phpbind_param有关问题，新手求教

php bind_param问题，新手求教
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。