我必须使用运算符从数据库中选择IN行。我想使用准备好的语句来做到这一点。这是我的代码：
<?php
$lastnames = array('braun', 'piorkowski', 'mason', 'nash');
$in_statement = '"' . implode('", "', $lastnames) . '"'; //"braun", "piorkowski", "mason", "nash"
$data_res = $_DB->prepare('SELECT `id`, `name`, `age` FROM `users` WHERE `lastname` IN (?)');
$data_res->bind_param('s', $in_statement);
$data_res->execute();
$result = $data_res->get_result();
while ($data = $result->fetch_array(MYSQLI_ASSOC)) {
...
}
?>
但是，尽管所有数据都存在于数据库中，但什么也没返回.
还有一个：如果我直接传递$in_statement来查询并执行它，数据将被返回。所以问题出现在准备.
我在谷歌中寻找这个问题，但它没有成功。我的代码出了什么问题？
感谢您的帮助！
我最近找到了我的问题的解决方案。也许这不是最好的方法，但它效果很好！证明我错了：）
<?php
$lastnames = array('braun', 'piorkowski', 'mason', 'nash');
$arParams = array();
foreach($lastnames as $key => $value) //recreate an array with parameters explicitly passing every parameter by reference
$arParams[] = &$lastnames[$key];
$count_params = count($arParams);
$int = str_repeat('i',$count_params); //add type for each variable (i,d,s,b); you can also determine type of the variable automatically (is_int, is_float, is_string) in loop, but i don't need it
array_unshift($arParams,$int);
$q = array_fill(0,$count_params,'?'); //form string of question marks for statement
$params = implode(',',$q);
$data_res = $_DB->prepare('SELECT `id`, `name`, `age` FROM `users` WHERE `lastname` IN ('.$params.')');
call_user_func_array(array($data_res, 'bind_param'), $arParams);
$data_res->execute();
$result = $data_res->get_result();
while ($data = $result->fetch_array(MYSQLI_ASSOC)) {
...
}
$result->free();
$data_res->close();
?>