I am totally confused right now.
我现在完全糊涂了。
What is a word that matches: ^.*(?=.*\\d)(?=.*[a-zA-Z])(?=.*[!@#$%^&]).*$
什么是词相匹配:^ . *(? = . * \ \ d)(? =。*[a-zA-Z])(? =。*[! @ # $ % ^ &])。* $
I tried at Regex 101 this 1Test@!. However that does not work.
我尝试了Regex 101这个1Test@!然而，这并不奏效。
I really appreciate your input!
非常感谢你的投入!
4 个解决方案
#1
3
What happens is that your regex seems to be in Java-flavor (Note the \\d)
你的正则表达式似乎是在java风格中(注意\\d)
that is why you have to convert it to work with regex101 which does not work with jave (only works with php, phyton, javascript)
这就是为什么要将它转换为与regex101一起工作的原因，而regex101与jave不兼容(只适用于php、phyton、javascript)
see converted regex:
看到转换正则表达式:
^.*(?=.*\d)(?=.*[a-zA-Z])(?=.*[!@#$%^&]).*$
which will match your string 1Test@!. Demo here: http://regex101.com/r/gE3iQ9
这将匹配您的字符串1Test@!演示:http://regex101.com/r/gE3iQ9
#2
1
You just want something that matches that regex?
你只是想要一些和正则表达式匹配的东西?
Here:
在这里:
a1a!
#3
1
This pattern matches
这种模式匹配
\dTest@!
\ dTest@ !
if u want a pattern which matches 1Test@! try this pattern
如果您想要一个匹配1Test@的模式!试试这个模式
^.(?=.\d)(?=.[a-zA-Z])(?=.[!@#$%^&]).*$
(? = ^。\ d)(? =。[a-zA-Z])(? =。[! @ # $ % ^ &])。*美元
#4
1
Your java string ^.*(?=.*\\d)(?=.*[a-zA-Z])(?=.*[!@#$%^&]).*$ encodes the regexp expression ^.*(?=.*\d)(?=.*[a-zA-Z])(?=.*[!@#$%^&]).*$.
您的java字符串^。*(? = . * \ \ d)(? =。*[a-zA-Z])(? =。*[! @ # $ % ^ &])。* ^ $编码regexp表达式。*(? = . * \ d)(? =。*[a-zA-Z])(? =。*[! @ # $ % ^ &])。*美元。
This is because the \ is an escape sequence.
这是因为\是一个转义序列。
The latter matches the string you specified.
后者匹配您指定的字符串。
If your original string was a regexp, rather than a java string, it would match strings such as \dTest@!
如果您的原始字符串是regexp，而不是java字符串，那么它将匹配字符串，比如\dTest@!
Also you should consider removing the first .*, doing so would make the regexp more efficient. The reason is that regexp's by default are greedy. So it will start by matching the whole string to the initial .*, the lookahead will then fail. The regexp will backtrack, matchine the first .* to all but the last character, and will fail all but one of the loohaheads. This will proceed until it hits a point where the different lookaheads succeed. Dropping the first .*, putting the lookahead immidiately after the start of string anchor, will avoid this problem, and in this case the set of strings matched will be the same.
此外，您应该考虑删除第一个。*，这样做会使regexp更有效。原因是regexp默认是贪婪的。因此，它将首先匹配整个字符串到初始值。regexp将会回溯，matchine第一个。*除了最后一个角色，而且将会失败除了一个loohaheads。这将继续下去，直到它达到一个不同的lookahead成功的点。删除第一个。*，在字符串锚开始后，将lookahead放在前面，将避免这个问题，在这种情况下，匹配的字符串集将是相同的。
#1
3
What happens is that your regex seems to be in Java-flavor (Note the \\d)
你的正则表达式似乎是在java风格中(注意\\d)
that is why you have to convert it to work with regex101 which does not work with jave (only works with php, phyton, javascript)
这就是为什么要将它转换为与regex101一起工作的原因，而regex101与jave不兼容(只适用于php、phyton、javascript)
see converted regex:
看到转换正则表达式:
^.*(?=.*\d)(?=.*[a-zA-Z])(?=.*[!@#$%^&]).*$
which will match your string 1Test@!. Demo here: http://regex101.com/r/gE3iQ9
这将匹配您的字符串1Test@!演示:http://regex101.com/r/gE3iQ9
#2
1
You just want something that matches that regex?
你只是想要一些和正则表达式匹配的东西?
Here:
在这里:
a1a!
#3
1
This pattern matches
这种模式匹配
\dTest@!
\ dTest@ !
if u want a pattern which matches 1Test@! try this pattern
如果您想要一个匹配1Test@的模式!试试这个模式
^.(?=.\d)(?=.[a-zA-Z])(?=.[!@#$%^&]).*$
(? = ^。\ d)(? =。[a-zA-Z])(? =。[! @ # $ % ^ &])。*美元
#4
1
Your java string ^.*(?=.*\\d)(?=.*[a-zA-Z])(?=.*[!@#$%^&]).*$ encodes the regexp expression ^.*(?=.*\d)(?=.*[a-zA-Z])(?=.*[!@#$%^&]).*$.
您的java字符串^。*(? = . * \ \ d)(? =。*[a-zA-Z])(? =。*[! @ # $ % ^ &])。* ^ $编码regexp表达式。*(? = . * \ d)(? =。*[a-zA-Z])(? =。*[! @ # $ % ^ &])。*美元。
This is because the \ is an escape sequence.
这是因为\是一个转义序列。
The latter matches the string you specified.
后者匹配您指定的字符串。
If your original string was a regexp, rather than a java string, it would match strings such as \dTest@!
如果您的原始字符串是regexp，而不是java字符串，那么它将匹配字符串，比如\dTest@!
Also you should consider removing the first .*, doing so would make the regexp more efficient. The reason is that regexp's by default are greedy. So it will start by matching the whole string to the initial .*, the lookahead will then fail. The regexp will backtrack, matchine the first .* to all but the last character, and will fail all but one of the loohaheads. This will proceed until it hits a point where the different lookaheads succeed. Dropping the first .*, putting the lookahead immidiately after the start of string anchor, will avoid this problem, and in this case the set of strings matched will be the same.
此外，您应该考虑删除第一个。*，这样做会使regexp更有效。原因是regexp默认是贪婪的。因此，它将首先匹配整个字符串到初始值。regexp将会回溯，matchine第一个。*除了最后一个角色，而且将会失败除了一个loohaheads。这将继续下去，直到它达到一个不同的lookahead成功的点。删除第一个。*，在字符串锚开始后，将lookahead放在前面，将避免这个问题，在这种情况下，匹配的字符串集将是相同的。