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mysql如何使用JSON_EXTRACT()取json值
2022-07-05 16:42:00 【1024问】
mysql取json字符串字段下的某个键的值
1.使用replace()做替换
2.使用 JSON_UNQUOTE()
mysql处理json字符串,JSON_EXTRACT()提取内容
MySQL自5.7之后开始支持json类型
mysql取json字符串字段下的某个键的值要求:mysql版本5.7及以上
SELECT JSON_EXTRACT('{"uid":"asas02234"}', "$.uid");
由于json的键值是带双引号。所以需要去掉双引号。
1.使用replace()做替换select replace(JSON_EXTRACT(infoJson,'$.uid'),'"','') uidfrom users
laravel 里使用JSON_EXTRACT
$applyList = DB::table('invoice') ->select('applied_at','total_amount','invoice_form','invoice_file','reject_reason') ->selectRAW('replace(JSON_EXTRACT(invoice_detail,"$.owner_name"),\'"\',\'\') as owner_name') ->where('uid',Auth::id())->get();
2.使用 JSON_UNQUOTE()SELECT JSON_UNQUOTE(JSON_EXTRACT('{"id":"3"}', "$.id"));
$applyList = DB::table('invoice') ->select('applied_at','total_amount','invoice_form','invoice_file','reject_reason') ->selectRAW('JSON_UNQUOTE(JSON_EXTRACT(invoice_detail,"$.owner_name")) as owner_name') ->where('uid',Auth::id())->get();
mysql处理json字符串,JSON_EXTRACT()提取内容MySQL自5.7之后开始支持json类型相应的解析函数主要是JSON_EXTRACT():
SELECT JSON_EXTRACT('{"ID":"1"}','$.ID');
执行结果:
JSON_EXTRACT('{"ID":"1"}','$.ID')
"1"
去除引号使用函数JSON_UNQUOTE():
SELECT JSON_UNQUOTE(JSON_EXTRACT('{"ID":"1"}','$.ID'));
执行结果:
JSON_UNQUOTE(JSON_EXTRACT('{"ID":"1"}','$.ID'))
1
以上为个人经验,希望能给大家一个参考,也希望大家多多支持软件开发网。
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