MySQL视图 – 非法混合排序规则

我会非常清楚：在MySQL中创建视图的解决方案是什么,没有非法的非法混合排序错误.

我的SQL代码是这样的(它有一些葡萄牙语单词),我的数据库默认排序规则是latin1_swedish_ci：

CREATE VIEW v_veiculos AS
SELECT
    v.id,
    v.marca_id,
    v.modelo,
    v.placa,
    v.cor,
    CASE v.combustivel
        WHEN 'A' THEN 'Álcool'
        WHEN 'O' THEN 'Óleo Diesel'
        WHEN 'G' THEN 'Gasolina'
        ELSE 'Não Informado'
    END AS combustivel,
    marcas.marca,
    /*I think that the CONCAT and COALESCE below causes this error, when the next line the view works fine*/
    CONCAT(marca, ' ', v.modelo, ' - Placa: ', v.placa, ' - Combustível: ', COALESCE(v.combustivel, 'Não informado')) AS info_completa
FROM veiculos v
LEFT JOIN
    marcas on(marcas.id = v.marca_id);

我认为错误原因是因为我正在使用coalesce和/或concat,因为完整错误的描述告诉我：非法混合排序(latin1_swedish_ci,IMPLICIT)和(utf8_general_ci,COERCIBLE)操作’coalesce’

解决方法:

您也可以使用CAST()将字符串转换为不同的字符集.语法是：

CAST(character_string AS character_data_type CHARACTER SET charset_name)

例如：

      SELECT CAST(_latin1'test' AS CHAR CHARACTER SET utf8);

替代方法：使用CONVERT(expr USING transcoding_name)

标签：mysql-error-1064,sql,mysql,collation
来源： https://codeday.me/bug/20190901/1783047.html