Question

我正在使用Unix Shell。如何删除两个特定字符串之间的换行符。

例如，输入为：

CASE when a in ('abcd','bdcdf') then
  Shng,
END as xyz

输出应为：

CASE when a in ('abcd','bdcdf') then Shng END as xyz,

Answer 1

解析行的文件行，并在看到CASE或END时记住下面的代码使用if语句的短语法和使用-n参数抑制\ n的回声。

incase=0
cat x.sql | while read -r line; do
   [[ ${line} = CASE* ]] && incase=1;
   [[ ${line} = END* ]] && incase=0
   [[ ${incase} = 0 ]] && echo "${line}"
   [[ ${incase} = 1 ]] && echo -n "${line} "                                               
done

编辑：
当你有嵌套的CASE（如CASE ... CASE ... END ... END）和所有 CASE从不同的行开始，您可以计算嵌套的深度。

incase=0
cat x.sql | while read -r line; do
   [[ ${line} = CASE* ]] && (( incase = incase + 1)) ;
   [[ ${line} = END* ]] && (( incase = incase - 1)) 
   [[ ${incase} = 0 ]] && echo "${line}"
   [[ ${incase} > 0 ]] && echo -n "${line} "                                               
done
# You might want an extra echo here so your last line will finish with a \n
echo

编辑2：通常你可以避免使用猫（寻找UUOC）。这里的代码更好

incase=0
cat x.sql | while read -r line; do
   [[ ${line} = CASE* ]] && incase=1;
   [[ ${line} = END* ]] && incase=0
   [[ ${incase} = 0 ]] && echo "${line}"
   [[ ${incase} = 1 ]] && echo -n "${line} "                                               
done

编辑：
当你有嵌套的CASE（如CASE ... CASE ... END ... END）和所有 CASE从不同的行开始，您可以计算嵌套的深度。

incase=0
while read -r line; do
   [[ ${line} = CASE* ]] && (( incase = incase + 1)) ;
   [[ ${line} = END* ]] && (( incase = incase - 1)) 
   [[ ${incase} = 0 ]] && echo "${line}"
   [[ ${incase} > 0 ]] && echo -n "${line} "                                               
done < x.sql
# You might want an extra echo here so your last line will finish with a \n
echo

删除两个特定字符串之间的换行符

1 个答案: