匹配使用grep并打印匹配的模式多种模式

Question

我有一个看起来像

..<long-text>..."field1":"some-value"...<long-text>...."field2":"some-value"... 
..<long-text>..."field1":"some-value"...<long-text>...."field2":"some-value"... 
..<long-text>..."field1":"some-value"...<long-text>...."field2":"some-value"... 

我想从在bash文件的每一行中提取出来field1和field2文件。我希望field1和field2出现在每行的同一行中。所以输出应该喜欢 -

"field1":"some-value" "field2":"some-value" 
"field1":"some-value" "field2":"some-value" 
"field1":"some-value" "field2":"some-value" 

我写了一个grep的表情像 -

grep -E '"field1":"[a-z]*".*"field2":"[a-z]*"' -o 

但由于.*之间，它产生这两个表达式之间所有的所有文本。我也试过

grep -E '"field1":"[a-z]*"|"field2":"[a-z]*"' -o 

但是，这会输出所有field1s在单独的行，然后在单独的行中输出所有field2s。

如何获得预期的输出？

Answer 1

您可以使用grep和awk来格式化结果：

grep -oE '"(field1|field2)":"[^"]*"' file | awk 'NR%2{p=$0; next} {print p, $0}' 

"field1":"some-value" "field2":"some-value" 
"field1":"some-value" "field2":"some-value" 
"field1":"some-value" "field2":"some-value" 

Answer 2

使用sed：

echo abcdef | sed 's/\(.\).*\(.\)/\1\2/' 
# yields: af 

您的具体情况：

sed 's/.*\("field1":"[a-z]*"\).*\("field2":"[a-z]*"\).*/\1 \2/' yourfile 

如果某些行不匹配的话，那么做你grep第一，例如，

grep -Eo '"field1":"[a-z]*".*"field2":"[a-z]*"' yourfile | 
    sed 's/.*\("field1":"[a-z]*"\).*\("field2":"[a-z]*"\).*/\1 \2/'