结合使用JSON_SEARCH和JSON_EXTRACT，我得到:“无效的JSON路径表达式". [英] Combining JSON_SEARCH and JSON_EXTRACT get me: "Invalid JSON path expression."

问题描述

我有一个名为广告系列"的表格.其中一列名为"filter_apps"，其类型为 JSON

I have a table names "campaigns". One of the columns is named "filter_apps" and his type is JSON

我有文件行，它们只包含令牌数组，如下所示:

I have file rows and they just contain array of tokens like so:

["be3beb1fe916ee653ab825fd8fe022", "c130b917983c719495042e31306ffb"]
["4fef3f1999c78cf987960492da4d2a"]
["106c274e319bdeae8bcf8daf515b1f"]
["2521f0df6cffb7487d527319674cf3"]
["c130b917983c719495042e31306ffb"]

示例:

SELECT JSON_SEARCH(filter_apps, 'one', 'c130b917983c719495042e31306ffb') FROM campaigns;

结果:

"$[1]"
null
null
null
"$[0]"

现在一切都正确了，匹配的列又回来了.如果我进行测试，我可以证明这一点:

Right now everything is correct, the matched columns come back. If I make a test I can prove it:

SELECT JSON_EXTRACT(filter_apps, '$[1]') FROM campaigns;

结果

"c130b917983c719495042e31306ffb"
null
null
null
null

因此，在这一点上，我认为我可以使用查询JSON_EXTRACT提取值:

So in this point I think I can extract the values using JSON_EXTRACT, my query:

SELECT JSON_EXTRACT(filter_apps, JSON_SEARCH(filter_apps, 'one', 'c130b917983c719495042e31306ffb')) FROM campaigns;

这导致我出现错误:

"[42000] [3143]无效的JSON路径表达式.错误在字符位置1附近."

"[42000][3143] Invalid JSON path expression. The error is around character position 1."

推荐答案

解决方案

就这么简单:

SELECT JSON_EXTRACT(filter_apps, JSON_UNQUOTE(JSON_SEARCH(filter_apps, 'one', 'c130b917983c719495042e31306ffb'))) FROM campaigns;

问题已解决！我将JSON_SEARCH包装在JSON_UNQUOTE方法中！

Problem resolved! I wrap JSON_SEARCH in a JSON_UNQUOTE method!

一个小技巧，我在这里找到了解决方案: https://dev.mysql.com/doc/refman/5.7/zh-CN/json-function-reference.html

A little tip, I found the solution here: https://dev.mysql.com/doc/refman/5.7/en/json-function-reference.html

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