BASH使用grep和awk一个变量中 [英] BASH using grep and awk inside a variable

问题描述

bash-3.2$ FNAME=$1
bash-3.2$ OLD_NO=$(grep "_version=" | awk -F '"' '{print $12}' $FNAME)

2号线似乎并不为我工作。我不能收/正确引用呢？
这似乎挂

Line 2 does not appear to be working for me. Am I not closing/quoting it correctly? It seems to hang

更新脚本，以反映低于建议

Updated the script to reflect below suggestions

echo $OLD_NO
OLD_NO=$(grep '_version=' "$FNAME" | awk -F '"' '{print $12}')
#Get the version of the
echo "What do you want to update release number to?"
REPLACEMENT="_version="$NEW_NO
echo $REPLACEMENT
sed -i ''s/$OLD_NO/$REPLACEMENT/g'' $FNAME

〜

得到一个新的错误

bash-3.2$ ./vu reader.xml 

What do you want to update release number to?
_version=
sed: -e expression #1, char 0: no previous regular expression

工作在bash虽然

Works in bash though

bash-3.2$ grep _version market_rules_cd.reader.xml | awk -F '"' '{print $12}'

14.8.21.1

14.8.21.1

推荐答案

行挂起，因为因为你没有通过的grep'_version ='将阻塞等待标准输入文件名参数。从挂传递文件名grep的，而不是AWK停止。 awk将随后处理的grep的输出：

The line hangs because grep '_version=' will blocking wait on stdin since you didn't passed a file name argument. To stop it from hanging pass the file name to grep, not to awk. awk will then process grep's output:

OLD_NO=$(grep '_version=' "$FNAME" | awk -F '"' '{print $12}')

---

顺便说一句：工作可以用 AWK 仅完成，你不需要的grep：

---

Btw: The job can be done with awk only, you don't need grep:

OLD_NO=$(awk -F '"' '/_version=/ {print $12}' "$FNAME")

请注意有以下形式awk程序：

Note that awk programs having the following form:

condition { action }  [; more of them ]

您可以省略情况，如果该行为应适用于每一行（这被广泛使用），但是在这种情况下，你可以使用正则表达式条件的动作仅限制含行 _version = ：

You can omit the condition if the action should apply to every line (this is widely used), however in this case you can use a regex condition to restrict the action only to lines containing _version=:

/_version=/ { print $12 }

这篇关于BASH使用grep和awk一个变量中的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！