用正则表达式匹配Word

Question

I have the following text file:

...
"somewords MYWORD";123123123123
"someother MYWORDOTHER";456456456456
"somedifferent MYWORDDIFFERENT";789789789
...

I need to match the word MYWORD, MYWORDOTHER, MYWORDDIFFERENT and then substitute the space before this word with ";". Someone can figure out a regex?

I have done something like that:

 +[^ ][^ ][^ ][^ ][^ ][^ ][^ ]";

but this works only with a specific word length. I need to modify to get any word of any length.

Any help?

Answer 1

why don't you str_replace() ?

$string = '"somewords MYWORD";123123123123
"someother MYWORDOTHER";456456456456
"somedifferent MYWORDDIFFERENT";789789789';

$replace = str_replace(' MYWORD', ';MYWORD', $string);
echo $replace;

Codepad Example

Answer 2

这未经测试，但是应该可以替换引号中最后一个单词之前的空格...

preg_replace('/(".+) (\w+";\d+)/',"$1;$2", $your_string);

Answer 3

preg_replace('/\s(\w+);\d/', ';$1', $text);

Answer 4

Maybe this:

$result = preg_replace('/([ ])(\w+)";/im', ';$2";', $subject);

in:

"somewords MYWORD";123123123123 
"someother MYWORDOTHER";456456456456 
"somedifferent MYWORDDIFFERENT";789789789

out:

"somewords;MYWORD";123123123123
"someother;MYWORDOTHER";456456456456
"somedifferent;MYWORDDIFFERENT";789789789

Answer 5

用这个 ：

preg_replace('#"(\w+)\s+(\w+)"#',"$1;$2",$text);

Answer 6

while($line=fgets($file))
{
    $str=preg_replace("/ (\w)/i",";$1",$line);//use this line if you want to replace every space
    $str=preg_replace("/ (\w+)\";(\d)/i",";$1\";$2",$line);//use this line if you only want to replace the last space
    echo $str;//or wherever you want to output
}

Edit :

Alright, I made a typo in the original answer.

Now corrected with a codepad: http://codepad.org/leQHTuFR