Question

好的，不知道如何提出这个问题，但是我试图将值集合到一个数组中，但是每个值可能在其余值上有重复（或附加）的字符串，我相信它们总是建立在彼此之上...

这就是我的意思。

如果我输出一个数组，它将是这样的：

[0] => This is a string.
[1] => here's another string. This is a string.
[2] => And now there's a third string.  here's another string. This is a string.
[3] => Here's a string that might not follow the pattern.  This is a string.

我想最终输出的是

[0] => This is a string.
[1] => here's another string.
[2] => And now there's a third string.
[3] => Here's a string that might not follow the pattern.

基本上，任何字符串都不能包含任何其他字符串中的重复文本。

不确定我是否有道理。

编辑：

以下是完整的故事 - 我通过PHP IMAP使用AJAX脚本收集电子邮件。我从每封电子邮件中获取了正文的文字，但不幸的是，无法通过所有电子邮件服务找到引用的文字。所以我所做的就是让每个电子邮件正文显示，并且我已经删除了任何额外的字符（例如行引号＆gt;＆gt;字符）。

我现在要做的是：从第一封电子邮件开始，检查其他电子邮件，然后删除第一个字符串，然后使用第二封电子邮件，检查其余电子邮件并删除那个字符串，依此类推。

它并不完美，但它会缩短很多信息。

Answer 1

在使用空字符串首次替换新数组中的每个字符串之后，使用旧数组中的每个字符串构建一个新数组。

filestore.allow({
    insert: function () {
        return true;
    },
    update: function () {
        return true;
    },
    remove: function () {
        return true;
    },
    download: function(){
        return true;
    }
});

Answer 2

var strings = [
    "This is a string.",
    "here's another string. This is a string.",
    "And now there's a third string.  here's another string. This is a string.",
    "Here's a string that might not follow the pattern.  This is a string."
];

var i, j, pos;
for (i = 0; i < strings.length; i += 1) {
    for (j = 0; j < strings.length; j += 1) {
        if (i === j) {
            continue;
        }
        pos = strings[i].indexOf(strings[j]);
        if (pos !== -1) {
            strings[i] = strings[i].substr(0, pos) + strings[i].substr(pos + strings[j].length);
        } 
    }
}

console.log(strings);
output: [
  "This is a string.",
  "here's another string. ",
  "And now there's a third string.  ",
  "Here's a string that might not follow the pattern.  "
];

请注意，此代码不会替换多次出现。如果你想要，添加一个while循环。

Answer 3

修改如果字符串具有特殊的正则表达式字符，这将导致问题。没有正则表达式的JS全局repalce可以利用join和var arr = [ 'This is a string.', "here's another string. This is a string.", "And now there's a third string. here's another string. This is a string.", "Here's a string that might not follow the pattern. This is a string." ]; for (var i = 0; i < arr.length; i++) { for (j = 0; j < arr.length; j++) { if (j !== i) { arr[i] = arr[i].split(arr[j]).join(""); } } } console.log(arr); [ "This is a string.", "here's another string.", "And now there's a third string.", "Here's a string that might not follow the pattern." ]，（http://www.adequatelygood.com/JS-Find-and-Replace-with-SplitJoin.html）见下文：

~~Javascript需要正则表达式来进行全局替换，所以像这样~~

{{1}}

它应该替换所有出现的

Answer 4

使用如下。使用jquery的.map函数迭代数组。代码如下：

var arr = [
           'This is a string.',
           "here's another string. This is a string.",
           "And now there's a third string.  here's another string. This is a string.",
           "Here's a string that might not follow the pattern.  This is a string."
         ];

     var out=$.map(arr,function (i,v){
         return i.substr(0,i.indexOf(".") + 1);
     });
     console.log("value is"+out);

Javascript / jQuery在数组的其余部分中查找并删除数组值

4 个答案: