属性抽取

batty123292

aa.word (990 Bytes, 下载次数: 19)

2013-12-06 00:39 上传

点击文件名下载附件

bb.word (1.08 KB, 下载次数: 19)

2013-12-06 00:43 上传

点击文件名下载附件

现在有很多个文件，暂且提供两个，aa.word，bb.word。后续还会有cc.word，dd.word等等。
我想把其中标签为data-src的属性值抽出来，并且我只希望抽每个文件其中的前3个。
期望输出为一行，一共4个字段。第一个字段为文件的名字aa/bb/cc，第二个字段为第一个data-src对应的值，
第三个字段为第二个data-src对应的值，第四个字段为第三个data-src对应的值。

输出到一个文件里面，样例为：
aa  http://t1.gstatic.com/sh/sh/a  http://t0.gstatic.com  http://t3.gstatic.com/images?q
bb  httphsh.gstatic.com/sh/sj       httpnsklal://hs           http://t3mss/aj

aa.word为：

1. aaaaaaaaaaaaaaaaaaaa
   
2. aaaaaaaaaaaasasas
   
3. target=_blank class=_l<img class=rg_i name=Veu1j2oKuKYzyM: data-src="http://t1.gstatic.com/sh/sh/a" data-sz=f onload="google.stb.csi.onTbn(1, this)"></a>gshjksk djlkl ksl oom=1" target=_blank class=rg_l<img class=rg_i name=4rq0eMtoD0OT6M: data-src="http://t0.gstatic.com" data-sz=f onload="google.stb.csi.onTbn(1, this)"></a>
   
4. hskjaj dklasj dkla dkl
   
5. target=_blank class=rg_l<img class=rg_i name=pM0PJS7WOnG-TM: data-src="http://t3.gstatic.com/images?q" data-sz=f  onload="google.stb.csi.onTbn(1, this)"></a>
   
6. sjk djlks dkls
   
7. djkas djka
   
8. gfwv4DoDA&amp;zoom=1" target=_blank class=rg_l<img class=rg_i name=fk2DM_qEo4qlcM: data-src="http://t3.gstatic.com/itbn:AN" data-sz=f  onload="google.stb.csi.onTbn(1, this)"></a>
   
9. sjk djkd kld kld
   
10. =1" target=_blank class=rg_l<img class=rg_i name=FAisdPNr43SpYM: data-src="http://t0.gstatic.com/images?q" data-sz=f  onload="google.stb.csi.onTbn(1, this)"></a>
    
11. hsj dj dkl djks djk
    
12. sjk djk djkl
    
13. the end

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bb.word为：

1. zh jsj dkl dkjld jk fkld
   
2. target=_blank class=rg_l<img class=rg_i name=Veu1j2oKuKYzyM: data-src="httphsh.gstatic.com/sh/sj" data-sz=f onload="google.stb.csi.onTbn(1, this)"></a>
   
3. gshjksk djlkl ksl
   
4. oom=1" target=_blank class=rg_l<img class=rg_i name=4rq0eMtoD0OT6M: data-src="httpnsklal://hs" data-sz=f onload="google.stb.csi.onTbn(1, this)"></a>
   
5. hskjaj dklasj dkla dkl
   
6. target=_blank class=rg_l<img class=rg_i name=pM0PJS7WOnG-TM: data-src="http://t3mss/aj" data-sz=f  onload="google.stb.csi.onTbn(1, this)"></a>
   
7. sjk djlks dkls
   
8. djkas djka
   
9. gfwv4DoDA&amp;zoom=1" target=_blank class=rg_l<img class=rg_i name=fk2DM_qEo4qlcM: data-src="http://t3jsjka:AN" data-sz=f  onload="google.stb.csi.onTbn(1, this)"></a>
   
10. sjk djkd kld kld
    
11. =1" target=_blank class=rg_l<img class=rg_i name=FAisdPNr43SpYM: data-src="http://t0.gstaticskaklsam/images?q" data-sz=f  onload="google.stb.csi.onTbn(1, this)"></a>
    
12. hsj dj dkl djks djk
    
13. target=_blank class=rg_l<img class=rg_i name=FAisdPNr43SpYM: data-src="http://kslataticskaklsam/images?q" data-sz=f  onload="google.stb.csi.onTbn(1, this)"></a>
    
14. hsj dj dkl djks djk

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yinyuemi

回复 1# batty123292

1. gawk4.0
   
2. 
   
