director","col2":"test","col3":"test","col4":"67.20%"},
{"name":"Baker , Ian Howard","col1":"Director","col2":"test","col3":"test","col4":"67.20%"}];
var new_foos = [];
var previous_foo = null;
for (var i = 0; i < foos.length; i++) {
if (previous_foo != foos[i].name)
{
new_foos.push(foos[i]);
}
previous_foo = foos[i].name;
}
Как упоминалось в комментариях, для этого требуется, чтобы foo был отсортирован. Это работает, если foo является несортированным:
var foos = [{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Baker , Ian Howard","col1":"Director","col2":"test","col3":"test","col4":"67.20%"},
{"name":"Ashmore , Abigail","col1":"Vice President","col2":"test","col3":"test","col4":"99.80%"},
{"name":"Baker , Ian Howard","col1":"Director","col2":"test","col3":"test","col4":"67.20%"},
{"name":"Aellen , Nick","col1":"Director","col2":"test","col3":"test","col4":"96.88%"},
{"name":"Armstrong , Peter","col1":"Director","col2":"test","col3":"test","col4":"95.15%"},{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Avent , Christopher","col1":"Vice President","col2":"test","col3":"test","col4":"90.11%"}];
var new_foos = [];
for (var i = 0; i < foos.length; i++) {
var new_foo = true;
for (var j = 0; j < new_foos.length; j++) {
if (foos[i].name == new_foos[j].name) {
new_foo = false;
break;
}
}
if (new_foo)
{
new_foos.push(foos[i]);
}
}