刚刚学到prepared这儿，敲了如下代码
$db = new mysqli(localhost, xx,xxxxxxx,books);
$insert_str = insert into customers(name,address,city) values(?,?,?);
$stmt = $db->prepare($insert_str);
$stmt->bind_param(sss,john,street one,beijing);
$stmt->execute();
echo $stmt->affected_rows;
然后运行，报错
cannot pass parameter 2 by reference
然后根据网上的代码改成
$db = new mysqli(localhost, xx,xxxxxx,books);
$insert_str = insert into customers(name,address,city) values(?,?,?);
$stmt = $db->prepare($insert_str);
$name =john;
$address = street one;
$city = beijing;
$stmt->bind_param(sss,$name,$address,$city);
$stmt->execute();
echo $stmt->affected_rows;
显示插入数据成功，就想问问第一个版本到底错在哪
------解决方案--------------------
bind_param 的第二个参数起传递的是引用
你直接写成字符串，这是在 php 5.3 及以后是不允许的
其实你可以不要
$stmt->bind_param(sss,john,street one,beijing);
而直接写成
$stmt->execute(array(john,street one,beijing));
------解决方案--------------------
引用手册中的话
引用mysqli_stmt::bind_param -- mysqli_stmt_bind_param — binds variables to a prepared statement as parameters
note that mysqli_stmt_bind_param() requires parameters to be passed by reference
你直接写成字符串是不能引用传递的。