26

嗨,我已经阅读了许多论坛和网站,它们告诉您如何将图像上传到服务器,并且我已经设法让它工作,我可以将文件上传到我的服务器,但存储文件名确实适用于我发现的以下示例而且我还需要创建一个允许将更多数据输入数据库的表单。我坚持这一点,因为以前做过很多 PHP。我已经尝试了不同的网站教程,但没有取得多大成功,任何人都可以帮助我!我需要为我正在做的项目完成它。

我基本上是在尝试制作一个 CMS,它允许用户上传乐队成员的照片并存储有关他们的信息,以便可以将其显示在网页上供公众查看。


我的表如下所示:

Field              Type             Null    Default     
id                 int(10)          No                   
nameMember         varchar(25)      No                   
bandMember         text             No                   
photo              varchar(30)      No                   
aboutMember        text             No                   
otherBands         text             No      

我想要的表格如下所示:

   <h1>Adding a new Band Member or Affiliate</h1>
      <form method="post" action="addMember.php" enctype="multipart/form-data">
       <p>
              Please Enter the Band Members Name.
            </p>
            <p>
              Band Member or Affiliates Name:
            </p>
            <input type="text" name="nameMember"/>
            <p>
              Please Enter the Band Members Position. Example:Drums.
            </p>
            <p>
              Member's Position:
            </p>
            <input type="text" name="bandMember"/>
            <p>
              Please Upload a Photo in gif or jpeg format. The file name should be named after the Members name. If the same file name is uploaded twice it will be overwritten!
            </p>
            <p>
              Photo:
            </p>
            <input type="file" name="filep" size=35 />
            <p>
              Please Enter any other information about the band member here.
            </p>
            <p>
              Other Member Information:
            </p>
<textarea rows="10" cols="35" name="aboutMember">
</textarea>
            <p>
              Please Enter any other Bands the Member has been in.
            </p>
            <p>
              Other Bands:
            </p>
            <input type="text" name="otherBands" size=30 />
            <br/>
            <br/>
            <input TYPE="submit" title="Add data to the Database" value="Add Member"/>
          </form>

仅将图像上传到服务器的示例,即:

<?

if ($_POST["action"] == "Load")
{
$folder = "images/";

move_uploaded_file($_FILES["filep"]["tmp_name"] , "$folder".$_FILES["filep"]["name"]);

echo "
<p align=center>File ".$_FILES["filep"]["name"]."loaded...";

$result = mysql_connect("localhost", "******", "*****") or die ("Could not save image name

Error: " . mysql_error());

mysql_select_db("project") or die("Could not select database");
mysql_query("INSERT into dbProfiles (photo) VALUES('".$_FILES['filep']['name']."')");
if($result) { echo "Image name saved into database

"; }

}

?>

我必须使用的示例表格是这样的:

<form action=addMember.php method=post enctype="multipart/form-data">
<table border="0" cellspacing="0" align=center cellpadding="3" bordercolor="#cccccc">
<tr>
<td>File:</td>
<td><input type="file" name="filep" size=45></td>
</tr>
<tr>
<td colspan=2><p align=center>
<input type=submit name=action value="Load">
</td>
</tr>
</table>
</form>

PS:图像文件已打开以供写入。

4

6 回答 6

26

Here is the answer for those of you looking like I did all over the web trying to find out how to do this task. Uploading a photo to a server with the file name stored in a mysql database and other form data you want in your Database. Please let me know if it helped.

Firstly the form you need:

    <form method="post" action="addMember.php" enctype="multipart/form-data">
    <p>
              Please Enter the Band Members Name.
            </p>
            <p>
              Band Member or Affiliates Name:
            </p>
            <input type="text" name="nameMember"/>
            <p>
              Please Enter the Band Members Position. Example:Drums.
            </p>
            <p>
              Band Position:
            </p>
            <input type="text" name="bandMember"/>
            <p>
              Please Upload a Photo of the Member in gif or jpeg format. The file name should be named after the Members name. If the same file name is uploaded twice it will be overwritten! Maxium size of File is 35kb.
            </p>
            <p>
              Photo:
            </p>
            <input type="hidden" name="size" value="350000">
            <input type="file" name="photo"> 
            <p>
              Please Enter any other information about the band member here.
            </p>
            <p>
              Other Member Information:
            </p>
<textarea rows="10" cols="35" name="aboutMember">
</textarea>
            <p>
              Please Enter any other Bands the Member has been in.
            </p>
            <p>
              Other Bands:
            </p>
            <input type="text" name="otherBands" size=30 />
            <br/>
            <br/>
            <input TYPE="submit" name="upload" title="Add data to the Database" value="Add Member"/>
          </form>

Then this code processes you data from the form:

   <?php

// This is the directory where images will be saved
$target = "your directory";
$target = $target . basename( $_FILES['photo']['name']);

// This gets all the other information from the form
$name=$_POST['nameMember'];
$bandMember=$_POST['bandMember'];
$pic=($_FILES['photo']['name']);
$about=$_POST['aboutMember'];
$bands=$_POST['otherBands'];


// Connects to your Database
mysqli_connect("yourhost", "username", "password") or die(mysqli_error()) ;
mysqli_select_db("dbName") or die(mysqli_error()) ;

// Writes the information to the database
mysqli_query("INSERT INTO tableName (nameMember,bandMember,photo,aboutMember,otherBands)
VALUES ('$name', '$bandMember', '$pic', '$about', '$bands')") ;

