1

嗨，我有以下代码；

if( ! empty( $post['search-bar'] ) ) {
    $search_data = preg_replace("#\s\s#is", '', preg_replace("#[^\w\d\s+]#is", '', $post['search-bar'] ) );
    $data_array = explode( " ", $search_data );
    $data_array = "'%" . implode( "%' OR '%", $data_array ) . "%'"; 
    $query =  "
SELECT CONCAT( PROFILE_PROFFESION, FIRST_NAME, LAST_NAME, DISPLAY_NAME) AS 'STRING'
FROM `" . ACCOUNT_TABLE . "` 
WHERE STRING LIKE ( " . $data_array . " ) 
  AND BUSINESS_POST_CODE LIKE '" . substr(P_BUSINESS_POST_CODE, 0, 4) . "%'";

    $q = mysql_query( $query, $CON ) or die( "_error_" . mysql_error() );   
    if( mysql_num_rows( $q ) != 0 ) {
        die();
    }
}

问题是我想在 where 子句中使用 temp col 'STRING' 但返回 'unknown coloumnSTRING任何人都可以指出我正确的方向，关于 Phil

我已经打印了查询；

SELECT PROFILE_PROFFESION, FIRST_NAME, LAST_NAME, DISPLAY_NAME 
FROM nnn_accounts 
WHERE 
  CONCAT( PROFILE_PROFFESION, FIRST_NAME, LAST_NAME, DISPLAY_NAME) 
  LIKE ( '%web%' OR '%design%' ) 
AND BUSINESS_POST_CODE LIKE 'NG19%'
4

4 回答 4

1

将 STRING 包裹在 ` 而不是 '

应该： ..AS `STRING`

然后你提到的任何地方 STRING 应该是`STRING`

编辑：这应该解决它：

SELECT CONCAT( PROFILE_PROFFESION, FIRST_NAME, LAST_NAME, DISPLAY_NAME) AS `STRING`
FROM `" . ACCOUNT_TABLE . "` 
WHERE BUSINESS_POST_CODE LIKE '" . substr(P_BUSINESS_POST_CODE, 0, 4) . "%'"
HAVING `STRING` LIKE ( " . $data_array . " ) ;
于 2010-04-20T05:42:55.960 回答
1

是的，看起来你只是不能在 where 子句中使用 temp col，只需按原样指定它：

WHERE CONCAT( PROFILE_PROFFESION, FIRST_NAME, LAST_NAME, DISPLAY_NAME) LIKE '%$data_array%')
于 2010-04-20T05:59:26.133 回答
0

去掉周围的引号'STRING'

于 2010-04-20T05:45:59.753 回答
0

更合适的语法是

SELECT CONCAT( PROFILE_PROFFESION, FIRST_NAME, LAST_NAME, DISPLAY_NAME) AS `STRING` FROM `tablename` WHERE BUSINESS_POST_CODE LIKE 'NG19%' HAVING `STRING` LIKE '%web%' OR `STRING` LIKE '%design%'
于 2010-04-20T06:51:32.333 回答