php bind_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

<br />$db = new mysqli("localhost", "xx","xxxxxxx","books");<br />    $insert_str = "insert into customers(name,address,city) values(?,?,?)";<br />    $stmt = $db->prepare($insert_str);<br />    $stmt->bind_param("sss","john","street one","beijing");<br />    $stmt->execute();<br />    echo $stmt->affected_rows;<br />

然后运行，报错
Cannot pass parameter 2 by reference
然后根据网上的代码改成

<br />$db = new mysqli("localhost", "xx","xxxxxx","books");<br />    $insert_str = "insert into customers(name,address,city) values(?,?,?)";<br />    $stmt = $db->prepare($insert_str);<br />    $name ="john";<br />    $address = "street one";<br />    $city = "beijing";<br />    $stmt->bind_param("sss",$name,$address,$city);<br />    $stmt->execute();<br />    echo $stmt->affected_rows;<br />

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。