Php 如何在mysql中从json中提取数据
我正在尝试从mysql中的json提取数据: 我的示例mysql查询是: 我已经尝试使用下面的代码,但它返回空值,但我在数据库中有值。 在下面的代码列中,列_name 1和er_trading_sales是列名,表_name是表的名称Php 如何在mysql中从json中提取数据,php,mysql,json,Php,Mysql,Json,我正在尝试从mysql中的json提取数据: 我的示例mysql查询是: 我已经尝试使用下面的代码,但它返回空值,但我在数据库中有值。 在下面的代码列中,列_name 1和er_trading_sales是列名,表_name是表的名称 SELECT column_name1 as column_name1, substring( er_trading_sales, locate('"C2":{"type":"label","value":"
SELECT column_name1 as column_name1,
substring(
er_trading_sales,
locate('"C2":{"type":"label","value":"":{"m":"label","k":"(a) to registered dealers","v":"}","C3": { "type":"textbox","value":"',er_trading_sales,locate('"Trading_B_I"',er_trading_s ales))+70,
(locate('"},"C4',er_trading_sales,locate('"C2":{"type":"label","value":"":{"m":"label","k":"(a) to registered dealers","v":"}","C3":{"type":"textbox","value":"',er_trading_sales,locate('"Trading_B_I"',er_trading_sales))) - (locate('"C2":{"type":"label","value":"":{"m":"label","k":"(a) to registered dealers","v":"}","C3": {"type":"textbox","value":"',er_trading_sales,locate('"Trading_B_I"',er_trading_ sales))+70))
) as TradingAmount
from table_name
//sample json
"R10" => array(
"col_data" => array(
"CR1" => array(
"C1" => array(
"type" => "label",
"value" => "8.",
),
"C2" => array(
"type" => "complex",
"value" => array(
"m" => "label",
"k" => "Turnover of sale of goods taxable @ ^d^%",
"v" => "4.5"
),
),
"C3" => array(
"type" => "textbox",
"value" => "",
"validation" => "decimal",
),
"C4" => array(
"type" => "textbox",
"value" => "",
"validation" => "decimal",
'htmlOptions' => array('readonly' => 'readonly'),
),
"C5" => array(
"type" => "",
"value" => "",
),
),
),
),
i want "value" of "C3" => array(
"type" => "textbox",
"value" => "",
"validation" => "decimal",
)
如果要存储JSON,请至少开始使用JSON扩展。如果要存储JSON,请至少开始使用JSON扩展