Warning: file_get_contents(/data/phpspider/zhask/data//catemap/0/vba/15.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
模式搜索的awk等价物_Awk - Fatal编程技术网

模式搜索的awk等价物

模式搜索的awk等价物,awk,Awk,我有以下文件: Schedule Name: Today Schedule Type: Standard Active: yes Effective date: 01/24/2014 11:17:05 Client Encrypt: no LC/CY/Custmr: EU NY Cindy BU CA Victor

我有以下文件:

Schedule Name:       Today

  Schedule Type:       Standard
  Active:              yes
  Effective date:      01/24/2014 11:17:05
  Client Encrypt:      no
  LC/CY/Custmr:  EU         NY  Cindy
                 BU         CA  Victor
                 GU         MI  Bob
  Include:
Schedule Name:       Tomorrow

  Schedule Type:       Standard
  Active:              yes
  Effective date:      01/26/2014 11:17:05
  Client Encrypt:      no
  LC/CY/Custmr:  MU         LA  Martha
                 EU         CA  Sam
  Include:
Schedule Name:       Yesterday

  Schedule Type:       Standard
  Active:              no
  Effective date:      01/21/2014 11:17:05
  Client Encrypt:      no
  LC/CY/Custmr:  NV         IL  Joe

  Include:
期望输出

Cindy    Today
Victor   Today
Bob      Today
Martha   Tomorrow
Sam      Tomorrow
现在,我想获得日程名称,即今天、明天以及客户名称,如果活动为是,这是第四个字段。因此,输出应为:

cat billing | 
    awk '/Schedule Name/ || /Active:/ || /LC/,/^$/' | 
    grep -v '^$'
在包含之前的LC之后是一个空行,所以我尝试获取所有数据,直到您找到一个空行,然后是grep-v空行,如果我尝试在不显示计划名称和活动的情况下工作,效果很好,但不适用于这两种模式的搜索

我使用下面的代码,这是相当缓慢

for pol in `cat /tmp/Active_Policies`
do
        count=`sudo /usr/openv/admin/pollist $pol -U | awk '/LC\/CY\/Custmr:/,/Include:/' | grep -v "Include:" | wc -l`
        if [ $count -gt 0 ]
        then
                first=`sudo /usr/openv/admin/pollist $pol -U | awk '/LC\/CY\/Custmr:/,/Include:/' | grep -v "Include:" | awk '{print $4}' | head -1`
                echo "$first    $pol" >> /tmp/Policies_$(date +%m-%d-%Y)
                counter=1
                for client in `sudo /usr/openv/admin/pollist $pol -U | awk '/LC\/CY\/Custmr:/,/Include:/' | grep -v "Include:" | awk '{print $3}' | sed '1d;$d'`
                do
                        ((counter = counter + 1))
                        if [ $counter -le $count ]
                        then
                                echo "$client   $pol" >> /tmp/Policies_$(date +%m-%d-%Y)
                        fi
                done
        fi
done

以下是解决问题的正确方法:

$ cat tst.awk
BEGIN { OFS="\t" }
/^[^[:space:]]/ { prt() }
NF {
    if ( /:/ ) {
        name = $0
        sub(/:.*/,"",name)
        gsub(/^[[:space:]]+|[[:space:]]+$/,"",name)
    }
    value = $0
    sub(/^[^:]+:/,"",value)
    gsub(/^[[:space:]]+|[[:space:]]+$/,"",value)

    n2v[name,++numVals[name]] = value
}
END { prt() }

function prt() {
    custFldName = "LC/CY/Custmr"
    if ( n2v["Active",1] == "yes" ) {
        sched = n2v["Schedule Name",1]
        for (valNr=1; valNr<=numVals[custFldName]; valNr++) {
            cust = n2v[custFldName,valNr]
            sub(/.*[[:space:]]/,"",cust)
            print cust, sched
        }
    }
    delete n2v
    delete numVals
}
关键是在上面创建一个数组n2v[],将名称映射到值,其中名称是每行第一个:之前的文本,关联的值是其后的任何内容:。然后,每次点击日程表行或输入文件末尾时,只需打印n2v数组中您关心的值

$ awk -f tst.awk file
Cindy   Today
Victor  Today
Bob     Today
Martha  Tomorrow
Sam     Tomorrow