Question

我有一个包含3列的简单表格：

testID是INT（11）自动递增主索引

name是VARCHAR文本字段

position是INT（11）字段

<?php
error_reporting(E_ALL);
require("mysqli.test.inc"); 

$name1="Rhonda Hotop"; $position1=456;
$name2="Perry Shafran"; $position2 = 789;
$sqlupdate = "UPDATE test SET name = ?, position =?"; 
$sqlinsert = "INSERT INTO test (name, position) VALUES (name = ?, position = ?)";
//create the connection to the database
$con = new mysqli($hostname,$username,$password,$database);
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
    }
//test an UPDATE
$statement = $con->prepare($sqlupdate);
$statement->bind_param("si", $name1, $position1);
$statement->execute();
$statement->close();
//test an INSERT
$statement = $con->prepare($sqlinsert);
$statement->bind_param('si', $name2, $position2);
$statement->execute();
$statement->close();
exit();

连接有效。

UPDATE有效。

然而，INSERT没有。它确实使用正确的testID字段创建了一条新记录，但name2和position2字段都设置为0。

非常感谢一些帮助。

Answer 1

更改

$sqlinsert = "INSERT INTO test (name, position) VALUES (name = ?, position = ?)";

到

$sqlinsert = "INSERT INTO test (name, position) VALUES (?, ?)";

Answer 2

修改此行

$sqlinsert = "INSERT INTO test (name, position) VALUES (?, ?)";

mysqli更新有效，但插入没有

2 个答案: