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我在使用 php 编码时遇到了一些问题。我想要做的是：

  1. 创建一个数组 ($rows) 并用 mysqli_query ($query1) 的结果填充它 --> OK
  2. 对于该数组中的每个元素，将某个键的值 (pilot_rule_id) 替换为另一个 mysqli_query ($query2) 的结果。（第二个查询将返回一行，因为 Pilot 表的 id 是主键）。

到目前为止我有

$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
4

1 回答 1

1

你可以这样：

$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，您确实应该学习 SQL 连接。使用 SQL 连接可以避免对数据库进行 N+1 次查询。

$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
}
于 2020-09-10T23:24:58.790 回答