正则表达式字边界打破' - '

I have a string in the following format.

我有一个以下格式的字符串。

"ad60 ad60-12 ad60 12-ad60"

“ad60 ad60-12 ad60 12-ad60”

now I want to find the matches only where "ad60" is written.

现在我想找到只写“ad60”的匹配项。

I started off with

我开始了

/\bad60\b/i

/ \ bad60 \ B / I

but regex finds that '-' is not part of the character string. which returns 4 matches. I tried many things but they all either return 4 matches or return nothing.

但正则表达式发现' - '不是字符串的一部分。返回4场比赛。我尝试了很多东西，但他们要么返回4场比赛，要么什么也不返回。

Any kind of help would be appreciated.

任何形式的帮助将不胜感激。

3 个解决方案

#1

You can use:

您可以使用：

var s = "ad60 ad60-12 ad60 12-ad60";
var r = s.replace(/(^|\s)ad60(?=\s|$)/g, "$1@@");
//=> @@ ad60-12 @@ 12-ad60

#2

this should find any word starting from letter

这应该找到从字母开始的任何单词

/[a-z][a-z\d]+/i

ADDED

添加

if you wish to find any words which are included ad60 string try this

如果你想找到包含ad60字符串的任何单词，试试这个

/[a-z\d\-\_]*ad60[a-z\d\-\_]*/i

but words separated by space will not get in result

但是用空格分隔的单词不会得到结果

#3

"ad60 ad60-12 ad60 12-ad60".match(/(^ad60\s+)|(\s+ad60$)|(\s+ad60\s+)|(^ad60$)/ig);

will result in ["ad60 ", " ad60 "] then you can trim the white spaces in matched elements.

将导致[“ad60”，“ad60”]然后您可以修剪匹配元素中的空白区域。

#1