1

我已经尝试了一个多小时才能将其正确插入到我的数据库中。即使查询与工作完全一样，我也无法终生解决这个问题。

页面上没有抛出错误，但它没有将数据插入数据库。

PHP 代码（变量正确发布到页面）：

<?php
error_reporting(E_ALL);
//start session
session_start();

//include database connection
include('../db_connect.php');

//import info
$client = $_POST['client'];
$name = $_POST['name'];
$company = $_POST['company'];
$email = $_POST['email'];
$phone = $_POST['phone'];
$address1 = $_POST['address1'];
$address2 = $_POST['address2'];
$address3 = $_POST['address3'];
$city = $_POST['city'];
$state = $_POST['state'];
$zip = $_POST['zip'];
$country = $_POST['country'];
$vehicle = $_POST['vehicle'];
$msrp = $_POST['msrp'];
$saleprice = $_POST['saleprice'];
$date = date("Y-m-d H:i:s");
$status = "Pending";
$notes = "Test notes are great!";


$sql = "INSERT INTO sales (salePrice, saleDate, saleStatus, saleNotes, saleName, saleCompany, salePhone, saleEmail, saleAddress1, saleAddress2, saleAddress3, saleCity, saleState, saleZipcode, saleCountry, clientFK) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)";

if ($stmt = $mysqli->prepare($sql)) {

/* Bind our params */
$stmt->bind_param('dssssssssssssssi', $saleprice , $date , $status , $notes , $name , $company , $phone , $email , $address1 , $address2 , $address3 , $city , $state , $zip , $country , $client);

/* Execute the prepared Statement */
$stmt->execute();

/* Echo results */
echo "Inserted user information into database.\n";

/* Close the statement */
$stmt->close();
}else{
    /* Error */
printf("Prepared Statement Error: %s\n", $mysqli->error);
}
?>

这是我的数据库中该表的布局：

CREATE TABLE IF NOT EXISTS `sales` (
  `idsale` int(11) NOT NULL AUTO_INCREMENT,
  `salePrice` double NOT NULL,
  `saleDate` datetime NOT NULL,
  `saleStatus` varchar(50) NOT NULL,
  `saleNotes` varchar(100) DEFAULT NULL,
  `saleName` varchar(70) DEFAULT NULL,
  `saleCompany` varchar(50) DEFAULT NULL,
  `salePhone` varchar(20) DEFAULT NULL,
  `saleEmail` varchar(255) DEFAULT NULL,
  `saleAddress1` varchar(35) DEFAULT NULL,
  `saleAddress2` varchar(35) DEFAULT NULL,
  `saleAddress3` varchar(35) DEFAULT NULL,
  `saleCity` varchar(20) DEFAULT NULL,
  `saleState` varchar(50) DEFAULT NULL,
  `saleZipcode` varchar(15) DEFAULT NULL,
  `saleCountry` varchar(50) DEFAULT NULL,
  `createdBy` varchar(30) DEFAULT NULL,
  `createdDate` datetime DEFAULT NULL,
  `modifiedBy` varchar(30) DEFAULT NULL,
  `modifiedDate` datetime DEFAULT NULL,
  `clientFK` int(11) NOT NULL,
  PRIMARY KEY (`idsale`),
  KEY `clientFK_INDEX` (`clientFK`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=2 ;

任何帮助将不胜感激！提前感谢您的帮助。

4

2 回答 2

1

您的查询未正确绑定变量，因为您在查询中传递了 16 个占位符并在绑定函数中传递了 17 个值。

(?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)  //16 values...

$stmt->bind_param('dssssssssssssssi', $saleprice , $date , $status , $notes , $name , $company , $phone , $email , $address1 , $address2 , $address3 , $city , $state , $zip , $country , $client); //17 variables

因此，您的查询会引发错误。在 php.ini 中启用错误以查看错误消息。

于 2013-04-22T03:36:27.013 回答
0

这现在看起来真的很愚蠢，但它指出：

//include database connection
include('../db_connect.php');

直到我将其更改为：

//include database connection
include('db_connect.php');

这是最奇怪的事情 ../ 如何使一切变得不同。谢谢大家的帮助。

于 2013-04-23T08:23:01.607 回答