Question

我有这样的输入（JSON格式）

｛“location”：[｛“id”：“1BCDEFGHIJKLM”，“somename”：“abcd”，“fewlame”：“xyzland”，“sid”：“”，“sname”：“，”regioname“：”Zee Whole“，”type“：”some“，”siteCode“：”“，”someCode“：“ROTXY”，”fewlCode“：HMOPRE“，”somename“：”abcd Junction“，”fewname“：”United States“，”sid“：”“，”sname“：”“USRTJ”，“fewlCode”：“US”，“pCode”：“USNWK”，“someid”：“7823YZHMOPRE”，“Fewlid”：“：”NL“，”pCode“：”“，”someid“：”799XYZHMOPRE“，”fewlid“：“OIUOWER348534”｝]｝

现在，我想使用正则表达式获取第一个“id”值，即1BCDEFGHIJKLM。我用

[^（｛“location”：[？｛“id”：“）].｛0,12｝但这是不完整的。有人能帮助我如何忽略值1BCDEFGHIJKLM之后的其余部分吗

Answer 1

Regex isn t the way to do this. Whatever platform you are using, it must have a JSON parser. That will be your best error-free solution.

假设您必须使用regex，则可以使用“id”：“（.*？）”获取所有id，并进行第一个匹配。

我发现了以下文章，这可能会对您有所帮助。

Answer 2

虽然很混乱，但您的正则表达式是如何不完整的？

它可以缩短为（“id”：“（[^”]+）”，这更具可读性，并且不将id限制在12个字符以内。如果这样做有益的话。

如果你的问题是得到多个结果，大多数语言都有一个“g”全局开关。

在javascript中，以下内容将返回“1BCDEFGHIJKLM”：

var firstID = str.match(/"id":"([^"]+)"/)[1]

As match（）返回一个数组，其中[0]是返回的整个字符串，[1]是第一个括号。

Answer 3

不必使用正则表达式。用你最喜欢的语言，用逗号分隔。然后浏览每个项目，检查“id”并在冒号（：）上拆分。获取最后一个元素。例如Python

>>> s
 {"location":[{"id":"1BCDEFGHIJKLM","somename":"abcd","fewname":"xyzland","sid":"","sname":"","regionname":"Zee-Whole","type":"some","siteCode":"","someCode":"ROTXY","fewCode":"NL","pCode":"ROTXY","someid":"1BCDEFGHIJKLM","fewid":"GIC8"},{"id":"7823XYZHMOPRE","somename":"abcd Junction","fewname":"United States","sid":"","sname":"","regionname":"New York","type":"some","siteCode":"","someCode":"USRTJ","fewCode":"US","pCode":"USNWK","someid":"7823XYZHMOPRE","fewid":"7823XYZLMOPRE"},{"id":"799XYZHMOPRE","somename":"abcd-Maasvlakte","fewname":"xyzland","sid":"","sname":"","regionname":"Zee-Whole","type":"some","siteCode":"","someCode":"XYROT","fewCode":"NL","pCode":"","someid":"799XYZHMOPRE","fewid":"OIUOWER348534"}]} 

>>> for i in s.split(","):
...     if  "id"  in i:
...         print i.split(":")[-1]
...         break
...
"1BCDEFGHIJKLM"

当然，理想情况下，您应该使用专用的JSON解析器。

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