小编Que*_*ter的帖子

SELECT *,DAYNAME(created_on) AS created_day FROM users_feedback WHERE created_day = 'wednesday'

当我执行上面的查询时,它会生成如下错误

Error Code : 1054
Unknown column 'created_day' in 'where clause'

Execution Time : 00:00:00:000
Transfer Time  : 00:00:00:000
Total Time     : 00:00:00:000

如何使用mysql在SubQuery结果中设置默认值VALUE

SELECT      
    p.`id`, p.`name`, p.`class_name`, cpd.`status_team`,
    cpd.`home`, cpd.`guest`, cpd.`mvp`, cpd.`oscar`,
    cpd.`wam`, cpd.`status`, cpd.`added_date`,
    (SELECT result FROM result_cards WHERE `id` = cpd.`result`) AS DEFAULT(`result`)
FROM `cron_players_data` cpd
INNER JOIN `players` p ON cpd.`player_id` = p.id
WHERE cpd.`added_date` = '2012-03-29' AND cpd.team_id = '15'

当我删除这个DEFAULT()时,Query将正常执行.其实我想默认结果值为0或者帮助肯定是值得赞赏的

嘿guyz我有一点codeigniter问题,我不知道如何解决这个问题.如果你对此有任何解决方案,那么回答我

SET @weekVideoCount := (SELECT COUNT(*) FROM videos v 
);

SELECT @weekVideoCount;

当我在Sqlyog上执行此查询时,结果将成功显示,但如果我通过这样的模型调用此查询

function getWeeklyUserData(){
        $query= $this->db->query("SET @weekVideoCount := (SELECT COUNT(*) FROM videos v);
                                SELECT @weekVideoCount;
                                ");
        return $query->result();
    }

错误已生成

这是我的MySQL查询,我在这个查询中有一周的问题.我不知道如何应用带条件的IF条件WHERE.

查询:

SELECT
*,
IFNULL((SELECT ur.user_rating FROM user_rating ur 
WHERE ur.vid_id = v.id AND ur.user_id = '1000'),'NULL') AS user_rating
FROM videos v
WHERE WEEK(v.video_released_date) = WEEK(NOW())
AND  
v.`is_done` = 1
ORDER BY v.admin_listing ASC; 

我想要OR(我如何在where子句中应用这个条件？)

IF( WEEK(v.video_released_date) = WEEK(NOW()) , WEEK(NOW()) , WEEK(NOW())-1)
=
IF( WEEK(v.video_released_date) = WEEK(NOW()) , WEEK(NOW()) , WEEK(NOW())-1)

简报

如果视频发布日期已过,但与当前周不匹配则适用上周

我

当我像这样尝试自己时,他们会将整个数据归还给我

SELECT
*,
IFNULL((SELECT ur.user_rating FROM user_rating ur 
WHERE ur.vid_id = v.id AND ur.user_id = '1000'),'NULL') AS user_rating
FROM videos v …

我有一个表,这个表包含多个列.我想知道如何更新除特定列中的条目之外的所有条目.我不想使用多个更新查询.

如果您对此有任何疑问,请与我分享.非常感谢您的帮助.

表结构

查询(我自己尝试过,但我对此也感到困惑)

UPDATE `table` SET STATUS = 0;
UPDATE `table` SET STATUS = 1 WHERE id = 4;

我的问题是如何在一个查询中组合这些查询

我只是想知道如何检查这个日期

2012-03-07与where子句一起到这个日期

这是我的Sql查询

SELECT u.`last_activity` FROM users u WHERE = '2012-03-07'

响应

空回应

撇号(')如何与WHEREMySQL中的子句一起使用？

我已经习惯mysql_real_escape_string了清理输入.

这是mySQL查询和数据库的截图

SELECT * FROM players WHERE `name` = 'Amar'e Stoudemire'

执行此查询会生成以下错误:

知道为什么吗？

我怎么做这样的if/else语句？用户可以选择是否添加图像.如果添加图像,我想要运行if,如果他们不希望别人运行,因为根据是否添加图像,他们将有两个不同的结束.

if (isset($_POST['formsubmitted'])) { 

if (isset($_POST['name'])){  }else{  }

if (isset($_POST['state'])){  }else{  }

if (isset($_FILES['images']['name'])) { echo 'images'; exit;} else {echo 'no images'; exit;}


} #end main form submitted

if(isset($ _ FILES ['images'] ['name']))dosnt工作,因为截至目前,即使没有提交图像,它仍然会提交图像.

html文件字段是:

<form>
<input type='file' name='images[]' id=''>
<input type='file' name='images[]' id=''>
<input type='file' name='images[]' id=''>
<input type = "submit">    
</form>

当我像这样执行这个查询时,他们需要花费很多执行时间,因为user_fans表包含10000个用户条目.我该如何优化它？

询问

SELECT uf.`user_name`,uf.`user_id`,
@post                := (SELECT COUNT(*) FROM post WHERE user_id = uf.`user_id`) AS post,
@post_comment_likes  := (SELECT COUNT(*) FROM post_comment_likes WHERE user_id = uf.`user_id`) AS post_comment_likes,
@post_comments       := (SELECT COUNT(*) FROM post_comments WHERE user_id = uf.`user_id`) AS post_comments,
@post_likes          := (SELECT COUNT(*) FROM post_likes WHERE user_id = uf.`user_id`) AS post_likes,

(@post+@post_comments) AS `sum_post`,
(@post_likes+@post_comment_likes) AS `sum_like`, 
((@post+@post_comments)*10) AS `post_cal`,      
((@post_likes+@post_comment_likes)*5) AS `like_cal`,
((@post*10)+(@post_comments*10)+(@post_likes*5)+(@post_comment_likes*5)) AS `total`  
FROM  `user_fans` uf  ORDER BY `total` DESC lIMIT 20

可能重复:
从PHP数组生成JavaScript数组

我只是想知道如何在PHP的帮助下创建JavaScript数组,并且非常感谢帮助

我有一个PHP结果到这个表单

Array
(
    [0] => stdClass Object
        (
            [id] => 11
            [name] => MLB
            [totalcount] => 4
        )

    [1] => stdClass Object
        (
            [id] => 35
            [name] => American Idol
            [totalcount] => 2
        )

) 

我想像这样将这个数组转换为JavaScript数组

<script type="text/javascript">
    var myData = new Array(['MLB', 4], ['American Idol', 2]);
</script>