我在 php 編碼方面遇到了一些問題。我想做的是：創建一個數組 ($rows) 并用 mysqli_query ($query1) 的結果填充它 --> OK對于該數組中的每個元素，將某個鍵 (pilot_rule_id) 的值替換為另一個 mysqli_query ($query2) 的結果。（第二個查詢將返回一行，因為 Pilot 表的 id 是主鍵）。到目前為止我已經$id = "96707ac6-ecae-11ea-878d-005056bbb446";$rows = array();$query1 = mysqli_query($con, "SELECT * FROM pilot_time_schedule WHERE pilot_id='$id'");while($r = mysqli_fetch_assoc($query1)) { $rows[] = $r;}foreach($rows as $pilotRuleId) { $pilotRuleId->$pilot_rule_id; $query2 = mysqli_query($con, "SELECT name FROM pilot_rule WHERE id='$piloteRuleId'"); while($r = mysqli_fetch_assoc($query2)) { $result[] = $r; }// Don't know how to continue from here
1 回答
aluckdog
TA貢獻1847條經驗 獲得超7個贊
你可以這樣：
$id = "96707ac6-ecae-11ea-878d-005056bbb446";
$stmt = $con->prepare('SELECT * FROM pilot_time_schedule WHERE pilot_id=?');
$stmt->bind_param('s', $id);
$stmt->execute();
$rows = $stmt->get_result()->fetch_all(MYSQLI_ASSOC);
foreach ($rows as $row) {
$stmt = $con->prepare('SELECT name FROM pilot_rule WHERE id=?');
$stmt->bind_param('s', $row['pilot_rule_id']);
$stmt->execute();
// replace with the `name` returned from the above statement.
$row['pilot_rule_id'] = $stmt->get_result()->fetch_row()[0] ?? null;
}
然而，您確實應該了解 SQL 連接。使用 SQL 連接，您可以避免對數據庫進行 N+1 查詢。
$id = "96707ac6-ecae-11ea-878d-005056bbb446";
$stmt = $con->prepare('SELECT pilot_time_schedule.*, pilot_rule.name
FROM pilot_time_schedule
JOIN pilot_rule ON pilot_rule.id=pilot_time_schedule.pilot_rule_id
WHERE pilot_id=?');
$stmt->bind_param('s', $id);
$stmt->execute();
$rows = $stmt->get_result()->fetch_all(MYSQLI_ASSOC);
foreach ($rows as $row) {
echo $row['name']; // contains the name from pilot_rule
}
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