PHP preg_replace function problems

PHP preg_replace function problems

$string = "april 15, 2003";$pattern = "/(/w+) (/d+), (/d+)/i";$replacement = "/${1}1,/$3";print preg_replace($pattern, $replacement, $string);/* output   ======april1,2003*/

$ Replacement = "/$ {1} 1,/$3 ";
"/$ {1} 1,/$3" what does this mean?

$patterns = array ("/(19|20)(/d{2})-(/d{1,2})-(/d{1,2})/",                   "/^/s*{(/w+)}/s*=/");$replace = array ("//3///4///1//2", "$//1 =");print preg_replace ($patterns, $replace, "{startdate} = 1999-5-27");

$ Replace = array ("// 3 // 4 /// 1 // 2", "$ // 1 ="); what does this mean?

Reply to discussion (solution)

$ {1} is the first matching, and $3 is the third matching.

$ {1} is the first matching, and $3 is the third matching.

The second code is wrong. how can I output $ startdate = 5/27/1999?

$ {1} is the first matching, and $3 is the third matching.

$ {1} is there an escape \ slash in front of it not the original output?

The first piece of code is wrong and should be written

$string = "april 15, 2003";$pattern = "/(\w+) (\d+), (\d+)/i";$replacement = "\${1}1,\$3";print preg_replace($pattern, $replacement, $string);

Or

$string = "april 15, 2003";$pattern = "/(\w+) (\d+), (\d+)/i";$replacement = '${1}1,$3';print preg_replace($pattern, $replacement, $string);

Similarly, the second paragraph of code should be written.

$patterns = array ("/(19|20)(\d{2})-(\d{1,2})-(\d{1,2})/",                   "/^\s*{(\w+)}\s*=/");$replace = array ("\\3/\\4/\\1/\\2", "$\\1 =");print preg_replace ($patterns, $replace, "{startdate} = 1999-5-27");

$ N or \ n are backward referenced marks, indicating the content in the nth parentheses

The first piece of code is wrong and should be written

$string = "april 15, 2003";$pattern = "/(\w+) (\d+), (\d+)/i";$replacement = "\${1}1,\$3";print preg_replace($pattern, $replacement, $string);

Or

$string = "april 15, 2003";$pattern = "/(\w+) (\d+), (\d+)/i";$replacement = '${1}1,$3';print preg_replace($pattern, $replacement, $string);

Similarly, the second paragraph of code should be written.

$patterns = array ("/(19|20)(\d{2})-(\d{1,2})-(\d{1,2})/",                   "/^\s*{(\w+)}\s*=/");$replace = array ("\\3/\\4/\\1/\\2", "$\\1 =");print preg_replace ($patterns, $replace, "{startdate} = 1999-5-27");

$ N or \ n are backward referenced marks, indicating the content in the nth parentheses

Thank you. $ {1} does the previous escape slash \ mean the original output?

In double quotes, php tries to interpret $ as the leading variable.
The string "$ {1}" will cause an undefined variable error.
So escape the code so that $ {1} can be handed over to prea_replace for processing.

In double quotes, php tries to interpret $ as the leading variable.
The string "$ {1}" will cause an undefined variable error.
So escape the code so that $ {1} can be handed over to prea_replace for processing.

Thank you.