[英]Replace value in array with result of mysqli_query
我在使用 php 編碼時遇到了一些麻煩。 我想做的是以下內容:
到目前為止我有
$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
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;
}
但是,您確實應該學習 SQL 連接。 使用 SQL 連接可以避免對數據庫的 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
}
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.