正則表達式不匹配的數據和日期

Question

我有一個SQL SELECT與許多行，每行看起來是這樣的轉儲：

07/11/2011 16:48:08,07/11/2011 16:48:08,'YD','MANUAL',0,1,'text','text','text','text',,,,'text',,,0,0, 

我想要做的兩件事情到每一行：

1. 替換所有日期Oracle的sysdate函數。日期也可以不帶小時（如07/11/2011）。
2. 與null串

這裏替換所有空值是我的嘗試：

$_ =~ s/,(,|\n)/,null$1/g;     # Replace no data by "null" 
$_ =~ s/\d{2}\/\d{2}\/d{4}.*?,/sysdate,/g; # Replace dates by "sysdate" 

但是，這將改變字符串：雖然我希望它是

07/11/2011 16:48:08,07/11/2011 16:48:08,'YD','MANUAL',0,1,'text','text','text','text',null,,null,'text',null,,0,0,null 

sysdate,sysdate,'YD','MANUAL',0,1,'text','text','text','text',null,null,null,'text',null,null,0,0,null 

我不明白爲什麼日期不匹配，爲什麼一些,,不被替換爲null。

任何見解歡迎，提前致謝。

Answer 1

\d{2}\/\d{2}\/d{4}.*?,沒有工作，因爲最後的d未被轉義。
如果,可以在任意一側，或開始/字符串的結尾，你可以做2個步驟：

第1步
s/(?:^|(?<=,))(?=,|\n)/null/g
擴展：

/ 
    (?:^   # Begining of line, ie: nothing behind us 
    | (?<=,)  # Or, a comma behind us 
) 
    # we are HERE!, this is the place between characters 
    (?= ,   # A comma in front of us 
    | \n   # Or, a newline in front of us 
) 
/null/g 
# The above regex does not consume, it just inserts 'null', leaving the 
# same search position (after the insertion, but before the comma). 

# If you want to consume a comma, it would be done this way: 
s/(?:^|(?<=,))(,|\n)/null$1/xg 
# Now the search position is after the 'null,' 

第2步
s/(?:^|(?<=,))\d{2}\/\d{2}\/\d{4}.*?(?=,|\n)/sysdate/g

或者，你可以將它們組合成一個單一的正則表達式，使用eval修改：
$row =~ s/(?:^|(?<=,))(\d{2}\/\d{2}\/\d{4}.*?|)(?=,|\n)/ length $1 ? 'sysdate' : 'null'/eg;

分佈看，它看起來像這樣

s{ 
    (?:^| (?<=,)) # begin of line or comma behind us 
    (    # Capt group $1 
     \d{2}/\d{2}/\d{4}.*?  # date format and optional non-newline chars 
    |       # Or, nothing at all 
    )     # End Capt group 1 
    (?= , | \n)  # comma or newline in front of us 
}{ 
    length $1 ? 'sysdate' : 'null' 
}eg 

如果不換行空格填充的機會

$row =~ s/(?:^|(?<=,))(?:([^\S\n]*\d{2}\/\d{2}\/\d{4}.*?)|[^\S\n]*)(?=,|\n)/ defined $1 ? 'sysdate' : 'null'/eg;

Answer 2

也許。*？太貪婪，嘗試：

$_ =~ s/\d{2}\/\d{2}\/d{4}[^,]+,/sysdate,/g; 

Answer 3

你可以這樣做：

$ cat perlregex.pl 
use warnings; 
use strict; 

my $row = "07/11/2011 16:48:08,07/11/2011 16:48:08,'YD','MANUAL',0,1,'text','text','text','text',,,,'text',,,0,0,\n"; 

print("$row\n"); 
while ($row =~ /,([,\n])/) { $row =~ s/,([,\n])/,null$1/; } 
print("$row\n"); 
$row =~ s/\d{2}\/\d{2}\/\d{4}.*?,/sysdate,/g; 
print("$row\n"); 

導致此：

$ ./perlregex.pl 
07/11/2011 16:48:08,07/11/2011 16:48:08,'YD','MANUAL',0,1,'text','text','text','text',,,,'text',,,0,0, 

07/11/2011 16:48:08,07/11/2011 16:48:08,'YD','MANUAL',0,1,'text','text','text','text',null,null,null,'text',null,null,0,0,null 

sysdate,sysdate,'YD','MANUAL',0,1,'text','text','text','text',null,null,null,'text',null,null,0,0,null 

這當然可以優化，但它跨越獲取點。

Answer 4

要替換的東西：，它可以作爲被寫入。通常向前看符號是這個更好的選擇：

$subject =~ s/(?<=,)(?=,|$)/null/g; 

說明：

" 
(?<=  # Assert that the regex below can be matched, with the match ending at this position (positive lookbehind) 
    ,   # Match the character 「,」 literally 
) 
(?=  # Assert that the regex below can be matched, starting at this position (positive lookahead) 
       # Match either the regular expression below (attempting the next alternative only if this one fails) 
     ,   # Match the character 「,」 literally 
    |   # Or match regular expression number 2 below (the entire group fails if this one fails to match) 
     \$   # Assert position at the end of the string (or before the line break at the end of the string, if any) 
) 
" 

Secodnly要更換日期：

$subject =~ s!\d{2}/\d{2}/\d{4}.*?(?=,)!sysdate!g; 

這幾乎與原來的正則表達式相同。只要用前視來替換最後一個。 （如果您不想替換它，請不要匹配。）

# \d{2}/\d{2}/\d{4}.*?(?=,) 
# 
# Match a single digit 0..9 «\d{2}» 
# Exactly 2 times «{2}» 
# Match the character 「/」 literally «/» 
# Match a single digit 0..9 «\d{2}» 
# Exactly 2 times «{2}» 
# Match the character 「/」 literally «/» 
# Match a single digit 0..9 «\d{4}» 
# Exactly 4 times «{4}» 
# Match any single character that is not a line break character «.*?» 
# Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?» 
# Assert that the regex below can be matched, starting at this position (positive lookahead) «(?=,)» 
# Match the character 「,」 literally «,»