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根據上一行更新行

tag-icon • tag-icon awk solaris csv
根據上一行更新行

我有一個問題，因為我不熟悉 awk。我將 sar -d 輸出產生的 csv 檔案轉換為 csv 樣式：

12:33:41        unix,restarts
12:35:00,lofi4096,0,0.0,0,0,0.0,0.0
,iscsi0,0,0.0,0,0,0.0,0.0
,scsi_vhc,0,0.0,0,0,0.0,0.0
,nfs1,0,0.0,0,0,0.0,0.0

12:45:00,lofi4096,0,0.0,0,0,0.0,0.0
,iscsi0,0,0.0,0,0,0.0,0.0
,scsi_vhc,0,0.0,0,0,0.0,0.0
,nfs1,0,0.0,0,0,0.0,0.0

我想轉換成這個

12:33:41        unix,restarts
12:35:00,lofi4096,0,0.0,0,0,0.0,0.0
12:35:00,iscsi0,0,0.0,0,0,0.0,0.0
12:35:00,scsi_vhc,0,0.0,0,0,0.0,0.0
12:35:00,nfs1,0,0.0,0,0,0.0,0.0

12:45:00,lofi4096,0,0.0,0,0,0.0,0.0
12:45:00,iscsi0,0,0.0,0,0,0.0,0.0
12:45:00,scsi_vhc,0,0.0,0,0,0.0,0.0
12:45:00,nfs1,0,0.0,0,0,0.0,0.0

我的嘗試，但由於 awk 逐行讀取，我不知道如何保留前一行的值。我希望我所擁有的應該有效。知道如何才能獲得我想要的結果。我嘗試使用 awk，但我想這應該可以使用 sed 或使用自訂 shell 腳本來實現（我試圖避免執行這部分）。

#!/usr/bin/awk -f
BEGIN {
        FS=",";
}
{
        print $1
        if ($1 != "") {
                mydate=$1;
                print $0;
        }
        else {
                print $mydate","$0;
        }
}

運行Solaris 11.1系統。

答案1

由於輸入中似乎有空行，因此它會有點長。以下內容可能對您有用：

awk -F'[, ]' '{if (NF!=0 && $1=="") {$1=prev} prev=$1}1' OFS=, inputfile

這個想法是分割字段,和空格（後者是為了處理第一行輸入）。檢查第一個欄位是否為空和場數不為零（處理空白行），然後用先前儲存的第一個欄位取代第一個欄位。

對於您的輸入，它會產生：

12:33:41        unix,restarts
12:35:00,lofi4096,0,0.0,0,0,0.0,0.0
12:35:00,iscsi0,0,0.0,0,0,0.0,0.0
12:35:00,scsi_vhc,0,0.0,0,0,0.0,0.0
12:35:00,nfs1,0,0.0,0,0,0.0,0.0

12:45:00,lofi4096,0,0.0,0,0,0.0,0.0
12:45:00,iscsi0,0,0.0,0,0,0.0,0.0
12:45:00,scsi_vhc,0,0.0,0,0,0.0,0.0
12:45:00,nfs1,0,0.0,0,0,0.0,0.0

答案2

和sed：

sed '/^[0-9]/{               # if line starts with digit
h                            # overwrite hold buffer with pattern space content
s/\([^,]*\),.*/\1/           # extract timestamp
x                            # exchange: put the original line back into pattern
}                            # space and the timestamp in hold space
/^,/{                        # if line starts with a comma
G                            # append hold space (timestamp) to pattern space
s/\(.*\)\n\(.*\)/\2\1/       # swap the initial line content and the timestamp 
}' infile

一行：

sed -e'/^[0-9]/{h;s/\([^,]*\),.*/\1/;x' -e\} -e'/^,/{G;s/\(.*\)\n\(.*\)/\2\1/' -e\} infile

答案3

其他sed：

sed '$!N;/\n,/s/\([^,]*\).*\n/&\1/;P;D' <in >out

!對於不是最後一個的每個輸入行$，sed會將Next 輸入行附加到模式空間，前面帶有\newline 字元。然後，它將嘗試進行s///替換，其中涉及將第一組可能的^,非逗號字元複製到緊跟在\newline 後面的逗號之前的空格。如果它不能這樣做，我想也沒什麼壞處。

sed然後將P列印到\n模式空間中的第一個 ewline 並D刪除相同的內容，然後從頂部使用下一對輸入行重新開始循環。

輸出

12:33:41        unix,restarts
12:35:00,lofi4096,0,0.0,0,0,0.0,0.0
12:35:00,iscsi0,0,0.0,0,0,0.0,0.0
12:35:00,scsi_vhc,0,0.0,0,0,0.0,0.0
12:35:00,nfs1,0,0.0,0,0,0.0,0.0

12:45:00,lofi4096,0,0.0,0,0,0.0,0.0
12:45:00,iscsi0,0,0.0,0,0,0.0,0.0
12:45:00,scsi_vhc,0,0.0,0,0,0.0,0.0
12:45:00,nfs1,0,0.0,0,0,0.0,0.0

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