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问正在上载文件，但值不会反映在数据库中
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Stack Overflow用户

提问于 2017-01-03 16:46:28

回答 3查看 48关注 0票数 0

<!doctype html>
<?php
mysql_connect("127.0.0.1","root@localhost","");
mysql_select_db("reco");

if(isset($_POST['submit'])) {
    $soundcard_title = $_FILES['file']['name'];
    $upload_tmp = $_FILES['file']['tmp_name'];
    $upload_type = $_FILES['file']['type'];

    $upload_name = preg_replace("#^a-z0-9.#i","", $soundcard_title);

    if(!$soundcard_title) {
        echo "No file selected";
    } else{
        move_uploaded_file( $upload_tmp, "soundcards/".$soundcard_title);
        echo "<br />".$soundcard_title." has been uploaded";
    }

    $path= "http://127.0.0.1/upload-playback/soundcards/$soundcard_title";
    $query = "INSERT INTO 'soundcards' VALUE('','','$soundcard_title','$path','','','')";
    $query1 = mysql_query($query);
}
?>
<html>
    <head>
        <meta charset="utf-8">
        <title>Upload screen</title>
    </head>
    <body>
        <form enctype="multipart/form-data" action="upload_screen.php" method="post">
            Browse:<br>
            <input name="file" type="file" size="80"> <br>
            <input type="submit" name="submit" id="upload_btn" value="Upload">
        </form>
    </body>
</html>

我想创建一个简单的上传播放在php使用mysql数据库。为此，我使用xampp创建apache服务器。现在我可以上传文件到指定的文件夹，但mysql数据库没有更新与上传的文件相关联的条目。因为它，我不能播放文件稍后。

我的数据库模式：

-- phpMyAdmin SQL Dump
-- version 4.5.1
-- http://www.phpmyadmin.net
--
-- Host: 127.0.0.1
-- Generation Time: Jan 03, 2017 at 08:49 AM
-- Server version: 10.1.13-MariaDB
-- PHP Version: 5.6.23

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 utf8mb4 */;

--
-- Database: `reco`
--

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

--
-- Table structure for table `soundcards`
--

CREATE TABLE `soundcards` (
  `s_id` int(10) NOT NULL,
  `u_id` int(11) NOT NULL,
  `soundcard_title` varchar(20) NOT NULL,
  `path` varchar(100) NOT NULL,
  `date-time` varchar(20) NOT NULL,
  `no_plays` int(11) NOT NULL,
  `no_reco` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

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

--
-- Table structure for table `user`
--

CREATE TABLE `user` (
  `u_id` int(11) NOT NULL,
  `name` varchar(50) NOT NULL,
  `username` varchar(20) NOT NULL,
  `email` varchar(50) NOT NULL,
  `password` varchar(50) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

--
-- Indexes for dumped tables
--

--
-- Indexes for table `soundcards`
--
ALTER TABLE `soundcards`
  ADD PRIMARY KEY (`s_id`);

--
-- Indexes for table `user`
--
ALTER TABLE `user`
  ADD PRIMARY KEY (`u_id`),
  ADD UNIQUE KEY `username` (`username`),
  ADD UNIQUE KEY `email` (`email`);

--
-- AUTO_INCREMENT for dumped tables
--

--
-- AUTO_INCREMENT for table `soundcards`
--
ALTER TABLE `soundcards`
  MODIFY `s_id` int(10) NOT NULL AUTO_INCREMENT, AUTO_INCREMENT=2;
--
-- AUTO_INCREMENT for table `user`
--
ALTER TABLE `user`
  MODIFY `u_id` int(11) NOT NULL AUTO_INCREMENT;
/*!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 */;

php

mysql

phpmyadmin

xampp

回答 3

Stack Overflow用户

发布于 2017-01-03 17:00:10

我认为你的sql写错了格式，插入的格式应该是这样的

INSERT INTO TABLE_NAME (column1,column2,column3,...columnN)
VALUES (value1,value2,value3,...valueN);

票数 0

Stack Overflow用户

发布于 2017-01-03 17:11:08

<!doctype html>
<?php
    mysql_connect("127.0.0.1","root@localhost","");
    mysql_select_db("reco");

    if(isset($_POST['submit'])){
        $soundcard_title = $_FILES['file']['name'];
        $upload_tmp = $_FILES['file']['tmp_name'];
        $upload_type = $_FILES['file']['type'];

        $upload_name = preg_replace("#^a-z0-9.#i","", $soundcard_title);

        if(!$soundcard_title){
            echo "No file selected";
        }
        else{
            move_uploaded_file( $upload_tmp, "soundcards/".$soundcard_title);
                echo "<br />".$soundcard_title." has been uploaded";
        }

        $path= "http://127.0.0.1/upload-playback/soundcards/$soundcard_title";
        $query = "INSERT INTO 'soundcards'  VALUES('','$soundcard_title','$path','','','')";
        $query1 = mysql_query($query);
    }
?>
<html>
<head>
<meta charset="utf-8">
<title>Upload screen</title>
</head>

<body>
<form enctype="multipart/form-data" action="upload_screen.php" method="post">
    Browse:<br>
    <input name="file" type="file" size="80"> <br>
    <input type="submit" name="submit" id="upload_btn" value="Upload">
</form>

</body>
</html>

票数 0

Stack Overflow用户

发布于 2017-01-03 17:24:51

您的插入查询中存在以下错误：

$path= "http://127.0.0.1/upload-playback/soundcards/$soundcard_title";
$query = "INSERT INTO 'soundcards' VALUE('','','$soundcard_title','$path','','','')";
$query1 = mysql_query($query);

解决方案：

$query = "INSERT INTO 'soundcards' **VALUES**('','','$soundcard_title','$path','','','')";
$query1 = mysql_query($query) or die('Error is='.mysql_error());

您使用了值而不是值。

票数 0

页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持

原文链接：

https://stackoverflow.com/questions/41439556

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EN