php bind_param有關問題，新手求教

php bind_param問題，新手求教
剛剛學到prepared這兒，敲了如下代碼

$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;

然後運行，報錯
Cannot pass parameter 2 by reference
然後根據網上的代碼改成

$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;

顯示插入資料成功，就想問問第一個版本到底錯在哪
------解決方案--------------------
bind_param 的第二個參數起傳遞的是引用
你直接寫成字串，這是在 php 5.3 及以後是不允許的

其實你可以不要
$stmt->bind_param("sss","john","street one","beijing");
而直接寫成
$stmt->execute(array("john","street one","beijing"));
------解決方案--------------------
引用手冊中的話

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

你直接寫成字串是不能引用傳遞的。

