我有一个简单的3列表:
testd是一个INT(11)自增主索引
name是一个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();
连接成功。
UPDATE有效。
但是,INSERT不这样做。它确实创建了一个具有正确testd字段的新记录,但是name2和position2字段都被设置为0。
变化
$sqlinsert = "INSERT INTO test (name, position) VALUES (name = ?, position = ?)";
$sqlinsert = "INSERT INTO test (name, position) VALUES (?, ?)";
修改这一行
$sqlinsert = "INSERT INTO test (name, position) VALUES (?, ?)";