使用 preg_replace 删除引号之间的内容以外的所有内容

Question

我有以下简码，我正在尝试运行 a preg_replace，以获取标题值。

[tab title="Emergency Contact" important="true"]

目前，我使用以下过滤器：

$sanitized_title = $tab_title ? preg_replace("/[\”\"\’\']*|(?:&#8217;|&#8221;)*\]/i", "", preg_replace("/\[tab *title=[\”\"\’\']*|(?:&#8217;|&#8221;)*/i", "", $tab_title[0])) : __("Open Tab");

这个returns"Emergency Contact important=true"，这不是我需要的。我基本上是想得到类似 $title = "Emergency Contact" 和 $important = "true" 的东西。

如何修改我的正则表达式字符串来执行此操作？我真的不知道我在用正则表达式做什么，很惊讶我已经做到了。

还有一点需要注意，并非每个简码都具有这两个值。一些替代示例：

[tab]
[tab title="Emergency Contact"]
[tab important="true"]

Answer 1

您是否在寻找类似的东西：

<?php

$string = 'some gibberish here [tab title="Emergency Contact" important="true"] some additional info here';

$regex = '~
            (?:\[tab\s         # looks for [tab + whitespace literally
            |                  # or
            (?!\A)\G\s)        # (?!\A) = negative lookahead
                               # to make sure the following is not the start
                               # of the string
                               # \G matches at the end of the previous match
            (?P<key>\w+)       # match a word character greedily to group "key"
            =
            "(?P<value>[^"]+)" # " + anything not " + ", save this to group "value"
         ~x';                  # verbose modifier
preg_match_all($regex, $string, $matches, PREG_SET_ORDER);
foreach ($matches as $match)
    echo "Key: {$match['key']}, Value: {$match['value']}\n";
/* output:
Key: title, Value: Emergency Contact
Key: important, Value: true
*/
?>

这将查找 [tab] 标签中的所有 key/value 对，参见 a demo on regex101.com。
额外 demo can be found on ideone.com.

Answer 2

您可以使用 "preg_match_all".

获取简码属性

<?php
//string
$qoute_string = '[tab title="Emergency Contact" important="true"]';

//pattern
$pattern = '/ (.*?)="(.*?)"/';

$tab_attr = array();
//get the attributes.
if(preg_match_all($pattern, $qoute_string, $data)) {
    foreach ($data[0] as $key => $val) {
        $variable = explode("=",$val);
        $tab_attr[trim($variable[0])] = str_replace('"', "", $variable[1]);
    }
}

//see the results
print_r($tab_attr);
echo '<br />';

//title
echo $tab_attr['title'];
echo '<br />';

//important
echo $tab_attr['important'];
?>

Answer 3

I'm basically trying to get something like $title = "Emergency Contact" and $important = "true".

尝试[a-z]+="[A-Za-z ]+"。

这是最简单的正则表达式，将匹配以下格式的模式。

Characters from a-z="Characters from A-Z and a-z including space"

Regex101 Demo

使用 preg_replace 删除引号之间的内容以外的所有内容

Using preg_replace to remove everything except content between quotes

php

regex

preg-replace