我试图用多个其他单词替换字符串中的多个单词。字符串是“我有一只猫,一只狗和一只山羊。”
然而,这并不会产生“我有一只狗、一只山羊和一只猫”,而是产生“我有一只猫、一只猫和一只猫”。是否有可能在JavaScript中同时用多个其他字符串替换多个字符串,以便产生正确的结果?
var str = "I have a cat, a dog, and a goat.";
str = str.replace(/cat/gi, "dog");
str = str.replace(/dog/gi, "goat");
str = str.replace(/goat/gi, "cat");
//this produces "I have a cat, a cat, and a cat"
//but I wanted to produce the string "I have a dog, a goat, and a cat".
以防有人想知道为什么原来海报上的解决方案不管用:
var str = "I have a cat, a dog, and a goat.";
str = str.replace(/cat/gi, "dog");
// now str = "I have a dog, a dog, and a goat."
str = str.replace(/dog/gi, "goat");
// now str = "I have a goat, a goat, and a goat."
str = str.replace(/goat/gi, "cat");
// now str = "I have a cat, a cat, and a cat."
var str = "I have a cat, a dog, and a goat.";
str = str.replace(/goat/i, "cat");
// now str = "I have a cat, a dog, and a cat."
str = str.replace(/dog/i, "goat");
// now str = "I have a cat, a goat, and a cat."
str = str.replace(/cat/i, "dog");
// now str = "I have a dog, a goat, and a cat."
我修改了本·麦考密克的答案以配合你的新测试用例。
我只是在正则表达式中添加了单词边界:
/\b(cathy|cat|catch)\b/gi
“运行代码片段”可以看到下面的结果:
var str = "我有一只猫,一个catch,和一个cathy.";
var mapObj = {
凯茜:“猫”,
猫:“抓”,
抓住:“凯蒂”
};
STR = STR .replace(/\b(cathy|cat|catch)\b/gi, function(matched){
返回mapObj(匹配);
});
console.log (str);
具体的解决方案
您可以使用一个函数来替换每一个。
var str = "I have a cat, a dog, and a goat.";
var mapObj = {
cat:"dog",
dog:"goat",
goat:"cat"
};
str = str.replace(/cat|dog|goat/gi, function(matched){
return mapObj[matched];
});
jsfiddle例子
概括它
如果您想动态地维护正则表达式,并且只是将未来的交换添加到映射中,您可以这样做
new RegExp(Object.keys(mapObj).join("|"),"gi");
生成正则表达式。就像这样
var mapObj = {cat:"dog",dog:"goat",goat:"cat"};
var re = new RegExp(Object.keys(mapObj).join("|"),"gi");
str = str.replace(re, function(matched){
return mapObj[matched];
});
要添加或更改任何替换,您只需编辑地图。
摆弄动态正则表达式
可重复使用
如果你想让它成为一般形式你可以把它变成这样一个函数
function replaceAll(str,mapObj){
var re = new RegExp(Object.keys(mapObj).join("|"),"gi");
return str.replace(re, function(matched){
return mapObj[matched.toLowerCase()];
});
}
然后你可以把str和你想要的替换的映射传递给函数它会返回转换后的字符串。
摆弄函数
确保对象。key适用于旧的浏览器,添加一个填充,例如从MDN或Es5。
注意!
如果您正在使用动态提供的映射,这里的解决方案都不够!
在这种情况下,有两种解决方法:(1)使用分割连接技术,(2)使用正则表达式和特殊字符转义技术。
这是一个分割连接技术,它比另一个快得多(至少快50%):
var str = "I have {abc} a c|at, a d(og, and a g[oat] {1} {7} {11."
var mapObj = {
'c|at': "d(og",
'd(og': "g[oat",
'g[oat]': "c|at",
};
var entries = Object.entries(mapObj);
console.log(
entries
.reduce(
// Replace all the occurrences of the keys in the text into an index placholder using split-join
(_str, [key], i) => _str.split(key).join(`{${i}}`),
// Manipulate all exisitng index placeholder -like formats, in order to prevent confusion
str.replace(/\{(?=\d+\})/g, '{-')
)
// Replace all index placeholders to the desired replacement values
.replace(/\{(\d+)\}/g, (_,i) => entries[i][1])
// Undo the manipulation of index placeholder -like formats
.replace(/\{-(?=\d+\})/g, '{')
);
这一个,是Regex特殊字符转义技术,它也有用,但慢得多:
var str = "I have a c|at, a d(og, and a g[oat]."
var mapObj = {
'c|at': "d(og",
'd(og': "g[oat",
'g[oat]': "c|at",
};
console.log(
str.replace(
new RegExp(
// Convert the object to array of keys
Object.keys(mapObj)
// Escape any special characters in the search key
.map(key => key.replace(/[-[\]{}()*+?.,\\^$|#\s]/g, '\\$&'))
// Create the Regex pattern
.join('|'),
// Additional flags can be used. Like `i` - case-insensitive search
'g'
),
// For each key found, replace with the appropriate value
match => mapObj[match]
)
);
后者的优点是,它也可以用于不区分大小写的搜索。