perl的 - 如何捕捉空單元格與正則表達式

Question

輸出：

id | status | name    | cluster | ip   | mac    | roles | pending_roles | online 
---|----------|------------------|---------|-------------|-------------------|-------|-----------------|------- 
11 | discover | Untitled (9a:3a) | 12  | 10.20.0.144 | c8:1f:66:ce:9a:3a |  | cinder   | True 
12 | discover | Untitled (9f:8d) | 12  | 10.20.0.186 | c8:1f:66:ce:9f:8d |  | cinder, compute | True 
10 | discover | Untitled (c7:f3) | None | 10.20.0.214 | c8:1f:66:ce:c7:f3 |  |     | True 
13 | discover | Untitled (9f:3d) | None | 10.20.0.233 | c8:1f:66:ce:9f:3d |  |     | True 
8 | discover | Untitled (74:8e) | 12  | 10.20.0.184 | c8:1f:66:ce:74:8e |  | controller  | True 
14 | discover | Untitled (75:4b) | None | 10.20.0.185 | c8:1f:66:ce:75:4b |  |     | True 
9 | discover | Untitled (76:23) | None | 10.20.0.213 | c8:1f:66:ce:76:23 |  |     | True 

我正則表達式：

\d+)\s+\|\s+(\w+)\s+\|\s+\w+\s+\((\S+)\)\s+\|\s+(\d+)\s+\|\s+(\S+)\s+\|\s+(\S+)\s+\|(.*?)\|(.*?)\|\s+(\w+) 

，但不能趕上空單元格！我嘗試了很多方法。

行示例：

13 | discover | Untitled (9f:3d) | None | 10.20.0.233 | c8:1f:66:ce:9f:3d |  |     | True 

Answer 1

不要試圖處理結構化數據非結構化線。您有管道分隔的數據，因此將其解析爲管道分隔的數據，然後檢查解析的內容。

請注意，我在單個單元格上使用正則表達式（/^\s*$/以查看它是否全部是空格），但不是在每一行上。

下面是一個例子：

#!/usr/bin/perl 

use strict; 
use warnings; 

while (my $line = <DATA>) { 
    chomp $line; 
    my @cells = split /\|/, $line, -1; 
    my $ncells = scalar @cells; 
    die "There should be 9 fields, but line $. has $ncells" unless $ncells == 9; 
    for my $i (1 .. $ncells) { 
     if ($cells[$i-1] =~ /^\s*$/) { 
      print "Cell #$i on line $. is empty\n"; 
     } 
    } 
} 

__DATA__ 
id | status | name    | cluster | ip   | mac    | roles | pending_roles | online 
---|----------|------------------|---------|-------------|-------------------|-------|-----------------|------- 
11 | discover | Untitled (9a:3a) | 12  | 10.20.0.144 | c8:1f:66:ce:9a:3a |  | cinder   | True 
12 | discover | Untitled (9f:8d) | 12  | 10.20.0.186 | c8:1f:66:ce:9f:8d |  | cinder, compute | True 
10 | discover | Untitled (c7:f3) | None | 10.20.0.214 | c8:1f:66:ce:c7:f3 |  |     | True 
13 | discover | Untitled (9f:3d) | None | 10.20.0.233 | c8:1f:66:ce:9f:3d |  |     | True 
8 | discover | Untitled (74:8e) | 12  | 10.20.0.184 | c8:1f:66:ce:74:8e |  | controller  | True 
14 | discover | Untitled (75:4b) | None | 10.20.0.185 | c8:1f:66:ce:75:4b |  |     | True 
9 | discover | Untitled (76:23) | None | 10.20.0.213 | c8:1f:66:ce:76:23 |  |     | True 

Answer 2

chomp(my $header = <>); 
chomp(my $sep = <>); 

my $pat = 
    join ' x3 ', 
     map "A".(length($_)-2), 
     "-$sep-" =~ /(-+)/g; 

my @headers = unpack($pat, $header); 
while (my $line = <>) { 
    my %row; @row{@headers} = unpack($pat, $line); 

    # Do whatever here. 
    print("Row id=$row{id} has no pending roles\n") 
     if !length($row{pending_roles}); 
} 

輸出：

Row id=10 has no pending roles 
Row id=13 has no pending roles 
Row id=14 has no pending roles 
Row id=9 has no pending roles 

Answer 3

如果必須使用正則表達式，只是儘量做到儘可能小。同樣假設你沒有|你的數據或任何東西..

my $r = 0; 
foreach my $row (@rows) { 
    my $c = 0; 
    print "Row $r\n"; 
    while($row =~ /([^|])*(\||$)/g) { 
     my $col = $1; 
     print " $c: $col\t"; 
     if ($col =~ /^\s+$/) { print "whitespace only!" } 
     print "\n"; 
     $c++; 
    } 
    $r++; 
}