php - 用###替换PHP中字符串中“”之间的文本中的空格

Question

我有一个这样的字符串text more text "empty space"。如何"empty space"用### 替换其中的空格并且仅替换这个空格？

Answer 1

$string = 'text more text "empty space"';
$search = 'empty space';
str_replace($search, 'empty###space', $string);

Answer 2

$somevar = "empty space";
$pattern = "/\s/";
$replacement = "###";
$somevar2 = preg_replace($pattern, $replacement, $somevar);
echo $somevar2;

Answer 3

$string = "My String is great";
$replace = " ";
$replace_with = "###";

$new_string = str_replace($replace, $replace_with, $string);

这应该为你做。http://www.php.net/manual/en/function.str-replace.php

Answer 4

评论后编辑

也许这不是最好的解决方案，但你可以这样做：

$string = 'text more text "empty space"';
preg_match('/(.*)(".*?")$/', $string, $matches);
$finaltext = $matches[1] . str_replace(' ', '###', $matches[2]);

Answer 5

这个怎么样，没有正则表达式：

$text = 'foo bar "baz quux"';
$parts = explode('"', $text);
$inQuote = false;

foreach ($parts as &$part) {
    if ($inQuote) { $part = str_replace(' ', '###', $part); }
    $inQuote = !$inQuote;
}

$parsed = implode('"', $parts);
echo $parsed;