t1.id AS item_id, t1.title AS item_name,t6.id AS topic_id, t6.title AS topic_name, t1.extra AS media_id, t1.biz_type from tem t1join component_item t2 on t1.id = t2.item_idjoin component t3 on t2.component_id = t3.idjoin drawer t4 on t4.id = t3.drawer_idjoin channel_drawer t5 on t5.drawer_id = t4.idjoin channel t6 on t6.id = t5.channel_idwhere t1.biz_type in ("JUMP_TO_SHOW","JUMP_TO_VIDEO")and t1....
mysql查询json的值的方法：首先打开命令窗口；然后执行SQL语句“SELECT REPLACE(json_extract(push_data,$.carRenewalInfoVo.licence)...”来查询json的值即可。推荐：《mysql视频教程》mysql查找json的某个字段SELECT json_extract(字段名,$.json结构) FROM 表名;如果json里有双引号，那这样取出来的数据也带双引号，要去掉就使用REPLACE函数例如：ps_push_data表里的push_data字段存的数据为：{"carRenewalInfoVo":{"licence":"浙...
表结构如下：id varchar(32) info json 数据：id = 1 info = {"age": "18","disname":"小明"}-------------------------------------------- 现在我需要获取info中disanme的值，查询方法有： 1.select t.id,JSON_EXTRACT(t.info,$.disname) as disname from tableName t where 1=1结果：id = 1， disname=“小明”以上sql查出的disname值是带有双引号的，有时我们不需要双引号，此时就需要用到下面这种方式。 2.select t.id,...
-- 要查找的值 SET @valueMapping = 17;-- 表字段：id, config -- config字段格式： /* {"fieldModels": [{"key": 0,"guid": "1","field": "Id","dataType": 1,"showName": "标识","textFormat": "","valueMapping": 17}, {"key": 1,"guid": "2","field": "orderid","dataType": 0,"showName": "orderid","textFormat": "","valueMapping": -1} } */ -- 需求：查找 config JSON字段（对象类型）中 fieldModels（数组类型）数组字段...
如何根据表格中字段的平均评级对查询进行排序,字段本身是JSON文本,结构如下：[{"Type":1,"Rating":5},{"Type":2,"Rating":5},{"Type":3,"Rating":5} ]我需要将我的查询按3个评级的平均值排序.总是只有3个值. 我目前的查询是：SELECT `Name`, `Town`, `Vehicle`, `Review`, `Rating`, `Pics`, `PostedOn` FROM `tbl_ShopReviews` WHERE `Approved` = 1 ORDER BY `PostedOn` DESC目前的结果：Name Town Vehicle Review Ratin...
我正在为MySQL中的JSON数组寻找类似forEach的东西. 我在MySQL JSON data type中的经理ID是这样的：[1,2,3,4,5],我想对列表中的每个项目执行操作. 一个简单的解决方案是使用从0开始的计数器执行WHILE循环,并在VAR_MANAGER_ID为空时结束.这是一个设计的例子,说明WHILE循环内部的外观：SET VAR_PATH = CONCAT('$[', COUNTER, ']'); SET VAR_MANAGER_ID = JSON_PARSE(VAR_MANAGER_IDS, PATH);# See if we've reached the end of the li...
我正在尝试创建一个JSON对象,然后将其中的值读入MySQL表.但是我面临着错误,而且我是JSON和MySQL的新手.SET @j = '{"key1": "value1", "key2": "value2"}'; CREATE TABLE Person (name int,id int); INSERT INTO Person (name,id) SELECT * FROM OPENJSON(@j) WITH (name int,id int);解决方法:在创建表时,将字段设置为JSON数据类型.CREATE TABLE `person` (`name` json DEFAULT NULL );并将JSON数据插入其中,INSERT INTO `person` ...
我有一个PHP脚本,有三个查询,每个查询返回一组结果.现在我想将三个查询的结果合并到一个JSON数组中. 谁知道如何实现这一目标？查询无法加入. 谢谢.解决方法:你可能想看看array_merge(). 一个简单的例子：<?php $result = array_merge($array_result_of_query1, $array_result_of_query2, $array_result_of_query3); echo json_encode($result); ?>
我有下表product_id product_name image_path misc ---------- -------------- ------------ ------1 flex http://firstpl... {"course_level_id":19,"group_id":"40067"}2 Android http://firstpl... {"course_level_id":20,"group_id":"40072"}那么如何检索product_name,image_path&只有“group_id”值类似“misc”...
我正在尝试将下面的mysql查询结果网格化为单个json对象,但不太确定如何正确地执行它.$id = $_POST['id'];$sql = "SELECT contracts.po_number, contracts.start_date, contracts.end_date, contracts.description, contracts.taa_required, contracts.account_overdue, jobs.id AS jobs_id, jobs.job_number, companies.id AS companies_id, companies.name AS companies_name FROM contracts LEFT JOIN jobs ON contracts.job_id ...