我正在尝试使用Prepared Statements方法将html表单中的数据保存到数据库wordpress
表wp_testdb
中。但是我在第mysqli_stmt_execute();
行出现了错误
警告:mysqli_stmt_execute()恰好期望1个参数,其中0给出 这是我的代码:
if(isset($_POST['BtnSubmit'])){
include_once 'dbConnection.php';
if($conn -> connect_error) {
die("connection failed:".$conn-> connect_error);
}
$date = $_POST['date'];
$select_bank = $_POST['select_bank'];
$entry_type = $_POST['entry_type'];
$income_cat = $_POST['income_cat'];
$amount = $_POST['amount'];
$sql = "INSERT INTO wp_testdb (data_one, data_two, data_three, data_four, data_five) VALUES (?, ?, ?, ?, ?);";
$stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt,$sql)) {
echo "SQL Error !!";
} else {
mysqli_stmt_bind_param($stmt, "sssss", $date, $select_bank, $entry_type, $income_cat, $amount);
mysqli_stmt_execute();
}
}
dbConnection.php具有以下数据:
<?php
$dbServername = "localhost";
$dbUsername = "root";
$dbPassword = "";
$dbName = "wordpress";
$conn= mysqli_connect($dbServername, $dbUsername, $dbPassword, $dbName);
我是php和mysql的新手。有什么建议可以解决这个问题吗?
答案 0 :(得分:0)
我将建议使用面向对象的接口而不是程序。两者都可以工作-但是,所有程序功能都需要将连接或语句对象作为第一个参数。
根据documentation,您需要将语句对象(在您的情况下为$stmt
)作为参数传递。
mysqli_stmt_execute($stmt);
此函数返回一个布尔值,您可以检查查询是否成功。
if (!mysqli_stmt_prepare($stmt,$sql)) {
echo "SQL Error !!";
error_log(mysqli_error($conn));
} else {
mysqli_stmt_bind_param($stmt, "sssss", $date, $select_bank, $entry_type, $income_cat, $amount);
if (mysqli_stmt_execute($stmt)) {
// Query succeeded!
// Do something
} else {
// Query failed. Log it, and do something?
error_log(mysqli_stmt_error($stmt));
}
}