3. awk -vRS="data-src=" -F\" 'BEGINFILE{printf FILENAME}FNR<=4&&$2{printf OFS $2}ENDFILE{print ""}' file1 file2 ...

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yestreenstars

1. awk '/data-src/{s=s"\t"gensub(/.*data-src="([^"]*).*/,"\\1",1);n++}n==3{s=FILENAME""s;sub(/\.[^\t]*/,"",s);print s;exit}' *.word

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yestreenstars

回复 2# yinyuemi
漏了个条件：前3个~

   

Herowinter

1. awk '{for(i=1;i<=NF;i++)if($i~/data-src/)src[++count]=gensub(/data-src=\"(.*)\"/,"\\1",1,$i);if(count==3){print "aa",src[1],src[2],src[3];exit}}' aa.word

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同时有个疑问，取出来的3个data-src字符串中，如果有两个值重复，
要去掉重复的在文件中取下一个吗？

yinyuemi

回复 4# yestreenstars


    我设了FNR<=4,应该没问题，帮我测试下,3ks

LikeLx

1. awk '{for(i=0;i++<NF;) if($i~/data-src/&&a[data-src]++<3) s=s?s"\t"gensub(/data-src="([^"]*)"/,"\\1",1,$i):gensub(/data-src="([^"]*)"/,"\\1",1,$i)}END{print FILENAME,s}'

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没看清题意，代码只能对一个文件进行操作

reyleon

1. awk -v RS='data-src="[^"]+"' 'FNR<=3{s=gensub(/[^"]+"([^"]+)"/,"\\1",1,RT);f=FILENAME;a[f]=a[f]?a[f]" "s:s}END{for(i in a)print i,a[i]}' file1 file2 file3 .....

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dn833

1. for i in $(ls *.word);do echo -n $i" ";grep -Pom3 '(?<=data-src=")[^"]+' $i|xargs;done

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reyleon

回复 9# dn833

1. 1. for i in $(ls *.mp3)
   
2. 
   
3. One of the most common mistakes BASH programmers make is to write a loop like this:
   
4. 
   
5. for i in $(ls *.mp3); do    # Wrong!
   
6.     some command $i          # Wrong!
   
7. done
   
8. 
   
9. for i in $(ls)              # Wrong!
   
10. for i in `ls`               # Wrong!
    
11. 
    
12. for i in $(find . -type f)  # Wrong!
    
13. for i in `find . -type f`   # Wrong!
    
14. Never use a CommandSubstitution -- of EITHER kind! -- around something that writes out filenames.
    
15. 
    
16. Why? This breaks when a file has a space in its name. Why? Because the output of the $(ls *.mp3) command substitution undergoes WordSplitting. Assuming we have a file named 01 - Don't Eat the Yellow Snow.mp3 in the current directory, the for loop will iterate over each word in the resulting file name:
    
17. 
    
18. some command 01
    
19. some command -
    
20. some command Don't
    
21. some command Eat
    
22. ...
    
23. You can't double-quote the substitution either:
    
24. 
    
25. for i in "$(ls *.mp3)"; do # Wrong!
    
26. This causes the entire output of the ls command to be treated as a single word. Instead of iterating once for each file name, the loop will only execute once, with all the filenames rammed together.
    
27. 
    
28. In addition to this, the use of ls is just plain unnecessary. It's an external command, which simply isn't needed to do the job. So, what's the right way to do it?
    
29. 
    
30. for i in *.mp3; do  # Better! and...
    
31.    some command "$i" # ...see Pitfall #2 for more info.
    
32. done
    
33. Let Bash expand the list of filenames for you. The expansion will not be subject to word splitting. Each filename that's matched by the *.mp3 glob will be treated as a separate word, and the loop will iterate once per filename.
    
34. 
    
35. Question: What to do if there are no files *.mp3-files in the current directory? Then the for loop is executed once, with i="*.mp3", which is not the expected behaviour!
    
36. 
    
37. Check the loop variable inside the loop:
    
38. for i in *.mp3; do
    
39.    [[ -f "$i" ]] || continue
    
40.    some command "$i"
    
41. done
    
42. Reading lines of a file with a for loop is also wrong. Doubly (or possibly triply) so if those lines are filenames.
    
43. 
    
44. Note the quotes around $i in the loop body. This leads to our second pitfall:

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@lkk_super http://bbs.chinaunix.net/thread-4113125-1-4.html  http://bash.cumulonim.biz/BashPitfalls.html