用空格替换双引号之间的换行符

Question

我想逐行读取数据,而且我发现双引号我想用空格替换新行字符,直到第二个双引号遇到像

090033ec82b13639,CPDM Initiated,Logistical,"There corrected.",Gul Y Serbest,Urology
090033ec82ae0c07,Initiated,NA,"To   local testing
Rohit  3 to 4.",Julienne B Orr,Oncology
090033ec82b35fd0,Externally Initiated,NA,regulatory agency requests,Kenneth A Lord,Oncology

就像上面的数据第二行一样,因为它在第3行中找到双引号(打开)和关闭双引号所以我们需要将这些行合并为单个空格,如下所示:

090033ec82b13639,CPDM Initiated,Logistical,"There corrected.",Gul Y Serbest,Urology
090033ec82ae0c07,Initiated,NA,"To   local testing Rohit  3 to 4.",Julienne B Orr,Oncology
090033ec82b35fd0,Externally Initiated,NA,regulatory agency requests,Kenneth A Lord,Oncology

Answer 1

你可以使用这个gnu-awk单线程:

awk -v RS='"[^"]*"' -v ORS= '{gsub(/\n/, " ", RT); print $0  RT}' file
090033ec82b13639,CPDM Initiated,Logistical,"There corrected.",Gul Y Serbest,Urology
090033ec82ae0c07,Initiated,NA,"To   local testing Rohit  3 to 4.",Julienne B Orr,Oncology
090033ec82b35fd0,Externally Initiated,NA,regulatory agency requests,Kenneth A Lord,Oncology

• RS='"[^"]*"' - 输入记录分隔符设置为正则表达式 '"[^"]*"'
• -v ORS= - 输出记录分隔符设置为空
• gsub(/\n/, " ", RT) - 在匹配的文本中用空格替换换行符 Input Record Separator

这是一个perl单行:

perl -0pe 's/"[^\n"]*"(*SKIP)(*F)|("[^"\n]*)\n([^"]*")/$1 $2/g' file
090033ec82b13639,CPDM Initiated,Logistical,"There corrected.",Gul Y Serbest,Urology
090033ec82ae0c07,Initiated,NA,"To   local testing Rohit  3 to 4.",Julienne B Orr,Oncology
090033ec82b35fd0,Externally Initiated,NA,regulatory agency requests,Kenneth A Lord,Oncology