Question

我有这个PHP代码 - for循环和每一步，每次增量搜索mysql表中的数据aktivnosti

PHP：

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);

正如您从上面的代码中看到的那样，我尝试获取每$i增量的数据并搜索id_activity on table aktivnosti = $i

我只得到一个结果，所以我得到第一个'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>

我怎么能这样做？一些想法？

对不起我的意见。感谢

Answer 1

正如上面的评论中指出的那样，你在这里采取了不好的做法。您应该能够在单个查询中获取所有这些数据。如果您希望每天都有与n个记录数相关的固定数量30天的概念，您可能还需要查看您的架构。我建议两张桌子

day_list

day_id  day_name (or any other day-related data fields)
1       ...
2       ...
...     ...
30      ...

days_records

record_id   day_id   other_data
1           1        ...
2           1        ...
3           3        ...
4           5        ...
...

然后您可以像下面这样查询：

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的含义。

然后您创建一个类似的查询：

$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
   }
}

Answer 2

如果您同时需要ID和activity：

$sql = <<<EOD
SELECT
    id_activity AS ID,
    CONCAT('<button>', naziv, '</button>') AS activity
FROM aktivnosti
WHERE id_activity BETWEEN 1 AND 30
ORDER BY id_activity
EOD;
$data = $db->query($sql)->fetchAll(PDO::FETCH_ASSOC);
$jsonTable = json_encode(compact('data'));

如果您只使用activity：

$sql = <<<EOD
SELECT CONCAT('<button>', naziv, '</button>')
FROM aktivnosti
WHERE id_activity BETWEEN 1 AND 30
ORDER BY id_activity
EOD;
$data = $db->query($sql)->fetchAll(PDO::FETCH_COLUMN, 0);
$jsonTable = json_encode(compact('data'));

Answer 3

试试这个......

 while($naz=$rs1->fetch(PDO::FETCH_OBJ))

{

echo $naz->column1;
echo $naz->column2;

}

而不是

 $naz = $rs1->fetchColumn();

使用php pdo从mysql数据库里面获取数据for循环

3 个答案: