是否有一种方法可以使用sed或grep删除所有不必要的文本并仅显示一个单词?

Question

有没有办法删除所有在code="在";>"之后在我的文件中，所以我留下了clearsky_night或cloudy，或sun等?

我已经尝试了grep -o -P '(?<=>).*(?=>)'，但得到一个错误消息，将未知选项保存到's'

我也试过grep - o - p"(? & lt;代码= =")。*(? =")’但这也不管用。这是我文件中的内容:

<symbol id="Sun" number="1" code="clearsky_night"></symbol>
<symbol id="Sun" number="1" code="clearsky_night"></symbol>
<symbol id="Sun" number="1" code="clearsky_night"></symbol>
<symbol id="Cloud" number="4" code="cloudy"></symbol>
<symbol id="PartlyCloud" number="3" code="partlycloudy_night"></symbol>
<symbol id="Cloud" number="4" code="cloudy"></symbol>
<symbol id="PartlyCloud" number="3" code="partlycloudy_night"></symbol>
<symbol id="PartlyCloud" number="3" code="partlycloudy_night"></symbol>
<symbol id="PartlyCloud" number="3" code="partlycloudy_night"></symbol>
<symbol id="PartlyCloud" number="3" code="partlycloudy_night"></symbol>
<symbol id="Cloud" number="4" code="cloudy"></symbol>
<symbol id="PartlyCloud" number="3" code="partlycloudy_night"></symbol>
<symbol id="Cloud" number="4" code="cloudy"></symbol>
<symbol id="PartlyCloud" number="3" code="partlycloudy_night"></symbol>
<symbol id="Cloud" number="4" code="cloudy"></symbol>
<symbol id="Cloud" number="4" code="cloudy"></symbol>
<symbol id="PartlyCloud" number="3" code="partlycloudy_night"></symbol>
<symbol id="Cloud" number="4" code="cloudy"></symbol>
<symbol id="PartlyCloud" number="3" code="partlycloudy_night"></symbol>
<symbol id="Cloud" number="4" code="cloudy"></symbol>
<symbol id="Cloud" number="4" code="cloudy"></symbol>
<symbol id="Cloud" number="4" code="cloudy"></symbol>
<symbol id="PartlyCloud" number="3" code="partlycloudy_night"></symbol>
<symbol id="PartlyCloud" number="3" code="partlycloudy_night"></symbol>
<symbol id="LightCloud" number="2" code="fair_night"></symbol>
<symbol id="PartlyCloud" number="3" code="partlycloudy_night"></symbol>
<symbol id="Sun" number="1" code="clearsky_night"></symbol>
<symbol id="PartlyCloud" number="3" code="partlycloudy_night"></symbol>
<symbol id="LightCloud" number="2" code="fair_night"></symbol>
<symbol id="LightCloud" number="2" code="fair_night"></symbol>
<symbol id="LightCloud" number="2" code="fair_night"></symbol>
<symbol id="PartlyCloud" number="3" code="partlycloudy_night"></symbol>
<symbol id="LightCloud" number="2" code="fair_night"></symbol>
<symbol id="Cloud" number="4" code="cloudy"></symbol>
<symbol id="Cloud" number="4" code="cloudy"></symbol>
<symbol id="Cloud" number="4" code="cloudy"></symbol>

Answer 1

这个怎么样:

grep -o -P '(?<=code=").+?(?=")' input_file.xml

我检查了lookaround(?<=...)和(?=...)的使用

或者，使用perlmy friend:

$ perl -pe 's:^.+code="(.+?)".+$:1:' <input_file.xml

解释:

• perl -pe:使用包含下一个字符串参数的命令运行perl。
• s:...:...:: substitution.
• "(.+?)":""内部的东西，non- gredy (?).
• ^.+code=":从第一行开始到code=".
• ".+$:从"到行尾的所有内容

当然，这是一个快速而肮脏的解决方案。XML解析器会更好。

(抱歉我的英语不好)

Answer 2

假设每个@cyrus注释都有有效的XML，可以通过xsltproc:

使用XSLT转换src.xslt

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text" />
<xsl:strip-space elements="*"/>
<xsl:template match="symbol">
<xsl:for-each select="@code">
<xsl:value-of select="concat(., '&#xA;')"/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>

使用xsltproc转换你的xml:

xsltproc src.xslt src.xml

输出:

clearsky_night
clearsky_night
clearsky_night
cloudy
partlycloudy_night
cloudy
partlycloudy_night
partlycloudy_night
partlycloudy_night
partlycloudy_night
cloudy
partlycloudy_night
cloudy
partlycloudy_night
cloudy
cloudy
partlycloudy_night
cloudy
partlycloudy_night
cloudy
cloudy
cloudy
partlycloudy_night
partlycloudy_night
fair_night
partlycloudy_night
clearsky_night
partlycloudy_night
fair_night
fair_night
fair_night
partlycloudy_night
fair_night
cloudy