// Writes the photo to the server
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target))
{

// Tells you if its all ok
echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory";
}
else {

// Gives and error if its not
echo "Sorry, there was a problem uploading your file.";
}
?> 

Code edited from www.about.com

于 2009-01-29T11:41:00.247 回答
3

每张照片都有一个 ID,所以我的建议是重命名照片。例如,您按日期重命名它

<?php
 $date = getdate();
 $name .= $date[hours];
 $name .= $date[minutes];
 $name .= $date[seconds];
 $name .= $date[year];
 $name .= $date[mon];
 $name .= $date[mday];
?>

注意:不要忘记文件的文件扩展名,或者您可以为照片生成随机字符串,但我不建议这样做。我还建议您在将文件上传到目录之前检查文件扩展名。

<?php 
if ((($_FILES["photo"]["type"] == "image/jpeg")
            || ($_FILES["photo"]["type"] == "image/pjpg"))
            && ($_FILES["photo"]["size"] < 100000000))
            {
                move_uploaded_file($_FILES["photo"]["tmp_name"], $target.$name);

                if(mysql_query("your query"))
                {
                    //success handling
                }
                else 
                {
                    //failed handling
                }
            }
            else
            {
                //error handling
            }
?>

希望这可能会有所帮助。

于 2012-12-30T11:47:12.953 回答
3
<form method="post" action="addMember.php" enctype="multipart/form-data">
    <p>
              Please Enter the Band Members Name.
            </p>
            <p>
              Band Member or Affiliates Name:
            </p>
            <input type="text" name="nameMember"/>
            <p>
              Please Enter the Band Members Position. Example:Drums.
            </p>
            <p>
              Band Position:
            </p>
            <input type="text" name="bandMember"/>
            <p>
              Please Upload a Photo of the Member in gif or jpeg format. The file name should be named after the Members name. If the same file name is uploaded twice it will be overwritten! Maxium size of File is 35kb.
            </p>
            <p>
              Photo:
            </p>
            <input type="hidden" name="size" value="350000">
            <input type="file" name="photo"> 
            <p>
              Please Enter any other information about the band member here.
            </p>
            <p>
              Other Member Information:
            </p>
<textarea rows="10" cols="35" name="aboutMember">
</textarea>
            <p>
              Please Enter any other Bands the Member has been in.
            </p>
            <p>
              Other Bands:
            </p>
            <input type="text" name="otherBands" size=30 />
            <br/>
            <br/>
            <input TYPE="submit" name="upload" title="Add data to the Database" value="Add Member"/>
          </form>

将其保存为 addMember.php

<?php

//This is the directory where images will be saved
$target = "your directory";
$target = $target . basename( $_FILES['photo']['name']);

//This gets all the other information from the form
$name=$_POST['nameMember'];
$bandMember=$_POST['bandMember'];
$pic=($_FILES['photo']['name']);
$about=$_POST['aboutMember'];
$bands=$_POST['otherBands'];


// Connects to your Database
mysql_connect("yourhost", "username", "password") or die(mysql_error()) ;
mysql_select_db("dbName") or die(mysql_error()) ;

//Writes the information to the database
mysql_query("INSERT INTO tableName (nameMember,bandMember,photo,aboutMember,otherBands)
VALUES ('$name', '$bandMember', '$pic', '$about', '$bands')") ;

//Writes the photo to the server
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target))
{

//Tells you if its all ok
echo "The file ". basename( $_FILES['photo']['name']). " has been uploaded, and your information has been added to the directory";
}
else {

//Gives and error if its not
echo "Sorry, there was a problem uploading your file.";
}
?>

在上面的代码中有一个小错误,我修复了那个错误。

于 2010-02-17T18:42:14.427 回答
1

添加以下内容可以避免文件名中的引号问题,例如

"freddy's pic.jpg"

这在某些操作系统上是可以接受的。

前:

$pic=($_FILES['photo']['name']);

后:

$pic=(mysql_real_escape_string($_FILES['photo']['name']));
于 2010-05-20T16:54:26.847 回答
0

如果要在表单中输入更多数据,只需通过 $_POST 访问提交的数据。

如果你有

<input type="text" name="firstname" />

你访问它

$firstname = $_POST["firstname"];

然后您可以更新您的查询行以读取

mysql_query("INSERT INTO dbProfiles (photo,firstname)
             VALUES('{$filename}','{$firstname}')");

注意:始终过滤和清理您的数据。

于 2009-01-16T17:19:43.820 回答
0

你的部分:

$result = mysql_connect("localhost", "******", "*****") or die ("Could not save image name

Error: " . mysql_error());

mysql_select_db("project") or die("Could not select database");
mysql_query("INSERT into dbProfiles (photo) VALUES('".$_FILES['filep']['name']."')");
if($result) { echo "Image name saved into database

";

没有多大意义,您的连接不应该命名为 $result 但这是一个命名问题而不是编码问题。

什么是编码问题是 if($result),如果您说无论插入查询失败还是成功都可以连接到数据库,您将输出“图像保存到数据库中”。

尝试添加做

$realresult = mysql_query("INSERT into dbProfiles (photo) VALUES('".$_FILES['filep']['name']."')");

并将 if($result) 更改为 $realresult

我怀疑您的查询失败了,也许您还有其他列或其他内容?

尝试复制/粘贴您的查询,替换“.$_FILES['filep']['name']。” 在您的查询浏览器中测试并运行它,看看它是否进入。

于 2009-01-16T17:39:09.053 回答