Question

我想从数组中提取数组。

在下面的小提琴中，我希望得到的数组从第一个“bob”出现开始，并在“bob”出现的最后一个结束。在小提琴中我放了项目的全名，而我希望它开始和结束时只看到“bob”。

var array = ["rrrrrrr-18082015-144751", "rrrrrrr-18082015-145552", "rrrrrrr-30072015-182930", "rrrrrrr-30072015-184607", "bob-04112015-172944", "bob-04112015-173503", "bob-06112015-175858", "bob-06112015-180111", "bob-09112015-092450", "bob-09112015-133119", "bob-09112015-135310", "bob-09112015-143033", "bob-10112015-094836", "bob-10112015-102459", "ggggggggggg-28082015-113014"];

var x = array.splice(array.indexOf("bob-04112015-172944"),array.lastIndexOf("bob-10112015-102459"));

所以我想要的是：

array.splice(array.indexOf("bob"),array.lastIndexOf("bob"));

有办法吗？

https://jsfiddle.net/jdsy6bnm/

提前谢谢。

Answer 1

我首先说“bob”出现不是“bob-04112015-172944”的第一次出现，并且最后一次出现。

如果目标是获取从第一个bob开始到最后一个bob结束的所有条目，包括中间的任何非bobs，那么找到索引没有真正的快捷方式，你有循环：

var first = -1, last = -1;
array.forEach(function(entry, index) {
    if (entry.indexOf("bob") !== -1) {
      if (first === -1) {
        first = index;
      }
      last = index;
    }
});
var bobs = array.splice(first, last - first + 1);

var array = ["rrrrrrr-18082015-144751", "rrrrrrr-18082015-145552", "rrrrrrr-30072015-182930", "rrrrrrr-30072015-184607", "bob-04112015-172944", "bob-04112015-173503", "bob-06112015-175858", "bob-06112015-180111", "bob-09112015-092450", "bob-09112015-133119", "bob-09112015-135310", "bob-09112015-143033", "bob-10112015-094836", "bob-10112015-102459", "ggggggggggg-28082015-113014"];

var first = -1, last = -1;
array.forEach(function(entry, index) {
    if (entry.indexOf("bob") !== -1) {
      if (first === -1) {
        first = index;
      }
      last = index;
    }
});
var bobs = array.splice(first, last - first + 1);

snippet.log("updated array: " + JSON.stringify(array));
snippet.log("bobs: " + JSON.stringify(bobs));

<!-- Script provides the `snippet` object, see http://meta.stackexchange.com/a/242144/134069 -->
<script src="//tjcrowder.github.io/simple-snippets-console/snippet.js"></script>

我在那里使用了splice，这将删除条目;您可以使用slice来保留它们：

var bobs = array.slice(first, last);

如果目标是查找所有bobs，而不是中间的条目，那么filter：

var bobs = array.filter(function(entry) {
    return entry.indexOf("bob") !== -1;
});

示例：

var array = ["rrrrrrr-18082015-144751", "rrrrrrr-18082015-145552", "rrrrrrr-30072015-182930", "rrrrrrr-30072015-184607", "bob-04112015-172944", "bob-04112015-173503", "bob-06112015-175858", "bob-06112015-180111", "bob-09112015-092450", "bob-09112015-133119", "bob-09112015-135310", "bob-09112015-143033", "bob-10112015-094836", "bob-10112015-102459", "ggggggggggg-28082015-113014"];

var bobs = array.filter(function(entry) {
    return entry.indexOf("bob") !== -1;
});

snippet.log(JSON.stringify(bobs));

<!-- Script provides the `snippet` object, see http://meta.stackexchange.com/a/242144/134069 -->
<script src="//tjcrowder.github.io/simple-snippets-console/snippet.js"></script>

在该示例中，不会更改

array。如果你想改变它，你可能想要使用forEach并构建两个新数组：

var bobs = [];
var nonbobs = [];
array.forEach(function(entry) {
    var a = entry.indexOf("bob") === -1 ? nonbobs : bobs;
    a.push(entry);
});

Answer 2

首先，您错误地使用splice：第一个参数是起始索引，第二个参数是序列的长度，而不是结束索引。因此，您应该执行类似

的操作

array.splice(start, end - start);

要查找以＆＃34; bob＆＃34;开头的第一个和最后一个项目，您可以从两侧迭代数组并检查每个项目是否符合您的标准，如果匹配，则保存其索引。

var array = ["rrrrrrr-18082015-144751", "rrrrrrr-18082015-145552", "rrrrrrr-30072015-182930", "rrrrrrr-30072015-184607", "bob-04112015-172944", "bob-04112015-173503", "bob-06112015-175858", "bob-06112015-180111", "bob-09112015-092450", "bob-09112015-133119", "bob-09112015-135310", "bob-09112015-143033", "bob-10112015-094836", "bob-10112015-102459", "ggggggggggg-28082015-113014"];

var first = -1, last = -1;
for (var i = 0; i < array.length; i++) {
    if (first < 0 && array[i].indexOf('bob') === 0)
        first = i;
    if (last < 0 && array[array.length - i - 1].indexOf('bob') === 0)
        last = array.length - i - 1;
}    

var x = array.splice(first, last - first);

Your updated JSFiddle

其他细节： 你可能会想要使用startsWith而不是indexOf这个函数，但这个浏览器并没有得到很好的支持，所以我强烈建议使用后者。

Answer 3

使用Array.prototype.slice。

更新了JSFiddle：https://jsfiddle.net/jdsy6bnm/1/

＆＃13;

var array = [
  "rrrrrrr-18082015-144751", "rrrrrrr-18082015-145552",
  "rrrrrrr-30072015-182930", "rrrrrrr-30072015-184607",
  "bob-04112015-172944", "bob-04112015-173503", 
  "bob-06112015-175858", "bob-06112015-180111", 
  "bob-09112015-092450", "bob-09112015-133119", 
  "bob-09112015-135310", "bob-09112015-143033", 
  "bob-10112015-094836", "bob-10112015-102459", 
  "ggggggggggg-28082015-113014"
];

var array = array.slice(
  array.indexOf('bob-04112015-172944'),
  array.lastIndexOf("bob-10112015-102459")
);

console.log(array);

＆＃13;

Answer 4

尝试使用while循环，RegExp.prototype.test()

var array = ["rrrrrrr-18082015-144751"
             , "rrrrrrr-18082015-145552"
             , "rrrrrrr-30072015-182930"
             , "rrrrrrr-30072015-184607"
             , "bob-04112015-172944"
             , "bob-04112015-173503"
             , "bob-06112015-175858"
             , "bob-06112015-180111"
             , "bob-09112015-092450"
             , "bob-09112015-133119"
             , "bob-09112015-135310"
             , "bob-09112015-143033"
             , "bob-10112015-094836"
             , "bob-10112015-102459"
             , "ggggggggggg-28082015-113014"];

var res = []
, i = 0
, re = /bob/;
while (i < array.length) {
  if (re.test(array[i]) && re.test(array[i + 1])) {
    res.push(array[i], array[i + 1])
  } else {
    if (/bob/.test(array[i]) && !/bob/.test(array[i + 1]) 
        && /bob/.test(array[i - 1])) {
      break;
    }
  }
  ++i;
}

console.log(res)

在javascript

4 个答案: