PHP正则表达式帮助 - 在另一个匹配中找到匹配并替换一些东西

时间:2010-03-03 03:17:51

标签: php regex

我有阵列,我必须匹配这种文本,然后删除单词之间的空格,

 Name:'lofse erbbnwq qweqw-qweqw' KKK
 Name:'lofsdsse erbsdsdbnwq sds sdsd sdqwsdseqw-qwsdseqw' KKK
 Name:'lofsse esdsdbnwq sds sds sddseqw-qwseqw' KKK



 i read somewhere that it will work like this, but i tried and its not working :(

$data = preg_replace_callback('%Name:\'(.*)\' kkk%',replace_within_tag, $data);
function replace_within_tag($groups) {return preg_replace('/\s/', '.', $groups[0]);}

输出应该是这样的

 Name:'lofse.erbbnwq.qweqw-qweqw' KKK
 Name:'lofsdsse.erbsdsdbnwq.sds.sdsd.sdqwsdseqw-qwsdseqw' KKK
 Name:'lofsse.esdsdbnwq.sds.sds.sddseqw-qwseqw' KKK

请为此我需要一些快速帮助,告诉我工作方式

2 个答案:

答案 0 :(得分:1)

$array=array("Name:'lofse erbbnwq qweqw-qweqw' KKK", "Name:'lofsdsse erbsdsdbnwq sds sdsd sdqwsdseqw-qwsdseqw' KKK","Name:'lofsse esdsdbnwq sds sds sddseqw-qwseqw' KKK");
foreach ($array as $k=>$v){
  if ( strpos($v,"Name:" ) !==FALSE) {
        $s = explode("'",$v);
        $s[1]=preg_replace("/\s+/",".",$s[1]);
        $array[$k]=implode("'",$s);
  }
}
print_r($array);

输出

$ php test.php
Array
(
    [0] => Name:'lofse.erbbnwq.qweqw-qweqw' KKK
    [1] => Name:'lofsdsse.erbsdsdbnwq.sds.sdsd.sdqwsdseqw-qwsdseqw' KKK
    [2] => Name:'lofsse.esdsdbnwq.sds.sds.sddseqw-qwseqw' KKK
)

答案 1 :(得分:0)

对匹配进行分组,然后在回调中使用$matches[1],仅替换引号之间文本部分中的空格。你有几种方法可以做到这一点。例如:

$output = preg_replace_callback("!(Name:')(.*?)(' KKK)!", 'replace_spaces', $input);

function replace_spaces($matches) {
  return $matches[1] . preg_replace('!\s+!', '.', $matches[2]) . $matches[3];
}

您需要执行此操作,因为您正在捕获前导和尾随字符串。另一种方法是捕获更少。例如:

$output = preg_replace_callback("!(?<=').*?(?=')!", 'replace_spaces', $input);

function replace_spaces($matches) {
  return preg_replace('!\s+!', '.', $matches[0]);
}

这是使用lookaheads and lookbehinds