Shellscript读取XML属性值 [英] Shellscript Read XML attribute value

问题描述

我们想从XML文件中读取XML属性.文件内容示例如下:

We want to read XML attributes from an XML file. Example of file content is as below:

<properties>
  <property name="abc" value="15"/>
  <property name="xyz" value="26"/>
</properties>

我们要使用shell脚本读取属性"abc"的值(即15).
请建议使用shell命令来实现此目的.

We want to read value (i.e. 15) for property "abc" using shell script.
Please suggest shell commands to achieve this.

推荐答案

您可以使用适当的XML解析器，例如xmllint.如果您的版本支持xpath，将很容易获取特定值.如果它不支持xpath，则可以使用--shell选项，如下所示:

You can use a proper XML parser like xmllint. If your version supports xpath, it will be very easy to grab specific values. If it doesn't support xpath, then you can use --shell option like so:

$ echo 'cat //properties/property[@name="abc"]/@value' | xmllint --shell myxml
/ >  -------
 value="15"
/ > 

然后可以使用awk或sed格式化并从输出中提取所需的字段.

You can then use awk or sed to format and extract desired field from output.

$ echo 'cat //properties/property[@name="abc"]/@value' | xmllint --shell myxmlfile | awk -F'[="]' '!/>/{print $(NF-1)}'
15

您可以使用命令替换通过说出以下内容来捕获变量中的输出:

You can use command substitution to capture the output in a variable by saying:

$ myvar=$(echo 'cat //properties/property[@name="abc"]/@value' | xmllint --shell myxml | awk -F'[="]' '!/>/{print $(NF-1)}')
$ echo "$myvar"
15

使用xmlparser以外的其他任何内容都容易出错，并且很容易出错.

Using anything else other than a xmlparser is prone to errors and will break easy.

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