php - 使用mysqli发布到数据库

Question

我正在学习 php 并尝试完成以下工作:

<?php
require_once("db_connect.php");

    // TODO - Check that connection was successful.

    $dname = $_POST["dname"];
    $daddress = $_POST["daddress"];


    $stmt = $mysqli->prepare("INSERT INTO test (dname, daddress) VALUES (?, ?)");

    // TODO check that $stmt creation succeeded

    // "s" means the database expects a string
    $stmt->bind_param("s", $dname, $daddress);

    $stmt->execute();

    $stmt->close();

    $mysqli->close();
?>

它仅适用于一个 bind_param 而不是 2。如果从代码中删除了 $daddress 则它会发布。该表单在数据库中有 26 个帖子，我目前正在处理 2 个以使其最少。

提交表单时出现以下错误。

Warning: mysqli_stmt::bind_param() [mysqli-stmt.bind-param]: Number of elements in type definition string doesn't match number of bind variables in /home/mymotorsportco/public_html/entry/actions/entry.php on line 15

Answer 1

根据 PHP manual :

types

A string that contains one or more characters which specify the types for the corresponding bind variables

i - corresponding variable has type integer

d - corresponding variable has type double

s - corresponding variable has type string

b - corresponding variable is a blob and will be sent in packets

您必须为要绑定(bind)的所有参数添加类型。所以如果第二个参数是一个字符串，你必须做

$stmt->bind_param("ss", $dname, $daddress);