MySQLi使用IN运算符编写语句

Question

我必须使用INoperator 从数据库中选择一些行.我想用准备好的声明来做.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,尽管数据库中存在所有数据,但不返回任

还有一个:如果我$in_statement直接传递查询并执行它,将返回数据.所以问题出现在准备上.

我在谷歌寻找这个问题,但并没有"成功".我的代码出了什么问题？
谢谢您的帮助!

Answer 1

我最近为我的问题找到了解决方案.也许这不是最好的方法,但它很好用!证明我是错的:)

<?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();
?>