Question

我有两个表：companies和countries，我想通过查看list_of_countries表中的companies字段来检索国家/地区。

SELECT id, name
FROM countries
WHERE name LIKE CONCAT('%', (SELECT list_of_countries FROM companies WHERE id = 1), '%')

这会返回零结果，并应返回两个值according to this SQLFiddle。

Answer 1

如果您可以重新格式化list_of_countries并删除多余的空格。

您可以将此类查询与FIND_IN_SET函数

一起使用

http://sqlfiddle.com/#!9/ad5ffd/2

SELECT c.id, c.name
FROM companies  comp
LEFT JOIN countries c
ON FIND_IN_SET(c.name, comp.list_of_countries)
WHERE comp.id = 1

Answer 2

使用INSTR函数

解决了。

$structure = imap_fetchstructure($connection, $msgno);
$encoding = $structure->encoding;
$body = imap_fetchbody ($connection, $msgno, 1);
$body=decodebody($encoding, $body);
echo $body;

function decodebody($encoding, $body)
{
switch ($encoding) 
{
    # 7BIT
    case ENC7BIT:
        echo "7BIT<br>";//
        return $body;
        break;
    # 8BIT
    case ENC8BIT:
        echo "8BIT<br>";//
        return quoted_printable_decode(imap_8bit($body));
        break;
    # BINARY
    case ENCBINARY:
        echo "BINARY<br>";//
        return imap_binary($body);
        break;
    # BASE64
    case ENCBASE64:
        echo "BASE64<br>";//
        return imap_base64($body);
        break;
    # QUOTED-PRINTABLE
    case ENCQUOTEDPRINTABLE:
        echo "QUOTED<br>";//
        return quoted_printable_decode($body);
        break;
    # OTHER
    case ENCOTHER:
        echo "OTHER<br>";//
        return $body;
        break;
    # UNKNOWN
    default:
        echo "UNKNOWN<br>";//
        return $body;
        break;
}
}

Answer 3

在这种情况下，您应该使用find_in_set。这是我的例子：

select *
from companies t1
left join countries t2 on find_in_set(t2.name, t1.list_of_countries) > 0

这里fiddle。

好像CONCAT没有返回值

3 个答案: