preg_replace（）的意外结果

Question

$subject = "'foo' = 'bar'";
$pattern = "/('foo' = ').*(')/";
$var = "123baz";
$replacement = "$1$var$2";
print_r(preg_replace($pattern, $replacement, $subject));

结果为23baz'而不是'foo' = '123baz'。为什么以及如何解决这个问题？

Answer 1

认为你正在寻找类似的东西：

$subject = "'foo' = 'bar'";
$pattern = "/(= ').*(')/";
$var = "123baz";
$replacement = "= '$var'";
print_r(preg_replace($pattern, $replacement, $subject));

输出：

'foo' = '123baz'