nivo滑块，是一种总是显示下一个和prev箭头的方法吗？(nivo slider, is a way to always show the next and prev arrows?)

 您的示例可以在字符串之后匹配空格： 
 
grep -E '\.pdf ' input.txt
 
 你所说的“字符串”与grep调用的“字”类似。 一个字是一个字母数字字符的运行。 单词的好处在于，您可以将单词末尾与特殊的\>匹配，该单词末尾匹配零个字符长度的行进。 这也匹配在行末。 但是字符不能改变，也不能包含标点符号，所以我们不能使用它。 
 
 如果您还需要在行尾进行匹配，在单词后面没有空格的情况下，请使用： 
 
grep -E '\.pdf |\.pdf$' input.txt
 
 要包括文件名后面的字符不是空格字符'  '，但是其他空格，比如制表符\t ，或者名称后面紧跟着一个以#开头的注释，请使用： 
 
grep -E '\.pdf[[:space:]#]|\.pdf$' input.txt
 
 我将说明单词边界的匹配，因为那将是一个完美的解决方案，除了我们不能在这里使用它，因为我们不能改变被看作单词的一部分的字符集。 
 
 输入包含foo作为单独的单词，并且作为更长的单词的一部分， foo不在单词的末尾，因此不在单词边界处： 
 
$ printf 'foo bar\nfoo.bar\nfoobar\nfoo_bar\nfoo\n'
foo bar
foo.bar
foobar
foo_bar
foo
 
 现在，要匹配单词的边界，我们可以使用\<开头， \>匹配结尾： 
 
$ printf 'foo bar\nfoo.bar\nfoobar\nfoo_bar\nfoo\n' | grep 'foo\>'
foo bar
foo.bar
foo
 
 请注意_如何匹配char字样 - 但除此之外，字符串只是字母数字[a-zA-Z0-9] 。 
 还要注意foo行尾是如何匹配的 - 在只包含foo的行中。 行结束时我们不需要特殊情况。 

Your example works with matching the space after the string also:
 
grep -E '\.pdf ' input.txt
 
What you call "string" is similar to what grep calls "word". A Word is a run of alphanumeric characters. The nice thing with words is that you can match a word end with the special \>, which matches a word end with a march of zero characters length. That also matches at the end of line. But the word characters can not be changed, and do not contain punctuation, so we can not use it.
 
If you need to match at the end of line too, where there is no space after the word, use:
 
grep -E '\.pdf |\.pdf$' input.txt
 
To include cases where the character after the file name is not a space character '', but other whitespace, like a tab, \t, or the name is directly followed by a comment, starting with #, use:
 
grep -E '\.pdf[[:space:]#]|\.pdf$' input.txt
 
I will illustrate the matching of word boundarys too, because that would be the perfect solution, except that we can not use it here because we can not change the set of characters that are seen as parts of a word.
 
The input contains foo as separate word, and as part of longer words, where the foo is not at the end of the word, and therefore not at a word boundary:
 
$ printf 'foo bar\nfoo.bar\nfoobar\nfoo_bar\nfoo\n'
foo bar
foo.bar
foobar
foo_bar
foo
 
Now, to match the boundaries of words, we can use \< for the beginning, and \> to match the end:
 
$ printf 'foo bar\nfoo.bar\nfoobar\nfoo_bar\nfoo\n' | grep 'foo\>'
foo bar
foo.bar
foo
 
Note how _ is matched as a word char - but otherwise, wordchars are only the alphanumerics, [a-zA-Z0-9].
 Also note how foo an the end of line is matched - in the line containing only foo. We do not need a special case for the end of line.