我正在尝试使用 php 将表单数据保存到 sql 表中。尽管我在提交时没有收到错误，但数据没有显示在表中。
我的提交按钮名称是input_submit
这是我的代码:
if(isset($_POST['input_submit'])){
include 'dbConnection.php';
include 'saveData.php';
}
dbConnection.php
<?php
$path = $_SERVER['DOCUMENT_ROOT'];
include_once $path . '/wp-load.php';
include_once $path . '/wp-config.php';
class ConnectDB{
private $servername;
private $username;
private $password;
private $dbname;
protected function connect(){
$this->servername ="localhost";
$this->username ="root";
$this->password ="";
$this->dbname ="testdb";
$conn = new mysqli($this->servername,$this->username,$this->password,$this->dbname);
if($conn -> connect_error) {
die("connection failed:".$conn-> connect_error);
}
return $conn;
}
}
?>
保存数据.php:
<?php
class saveinput extends ConnectDB {
public function Savein(){
$date = $_POST['date'];
$entry_type = $_POST['entry_type'];
$amount = $_POST['amount'];
$sql = $conn->prepare("INSERT INTO wp_myexpenses (date, entry_type, amount)
VALUES(?, ?, ?)");
$sql->bind_param("sss",$date, $entry_type, $amount);
$sql->execute();
if ($sql->execute()) {
echo "success";
} else {
echo "failed";
}
}
}
?>
提交时，表单正在提交。但是当我检查数据库表时，什么也没有显示。我不明白这里出了什么问题。有人可以指导我吗？
最佳答案
您应该在 var 中的 Savein 方法中调用“connect”方法。因此，您的 Savein 方法应该是:
public function Savein(){
$conn = parent::connect(); // This is the only thing i've added
$date = $_POST['date'];
$entry_type = $_POST['entry_type'];
$amount = $_POST['amount'];
$sql = $conn->prepare("INSERT INTO wp_myexpenses (date, entry_type, amount)
VALUES(?, ?, ?)");
$sql->bind_param("sss",$date, $entry_type, $amount);
if ($sql->execute()) {
echo "success";
} else {
echo "failed";
}
}
关于php - 无法使用php将表单数据保存到sql表，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/53386457/