[英]store more than one value into a db field to be retrieved as separate values
[英]If more than one result, rank values from one field in db and return only the highest/lowest ranked value
我正在尝试找出执行此操作的最佳方法,但不确定如何操作。 有人可以帮我吗?
假设我有4个标记-金,银,铜和发展中标记之一就是来自数据库的$ gsb ['gsb']值。
下面的代码获取课程,并回显img和链接。 我苦苦挣扎的一点是如何对上面的4个值进行排名,因此,如果返回的结果超过一个,它将根据选择的排名获得最高或最低值,并且仅回显该结果/ img
希望这很清楚。 如有任何帮助/指导,我将不胜感激。
$metacourses = mysql_query("SELECT * FROM mdl_course_meta where parent_course = $courseid");
while($parentcourse = mysql_fetch_assoc($metacourses)){
//Select GSB for course
$parentcourseid = $parentcourse['child_course'];
$gsb = mysql_fetch_array(mysql_query("SELECT * FROM mdl_gsb_content where course = $parentcourseid"));
//Used for $img name
if ($gsb['gsb']=="") {$thegsbscore = "in_development";}
else {$thegsbscore = $gsb['gsb'];}
if ($viewgsb == 'Yes'){
$img = '<div align="center"><img src="'.$CFG->wwwroot.'/blocks/gsb/images/'.strtolower($thegsbscore).'.png" width="90" height="98"></div>';
$link = '<p align="center"><b><a href="'.$CFG->wwwroot.'/blocks/gsb/gsb_explained.htm" target="_blank">How can I improve my course medal?</a></b></p>';
}
else {
$img = '';
$message = '';
$viewgsb = '';
$link = '';
}
}
$this->content = new stdClass;
$this->content->text = $message . $img . $link;
$this->content->footer = '';
return $this->content;
return $this->content->text;
将权重分配给您的值,最好在表格中。
像这样使用:
TABLE_WEIGHT (It's a better idea to store these in a separate table, that way you can always add/delete more scores, irrespective of other tables)
ID(PRIMARY KEY) SCORE_NAME(VARCHAR) SCORE_WEIGHTAGE(INT/ENUM)
1 In Development 1
2 Bronze 2
3 Silver 3
4 Gold 4
现在,您始终可以在第二个查询中使用SQL JOIN对数据进行排序。 另外,如果有多个结果,则可以使用PHP的usort过滤结果。
希望这可以帮助。
您可以通过多种方式执行此操作,通过映射权重的文件进行排序,连接到具有权重的另一个表并获取权重。 pushpesh是您将要做的最好的解决方案
SELECT * FROM mdl_gsb_content
where course = $parentcourseid
join table_weight
on table_weight.score_name = mdl_gsb_content.gsb
order by table_weight.SCORE_WEIGHTAGE.
(假设gsb是他的得分)
未经测试,但应该可以解决。
测试后编辑方式:
SELECT * FROM mdl_gsb_content
JOIN mdl_gsb_scores on mdl_gsb_scores.score = mdl_gsb_content.gsb
WHERE mdl_gsb_content.courseid = $parentcourseid
ORDER by mdl_gsb_scores.rank ASC
LIMIT 1
在实现Pushpeshs权重表并解决编码SQL查询后,我终于想出了正确的方法。
$metacourses = mysql_query("SELECT * FROM mdl_course_meta where parent_course = $courseid");
//Select GSB for course
$types = array();
while(($row = mysql_fetch_assoc($metacourses))) {
$types[] = $row['child_course'];
}
$theids = implode(',',$types);
$gsb = mysql_query("
SELECT * FROM mdl_gsb_content
JOIN mdl_gsb_scores on mdl_gsb_scores.score = mdl_gsb_content.gsb
WHERE mdl_gsb_content.courseid IN ($theids)
ORDER by mdl_gsb_scores.rank ASC
LIMIT 1");
while($score = mysql_fetch_assoc($gsb))
//Work out img
$thegsbscore = $score['gsb'];
if ($viewgsb == 'Yes'){
$img = '<div align="center"><img src="'.$CFG->wwwroot.'/blocks/gsb/images/'.strtolower($thegsbscore).'.png" width="90" height="98"></div>';
$link = '<p align="center"><b><a href="'.$CFG->wwwroot.'/blocks/gsb/gsb_explained.htm" target="_blank">How can I improve my course medal?</a></b></p>';
$message = '<div align="center">Your Unit VLE is:'.print_r($types).'</div><br />';
}
else {
$img = '';
$message = '';
$viewgsb = '';
$link = '';
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.