我有这个php代码for循环,在每一步中,每一个增量都在mysql表中搜索数据aktivnosti
菲律宾比索:
for ($i=1; $i<=30; $i++;){
$temp = array();
$temp['ID'] = $i;
// ATTEMP TO GET DATA FROM aktivnosti WHERE id_activity = $i
$rs1 = $db->prepare('SELECT naziv FROM aktivnosti WHERE id_activity=:idd');
$rs1->bindParam(':idd', $i);
$rs1->execute();
$naz = $rs1->fetchColumn();
$temp['activity'] = '<button>'.$naz.'</button>';
$output['data'][] = $temp;
}
$jsonTable = json_encode($output);
从上面的代码中可以看到,我试图获取每个
$i
增量的数据,并搜索是否id_activity on table aktivnosti = $i
我只得到一个结果,所以我只得到第一个
'naziv'
,我需要从表aktivnosti获取所有naziv
数据,其中id_activity=$I并创建:<button>$naz[0]<button>
<button>$naz[1]<button>
<button>$naz[2]<button>
<button>$naz[how many times id_activity = $i]<button>
我该怎么做?有什么想法?
对不起我的英语。谢谢
最佳答案
正如上面的评论所指出的,您在这里采取了一种糟糕的做法。您应该能够在一个查询中获取所有这些数据。如果您想要有一个固定的30天的概念,并且每一天都与n
记录数相关,那么您可能还需要查看您的模式。我建议两张桌子
日清单
day_id day_name (or any other day-related data fields)
1 ...
2 ...
... ...
30 ...
天记录
record_id day_id other_data
1 1 ...
2 1 ...
3 3 ...
4 5 ...
...
然后您可以这样查询:
SELECT
d.day_id AS day_id
dr.record_id AS record_id
dr.other_date AS other_data
FROM day_list AS d
LEFT JOIN day_records AS dr
ON d.day_id = dr.day_id
很抱歉更改了表名,因为不知道您的数据库架构在现实世界中代表什么。
然后进行一个查询,如下所示:
$query = <<<EOT
SELECT
d.day_id AS day_id
dr.record_id AS record_id
dr.other_date AS other_data
FROM day_list AS d
LEFT JOIN day_records AS dr
ON d.day_id = dr.day_id
EOT;
$rs1 = $db->execute($query);
if (false === $rs1) {
// something went wrong. perhaps log an error
} else {
while($row = $rs1->fetch(PDO::FETCH_ASSOC)) {
$temp = $row;
// check to see if this date has a record
if (empty($temp['record_id'])) {
// this is a day with no associated record.
// do something
}
// not shown - continue to manipulate your $temp as desired
// then add to output array
$output['data'][] = $temp
}
}