在字符串中查找JSON字符串 [英] Find JSON strings in a string

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本文介绍了在字符串中查找JSON字符串的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在寻找一种在字符串中查找JSON数据的方法.像wordpress简码一样思考它.我认为最好的方法是使用正则表达式.我不想解析JSON,只需查找所有出现的情况即可.

I am looking for a way to find JSON data in a string. Think about it like wordpress shortcodes. I figure the best way to do it would be a regular Expression. I do not want to parse the JSON, just find all occurences.

正则表达式中是否有一种方法可以使括号的数量匹配?目前,当我嵌套对象时遇到了这个问题.

Is there a way in regex to have matching numbers of parentheses? Currently I run into that problem when having nested objects.

简单的演示示例:

This is a funny text about stuff,
look at this product {"action":"product","options":{...}}.
More Text is to come and another JSON string
{"action":"review","options":{...}}

因此,我想拥有两个JSON字符串. 谢谢!

As a result i would like to have the two JSON strings. Thanks!

推荐答案

从给定文本中提取JSON字符串

由于您正在寻找一种简单的解决方案,因此可以使用以下使用递归的正则表达式来解决匹配括号集的问题.它递归地匹配{}之间的所有内容.

Extracting the JSON string from given text

Since you're looking for a simplistic solution, you can use the following regular expression that makes use of recursion to solve the problem of matching set of parentheses. It matches everything between { and } recursively.

尽管,您应该注意,不能保证此方法在所有可能的情况下都适用.它仅用作快速JSON字符串提取方法.

Although, you should note that this isn't guaranteed to work with all possible cases. It only serves as a quick JSON-string extraction method.

$pattern = '
/
\{              # { character
    (?:         # non-capturing group
        [^{}]   # anything that is not a { or }
        |       # OR
        (?R)    # recurses the entire pattern
    )*          # previous group zero or more times
\}              # } character
/x
';

preg_match_all($pattern, $text, $matches);
print_r($matches[0]);

输出:

Array
(
    [0] => {"action":"product","options":{...}}
    [1] => {"action":"review","options":{...}}
)

Regex101演示

在PHP中,了解JSON字符串是否有效的唯一方法是应用json_decode().如果解析器理解JSON字符串并且符合定义的标准,则json_decode()将创建JSON字符串的对象/数组表示形式.

In PHP, the only way to know if a JSON-string is valid is by applying json_decode(). If the parser understands the JSON-string and is according to the defined standards, json_decode() will create an object / array representation of the JSON-string.

如果您想过滤掉无效的JSON,则可以使用 array_filter() 具有回调功能:

If you'd like to filter out those that aren't valid JSON, then you can use array_filter() with a callback function:

function isValidJSON($string) {
    json_decode($string);
    return (json_last_error() == JSON_ERROR_NONE);
}

$valid_jsons_arr = array_filter($matches[0], 'isValidJSON');

在线演示

这篇关于在字符串中查找JSON字符串的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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