使用忽略引用字符串的 PHP preg_replace 函数替换空格

Question

看下面的字符串：

SELECT
    column1 ,
    column2, column3
FROM
    table1
WHERE
    column1 = 'text, "FROM" \'from\\\' x' AND
    column2 = "sample text 'where' \"where\\\" " AND
    ( column3 = 5 )

我需要从字符串中转义不必要的空白字符，例如：

• 从 、( ) 等的开始和结束位置删除空格
• 删除换行符 (\r\n) 和制表符 (\t)

但有一件事。删除过程无法从引用的字符串中删除空格，例如：

• 'text, "FROM" \'from\\' x'
• "示例文本 'where' \"where\\" "

等等

我需要使用PHP函数：preg_replace($pattern, $replacement, $string);

那么 $pattern 和 $replacement 的值是多少，其中 $string 的值是给定的 SQL。

Answer 1

单个正则表达式模式和替换字符串字符串将不起作用。您的第一步可能是对输入字符串进行标记：首先尝试匹配 cmets 和字符串文字，然后尝试匹配空白字符，最后尝试匹配非空格字符。快速演示：

$text = <<<BLOCK
SELECT
    column1 ,
    column2, column3
FROM
    table1
-- a comment line ' " ...
WHERE
    column1 = 'text, "FROM" \\'from\\\\\\' x' AND
    column2 = "sample text 'where' \\"where\\\\\\" " AND
    ( column3 = 5 )
BLOCK;

echo $text . "\n\n";

preg_match_all('/
    --[^\r\n]*                # a comment line
    |                         # OR
    \'(?:\\\\.|[^\'\\\\])*\'  # a single quoted string
    |                         # OR
    "(?:\\\\.|[^"\\\\])*"     # a double quoted string
    |                         # OR
    `[^`]*`                   # a string surrounded by backticks
    |                         # OR
    \s+                       # one or more space chars
    |                         # OR
    \S+                       # one or more non-space chars
/x', $text, $matches);

print_r($matches);

产生：

SELECT
    column1 ,
    column2, column3
FROM
    table1
-- a comment line ' " ...
WHERE
    column1 = 'text, "FROM" \'from\\\' x' AND
    column2 = "sample text 'where' \"where\\\" " AND
    ( column3 = 5 )

Array
(
    [0] => Array
        (
            [0] => SELECT
            [1] => 

            [2] => column1
            [3] =>  
            [4] => ,
            [5] => 

            [6] => column2,
            [7] =>  
            [8] => column3
            [9] => 

            [10] => FROM
            [11] => 

            [12] => table1
            [13] => 

            [14] => -- a comment line ' " ...
            [15] => 

            [16] => WHERE
            [17] => 

            [18] => column1
            [19] =>  
            [20] => =
            [21] =>  
            [22] => 'text, "FROM" \'from\\\' x'
            [23] =>  
            [24] => AND
            [25] => 

            [26] => column2
            [27] =>  
            [28] => =
            [29] =>  
            [30] => "sample text 'where' \"where\\\" "
            [31] =>  
            [32] => AND
            [33] => 

            [34] => (
            [35] =>  
            [36] => column3
            [37] =>  
            [38] => =
            [39] =>  
            [40] => 5
            [41] =>  
            [42] => )
        )

)

然后您可以遍历标记化的 $matches 数组并替换您认为合适的空间匹配。

但正如您可能在我已删除的评论中看到的那样，更好的选择是使用一些专用的 SQL 解析器来执行此标记化：我不精通 SQL，但我很确定我上面的演示很容易被破坏.