简体   繁体   中英

Combine two rows into one string

i have dynamic database, and i want to search in this table

FieldID | Content | TypeID
--------------------------
ABC-123 | jon     | 1
EFG-456 | doe     | 1
HIJ-789 | man     | 1

So my SELECT query looks something like this:

SELECT 
    GROUP_CONCAT(fieldsContent.Content SEPARATOR '|*|') AS Content, 
    GROUP_CONCAT(fieldsContent.FieldID SEPARATOR '|*|') AS FieldID, 
FROM (`customers`) 
LEFT OUTER JOIN `fieldsContent` ON `fieldsContent`.`TypeID` = `customers`.`ID` 
GROUP BY `fieldsContent`.`TypeID` 
ORDER BY `customers`.`Created` DESC

And the result looks like this

Array
(
    [0] => stdClass Object
        (
            [Content] => jon|*|doe|*|man
            [FieldID] => ABC-123|*|EFG-456|*|HIJ-789
        )
)

And when i'm adding HAVING and searching only for jon , it will return me the result

HAVING Content LIKE "%jon%" 

But, when i'm trying to search jon doe it will return empty result

HAVING Content LIKE "%jon doe%"

Because jon doe don't exists in the string, only jon|*|doe

So how can i combine these two rows to one string without the SEPARATOR for me the search in them jon doe ?

BUT!! keep in mind that i need to obtain the SEPARATOR because i need to combine the data to be used in php.

ex:

$field = explode('|*|',$data->FieldID);
$content = explode('|*|',$data->Content);

foreach($field as $k => $FieldID){
    switch($FieldID){
        case 'ABC-123':
            $res['first_name'] = $content[$k];
        break;
        case 'EFG-456':
            $res['last_name'] = $content[$k];
        break;
        case 'HIJ-789':
            $res['gender'] = $content[$k];
        break;      
    }
}

Any ideas will be appreciated :)

TRY THIS #1

HAVING Content LIKE "%jon%doe%"

TRY THIS #2

HAVING Content LIKE "%jon" AND Content LIKE "%doe%"

TRY THIS #3

HAVING REPLACE(Content,'|*|',' ') LIKE "%jon doe%"

GIVE IT A TRY !!! :-)

Use:-

HAVING Content LIKE "%jon%" AND Content LIKE "%doe%"

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM