Question

我的字符串具有以下模式。 Some text on some more text before some text。

我如何找到

“之前”一词的位置，然后将其删除，
以及“on”这个词的位置并在之前删除它？

Answer 1

正则表达式对此有些过分。只需从你找到“之前”的地方做一个子串：

$str = 'Some text on some more text before some text';

// Find " on "
$start = strpos($str, " on ");

// Check for errors
// Note the 3 ='s
if($start === FALSE) {
    // Error occurred, do something
}
$start += 4; // Go after "on"

// Find " before "
$end = strpos($str, " before ", $start);

// Check for errors
// Note the 3 ='s
if($end === FALSE) {
    // Error occurred, do something
}

// Get just the part we want
$length = $end - $start;
$result = substr($str, $start, $end - $start);

Answer 2

preg_match('~\bon\b(.+)\bbefore\b~',$text,$match);
//use $match[1]; e.g $text=$match[1];

Answer 3

考虑使用正则表达式：

$text= "Some text on some more text before some text";
$pattern = '/^.* on(.*)before.*$/';

preg_match($pattern, $text, $matches);
if($matches) {
    print $matches[1];
}

Answer 4

试试这个：

<?php

$string = 'Some text on some more text before some text';

$newString = preg_replace('/(.*)on(.*)before(.*)/', '$2', $string);

echo $newString;

正则表达式：http://regexr.com?2uqq7

PHP演示：http://codepad.org/uXBL8IaA

Answer 5

    $var="Some text on some more text before some text";
preg_match('/\bon\b(.*)\bbefore\b/',$var,$match);
//Edit: You can use `(.+)` too if, by any chance, there is no text between 'on' and 'before'.
print_r ($match);
//$match[1] has "some more text"

这是你想要的吗？

至于位置事物，你可以echo strrpos($var, " on ");如果我正确地得到这个（正如你在评论中说的那样“on”周围有空格。echo strrpos($var, "before");这将会返回第一个角色的位置。

Answer 6

$str = 'Some text on some more text before some text';
$str = str_replace(array('on','before'), '', $str);

如何正确破坏这个字符串

6 个答案: