Question

我有一个表numbers，其中包含3列id，number和ts

我想随机选择其中ts为NULL的10条记录来获取它们，同时用当前时间戳更新它们的ts，以便下一次选择我得不到相同的结果，并知道何时每个number都已被选中。

$ts = time();

$query = 'UPDATE numbers SET ts = ' . $ts . ' WHERE number IN (SELECT number FROM (SELECT number FROM numbers WHERE ts IS NULL ORDER BY RAND() LIMIT 10) AS t)';
if (!$stmt = $mysqli->prepare($query)) {
    echo $mysqli->error;
}
$stmt->execute();
$stmt->bind_result($number);
$stmt->store_result();

while ($stmt->fetch()) {
    echo $number . '<br>';
}

这里我不需要任何bind_param（）因为没有外部输入。

现在，此查询返回

警告：mysqli_stmt :: bind_result（）：绑定变量数与...中预准备语句中的字段数不匹配

Answer 1

只需使用此查询

$query = 'UPDATE numbers SET ts = " . $ts . " WHERE number IN (SELECT number FROM numbers WHERE
          ts IS NULL ORDER BY RAND() LIMIT 10)';

修饰： -
我只是删除额外的选择语句并更正引号。

Answer 2

我找到了，更新，然后选择更新的行，其中ts = $ ts

$ts = time();

$query = 'UPDATE numbers SET ts = ' . $ts . ' WHERE ts IS NULL ORDER BY RAND() LIMIT 10';
if (!$stmt = $mysqli->prepare($query)) {
    echo $mysqli->error;
}
$stmt->execute();
$stmt->close();

$query = 'SELECT number FROM numbers WHERE ts = ' . $ts . ' ORDER BY RAND()';
if (!$stmt = $mysqli->prepare($query)) {
    echo $mysqli->error;
}
$stmt->execute();
$stmt->bind_result($number);
$stmt->store_result();

while ($stmt->fetch()) {
    echo $number . '<br>';
}

mySQLi同时选择和更新

2 个答案: