【发布时间】:2020-05-29 06:34:35
【问题描述】:
$type = array('i','i');
$param = array("1","1");
$stmt = $mysqli->prepare("SELECT par1, par2 FROM table WHERE par3 = ? AND par4 = ?");
$refs = array();
foreach($param as $key => $value) {
$refs[$key] = &$param[$key];
}
$result_params = array_merge($type,$refs);
call_user_func_array(array($stmt, 'bind_param'), $result_params);
$stmt->execute();
var_dump($result_params):
array(4) {
[0]=>
string(1) "i"
[1]=>
string(1) "i"
[2]=>
&string(1) "1"
[3]=>
&string(1) "1"
}
当我们使用代码时,我们会得到错误:
mysqli_stmt::bind_param() 的参数 2 应为参考, 在...中给出的值
为什么我们会收到此错误以及如何解决此问题?
【问题讨论】:
标签: php mysqli prepared-statement