PHP 5.3 eregi_replace到的preg_replace

Question

可能重複：
Converting ereg expressions to preg

我試圖解決一些舊代碼，使其與PHP 5.3的作品。 希望第一次轉換到str_replace是可以的，但我完全不確定第二次轉換爲preg_replace。

任何幫助將非常感激。由於

$this->Query = str_replace(" where "," WHERE ", $this->Query); 
$this->Query = str_replace(" select "," SELECT ", $this->Query); 
$this->Query = str_replace(" from "," FROM ", $this->Query); 
$this->Query = str_replace(" as "," AS ", $this->Query); 

// $this->Query = eregi_replace(" WHERE ", " where ", $this->Query); 
// $this->Query = eregi_replace("SELECT ", "select ", $this->Query); 
// $this->Query = eregi_replace(" FROM ", " from ", $this->Query); 
// $this->Query = eregi_replace(" AS ", " as ", $this->Query); 

$TempQuery = eregi_replace("^select .* from ", "select count(1) from ", $this->Query); 
$TempQuery = eregi_replace(" order by .*$", "", $TempQuery); 

Answer 1

至於說，SQL語法不必是大寫......不過無論如何回答你的問題：

$TempQuery = preg_replace("/^select .*? from /i", "select count(1) from ", $this->Query); 
$TempQuery = preg_replace("/ order by .*?$/i", "", $TempQuery); 

這應該工作。我對正則表達式不區分大小寫，因爲你似乎遇到了大寫和小寫的問題。此外，我通過添加一個問號以避免額外的複雜性而使得.*不明確。

Answer 2

* eregi_replace *必須是*的preg_replace *代替

例如：

$string = $this->Query; 
$pattern = '/^select .* from /i'; 
$replacement = "select count(1) from "; 
$TempQuery = preg_replace($pattern, $replacement, $string);