grep：提取前缀和后缀之间的内容

Question

我有这样的文件内容：

Listening for transport dt_socket at address: 8000
------------------------------------------------------------
 ^[[1m HAPI FHIR^[[22m 5.4.0 - Command Line Tool
------------------------------------------------------------
Process ID                      : 21719@psgd
Max configured JVM memory (Xmx) : 3.2GB
Detected Java version           : 11.0.7
------------------------------------------------------------
^[[32m2021-07-01^[[0;39m ^[[1;32m12:27:40.79^[[0;39m ^[[37m[main]^[[0;39m ^[[37mWARN ^[[0;39m ^[[1;34mo.f.c.i.s.c.ClassPathScanner^[[0;39m ^[[1;37mUnable to resolve location classpath:db/migration. Note this warning will become an error in Flyway 7.
^[[0;39m^[[32m2021-07-01^[[0;39m ^[[1;32m12:27:42.641^[[0;39m ^[[37m[main]^[[0;39m ^[[37mWARN ^[[0;39m ^[[1;34mo.f.c.i.s.c.ClassPathScanner^[[0;39m ^[[1;37mUnable to resolve location classpath:db/migration. Note this warning will become an error in Flyway 7.
^[[0;39m^[[32m2021-07-01^[[0;39m ^[[1;32m12:27:44.693^[[0;39m ^[[37m[main]^[[0;39m ^[[37mINFO ^[[0;39m ^[[1;34mc.u.f.j.m.t.InitializeSchemaTask^[[0;39m ^[[1;37m3_3_0.20180115.0: Initializing ORACLE_12C schema for HAPI FHIR
^[[0;39m^[[32m2021-07-01^[[0;39m ^[[1;32m12:27:44.848^[[0;39m ^[[37m[main]^[[0;39m ^[[37mINFO ^[[0;39m ^[[1;34mc.u.f.j.m.t.BaseTask^[[0;39m ^[[1;37m3_3_0.20180115.0: SQL "create sequence SEQ_BLKEXCOL_PID start with 1 increment by  50" returned 0
^[[0;39m^[[32m2021-07-01^[[0;39m ^[[1;32m12:27:44.918^[[0;39m ^[[37m[main]^[[0;39m ^[[37mINFO ^[[0;39m ^[[1;34mc.u.f.j.m.t.BaseTask^[[0;39m ^[[1;37m3_3_0.20180115.0: SQL "
create sequence SEQ_BLKEXCOLFILE_PID start with 1 increment by  50" returned 0
^[[0;39m^[[32m2021-07-01^[[0;39m ^[[1;32m12:27:47.573^[[0;39m ^[[37m[main]^[[0;39m ^[[37mINFO ^[[0;39m ^[[1;34mc.u.f.j.m.t.BaseTask^[[0;39m ^[[1;37m3_3_0.20180115.0: SQL "

    create table HFJ_BINARY_STORAGE_BLOB (
       BLOB_ID varchar2(200 char) not null,
        BLOB_DATA blob not null,
        CONTENT_TYPE varchar2(100 char) not null,
        BLOB_HASH varchar2(128 char),
        PUBLISHED_DATE timestamp not null,
        RESOURCE_ID varchar2(100 char) not null,
        BLOB_SIZE number(10,0),
        primary key (BLOB_ID)
    )" returned 0

我只需要提取之间的内容SQL "并" returned 0修剪所有空格。

有任何想法吗？

我尝试使用以下方法减少问题：

$ echo 'sdf SQL" sdf sdf" returned 0' | grep 's/SQL"\(.*\)" returned 0/\1/' -

但是越来越空了。

我的预期输出是：

create sequence SEQ_BLKEXCOL_PID start with 1 increment by  50;
create sequence SEQ_BLKEXCOLFILE_PID start with 1 increment by  50;
create table HFJ_BINARY_STORAGE_BLOB (
       BLOB_ID varchar2(200 char) not null,
        BLOB_DATA blob not null,
        CONTENT_TYPE varchar2(100 char) not null,
        BLOB_HASH varchar2(128 char),
        PUBLISHED_DATE timestamp not null,
        RESOURCE_ID varchar2(100 char) not null,
        BLOB_SIZE number(10,0),
        primary key (BLOB_ID)
    );

我试图执行：

cat test.log | sed -E 's/.* SQL"(.*)" returned 0/\1/'

它正在返回我所有的文件内容......

使用awk，它返回空：

$ awk -v RS='SQL "[[:space:]]+?\n\n+.*returned 0' '                                                                                                  
RT{
 gsub(/^SQL "\n+|\n+$/,"",RT)
 sub(/" returned 0[[:space:]]+?\n*$/,"",RT)
 print RT";"
}
' test.log

Answer 1

这可以使用自定义来完成RS，gnu-awk因为它在SQL "..."文本块上拆分数据，然后在操作块内提取引号之间的文本而没有前导空格。

awk -v RS=' SQL "[^"]+"' 'RT {
gsub(/^[^"]*"[[:space:]]*|"[^"]*$/, "", RT); print RT ";"}' file.sql

create sequence SEQ_BLKEXCOL_PID start with 1 increment by  50;
create sequence SEQ_BLKEXCOLFILE_PID start with 1 increment by  50;
create table HFJ_BINARY_STORAGE_BLOB (
       BLOB_ID varchar2(200 char) not null,
        BLOB_DATA blob not null,
        CONTENT_TYPE varchar2(100 char) not null,
        BLOB_HASH varchar2(128 char),
        PUBLISHED_DATE timestamp not null,
        RESOURCE_ID varchar2(100 char) not null,
        BLOB_SIZE number(10,0),
        primary key (BLOB_ID)
    );