Question

我可以将数据插入上一个受影响的行吗首先，我使用data.php在'198'行插入数据第二个使用image.php更新行'198'本身的url_path，但是添加为下一行。

PHP代码 -

    $file = $_FILES['image']['tmp_name'];

    $con=mysqli_connect("host","user","password","db");
    // Check connection
    if (mysqli_connect_errno()) {
    $response["success"] = 0;
    $response["message"] = "Database Error!";
    die(json_encode($response));
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

   try{
   if (!isset($_FILES['image']['tmp_name'])) {
   echo "Image not recognised";
    }else{
   $file=$_FILES['image']['tmp_name'];
   $image= addslashes(file_get_contents($_FILES['image']['tmp_name']));
   $image_name= addslashes($_FILES['image']['name']);
        //folder name 'images'
        move_uploaded_file($_FILES["image"]["tmp_name"],"images/" . $_FILES["image"]["name"]);

        $location="http://localhost/folder/images/" . $_FILES["image"]["name"];

  }
  $result = mysqli_query($con, "INSERT INTO products (pro_img_path) VALUES ('$location')"); 

  $response["success"] = 1;
  $response["message"] = "Image Added Successful!";
  die(json_encode($response));
  }
  catch(Exception $e){
  $response["success"] = 0;
  $response["message"] = "Image Not Added";
  die(json_encode($response)); 
  }

  mysqli_close($con); 
  ?>

任何帮助

已编辑的代码

    try{
    if (!isset($_FILES['image']['tmp_name'])) {
    echo "Image not recognised";
    }else{
    $file=$_FILES['image']['tmp_name'];
    $image= addslashes(file_get_contents($_FILES['image']['tmp_name']));
    $image_name= addslashes($_FILES['image']['name']);
        //folder name 'images'
        move_uploaded_file($_FILES["image"]["tmp_name"],"images/" . $_FILES["image"]["name"]);

        $location="http://localhost/folder/images/" . $_FILES["image"]["name"];

    }
    $result = mysqli_query($con, "INSERT INTO products (pro_img_path) VALUES ('$location')");
    $lastid=mysqli_insert_id($con);
    $result = mysqli_query($con, "UPDATE products SET pro_img_path='$location' WHERE time='$lastid' LIMIT 1");

    $response["success"] = 1;
    $response["message"] = "Image Added Successful!";
    die(json_encode($response));
    }

  **Table Structure**
   -- phpMyAdmin SQL Dump
   -- version 4.1.6
   -- http://www.phpmyadmin.net
   --
   -- Host: 127.0.0.1
   -- Generation Time: Sep 15, 2014 at 08:40 PM
   -- Server version: 5.6.16
   -- PHP Version: 5.5.9

   SET SQL_MODE = "NO_AUTO_VALUE_ON_ZERO";
   SET time_zone = "+00:00";


   /*!40101 SET @OLD_CHARACTER_SET_CLIENT=@@CHARACTER_SET_CLIENT */;
   /*!40101 SET @OLD_CHARACTER_SET_RESULTS=@@CHARACTER_SET_RESULTS */;
   /*!40101 SET @OLD_COLLATION_CONNECTION=@@COLLATION_CONNECTION */;
   /*!40101 SET NAMES utf8 */;

   --
   -- Database: `db` 
   --

   -- --------------------------------------------------------

   --
   -- Table structure for table `products`
   --

   CREATE TABLE IF NOT EXISTS `products` (
   `pid` int(100) NOT NULL AUTO_INCREMENT,
   `category` varchar(20) COLLATE utf8_bin NOT NULL,
   `subcategory` varchar(20) COLLATE utf8_bin NOT NULL,
   `product_name` varchar(100) COLLATE utf8_bin NOT NULL,
   `product_descrip` varchar(255) COLLATE utf8_bin NOT NULL,
   `product_price` varchar(100) COLLATE utf8_bin NOT NULL,
   `more_info` varchar(255) COLLATE utf8_bin NOT NULL,
   `pro_img_path` varchar(255) COLLATE utf8_bin NOT NULL,
   `time` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
   PRIMARY KEY (`pid`)
   ) ENGINE=InnoDB  DEFAULT CHARSET=utf8 COLLATE=utf8_bin COMMENT='Product Details' AUTO_INCREMENT=185 ;

   --
   -- Dumping data for table `products`
   --

   INSERT INTO `products` (`pid`, `category`, `subcategory`, `product_name`, `product_descrip`,                                                  `product_price`, `more_info`, `pro_img_path`, `time`) VALUES
   (176, 'Electronics', 'Mobile Accessories', 'Samsung Charger', 'Model G-S5830', '0890',     'MegaWat 10', '', '2014-09-15 14:19:11'),
   (182, 'Category1', 'Subcategory1', 'Electronics', 'Camera', 'Sony', 'ZeroShot', '', '2014-09-   15   17:35:42');

    /*!40101 SET CHARACTER_SET_CLIENT=@OLD_CHARACTER_SET_CLIENT */;
    /*!40101 SET CHARACTER_SET_RESULTS=@OLD_CHARACTER_SET_RESULTS */;
    /*!40101 SET COLLATION_CONNECTION=@OLD_COLLATION_CONNECTION */;

Answer 1

如果要在插入数据后更新数据，请执行以下操作：

代码

之后

$result = mysqli_query($con, "INSERT INTO products (pro_img_path) VALUES ('$location')");

找出插入ID并将其分配给变量http://cz2.php.net/manual/en/mysqli.insert-id.php

$lastid=mysqli_insert_id($con);

然后运行类似于此的更新查询

UPDATE products SET column=value WHERE id='$lastid' LIMIT 1

请注意limit 1子句，如果您是php和mysql的新手，可以防止大量错误。如果您计划更新一行，则将查询限制为一行，以防您搞乱WHERE子句。

以下是完整的更新语法： http://dev.mysql.com/doc/refman/5.0/en/update.html 这是mysqli文档： http://php.net/manual/en/book.mysqli.php

Answer 2

你可以使用：mysqli_insert_id来获取最后插入的id

 $last_inserted_id=mysqli_insert_id($con) ;

然后只使用您的值SET

进行更新

UPDATE products SET column1=value1,column2=value2 WHERE id='$last_inserted_id';

如何将数据插入到上一个受影响的行中

2 个答案: