Question

我正在使用php进行一个项目，并试图将一些数据插入到MySQL数据库中。当我在线运行代码并尝试将数据插入数据库时，我遇到致命的错误。从错误消息我认为可能与数据库使用时间戳的事实有关。我试图使用NOW（）来插入当前的日期和时间，但它没有工作。如果有人能指出我正确的方向，我们将不胜感激。

The image is of the structure of the table (frs_Payment). I am trying to insert the data into this table

这是错误消息

Fatal error: Uncaught exception 'Exception' with message 'Database Error [1064] You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near '12:41:28, NIN012, 12, 1, 1)' at line 1' in E:\xampp\htdocs\CIT2318\Relational database and web integration\models\DAO.php:23 Stack trace: #0 E:\xampp\htdocs\CIT2318\Relational database and web integration\models\LoanModel.php(15): DAO->query('INSERT INTO frs...') #1 E:\xampp\htdocs\CIT2318\Relational database and web integration\controllers\LoanController.php(24): LoanModel->insertRental('104', '10.50', '2015-05-22 12:4...', 'NIN012', '12', '1', '1') #2 {main} thrown in E:\xampp\htdocs\CIT2318\Relational database and web integration\models\DAO.php on line 23

更新，新错误消息

INSERT INTO frs_Payment (payid, amount, paydatetime, empnin, custid, pstatusid, ptid) VALUES (104, 10.50, '2015-05-22 12:41:28', 'NIN012', 1, 1, 1)
Fatal error: Call to a member function fetch_assoc() on boolean in E:\xampp\htdocs\CIT2318\Relational database and web integration\controllers\LoanController.php on line 25

我现在能够将数据插入数据库，但现在出现了这条新的错误消息。关于如何解决这个问题的任何想法。

这是贷款控制方

<?php

session_start();
require_once("../models/LoanModel.php");
require_once("../views/LoanView.php");

$error = "";

if (isset($_POST["Submit"])) {
$payid=($_POST["payid"]);
$amount=($_POST["amount"]);
$paydatetime=($_POST["paydatetime"]);
$empnin=($_POST["empnin"]);
$custid=($_POST["custid"]);
$pstatusid=($_POST["pstatusid"]);
$ptid=($_POST["ptid"]);

if (empty($payid)) {
    //header('Location: ../views/LoanView.php?error=1');
    $error = "Payment ID is required";
} else {

$lgi = new LoanModel;
$result = $lgi->insertRental($payid, $amount, $paydatetime, $empnin,     $custid, $pstatusid, $ptid);
$row = $result->fetch_assoc();
$error = "";
$_SESSION['payid'] = $row["payid"];
$_SESSION['amount'] = $row["amount"];
$_SESSION['paydate'] = $row["paydatetime"];
$_SESSION['employeeid'] = $row["empnin"];
$_SESSION['customerid'] = $row["custid"];
$_SESSION['pstatus'] = $row["pstatusid"];
$_SESSION['ptype'] = $row["ptid"];

 header('Location: ../views/MenuView.php');
}
 echo "<hr>" . $error;
}

?>

这是贷款模式 已更新

<?php
 require_once('DAO.php');

 class LoanModel extends DAO{

protected $target = "frs_Payment";

public function __construct(){
  parent::__construct();
}

public function insertRental($payid, $amount, $paydatetime, $employeeid, $customerid, $pstatusid, $ptid){
  $sql = "INSERT INTO frs_Payment (payid, amount, paydatetime, empnin, custid, pstatusid, ptid) VALUES ($payid, $amount, '$paydatetime', '$employeeid', $customerid, $pstatusid, $ptid)";
echo $sql;
  return parent::query($sql);
}
}
?>

这是贷款视图

<html>
<head>
</head>
<body>
<h1>Add a new rental</h1>
<form action="../controllers/LoanController.php" method="POST">

  <label for="payid">Payment ID</label>
  <input type="text" id="payid" name="payid">
    <p> </p>
    <label for="amount">Amount</label>
    <input type="text" id="amount" name="amount">
    <p> </p>
    <label for="paydatetime">Payment date/time</label>
    <input type="text" id="paydatetime" name="paydatetime">
    <p> </p>
    <label for="empnin">Empnin</label>
    <input type="text" id="empnin" name="empnin">
    <p> </p>
    <label for="custid">Customer ID</label>
    <input type="text" id="custid" name="custid">
    <p> </p>
    <label for="pstatusid">Payment Status</label>
    <input type="text" id="pstatusid" name="pstatusid">
    <p> </p>
    <label for="ptid">Payment Type</label>
    <input type="text" id="ptid" name="ptid">
    <p> </p>
  <input type="submit" name="Submit" value="Add Rental"> 
    <br></br>
    <a href="../views/MenuView.php">Menu Page</a>
    <br></br>
</form>

Answer 1

你需要用撇号包装日期变量。

$sql = "INSERT INTO frs_Payment (payid, amount, paydatetime, empnin, custid, pstatusid, ptid) VALUES ($payid, $amount, '$paydatetime', $employeeid, $customerid, $pstatusid, $ptid)";

确保格式正确：尝试$ paydatetime = date（＆＃34; Y-m -d H：i：s＆＃34;）;

试图将一些数据插入MySQL数据库

1 个答案: