首次尝试上传图像并存储到db [英] First attempt at uploading image and storing to db
问题描述
我正在尝试上传图片，根据该图片创建一个新文件
然后将base64编码的图像数据存储到一个数据库。
我真的不知道我的代码出错了所以我只会在下面显示
：
以下代码包含处理上传文件的脚本，
上传文件的表单，然后是一些链接，用于查看来自
a php脚本的文件simpleimageviewer.php：
<？php
//此文件包含函数''dbconn（）''，允许你要
//连接到数据库
include（''functions.php''）;
？>
< html>
< body>
<？php
//
if（isset（$ _ GET [''action'']））{
if（isset（$ _ GET [''action'']）=="添加）{
$ tmpfilesize = $ _FILES [''imgfile''] ['s ize''];
$ tmpfilename = $ _FILES [''imgfile''] [''tmp_name''];
$ tmpfiletype = $ _FILES [''imgfile ''] [''type''];
if（$ tmpfilesize> 0）{
if（substr（$ tmpfiletype，0,6）==''image /''）{
//从上传的图片创建图片
开关（$ tmpfiletype）
{
case" image / jpeg"：
case" image / pjpeg" ：
$ img = imagecreatefromjpeg（$ tmpfilename）;
break;
case" image / gif"：
$ img = imagecreatefromgif（$ tmpfilename）;
break;
case" image / png"：
$ img = imagecreatefrompng（ $ tmpfilename）;
休息;
}
//创建新文件
$ newfile = tempnam（" ; / tmp"，" img_"）;
//创建jpeg并另存为新文件
imagejpeg（$ img，$ newfile，80）;
//打开新文件
$ nf = fopen（$ newfile，" r"）;
//获取已打开文件的内容并存储在var中
$ filecontents = fread（$ nf，filesize（$ newfile））;
//销毁文件
fclose（$ nf）;
unlink（$ newfile） ;
if（$ filecontents）{
$ insert =" INSERT INTO imglib（imgid，imgdata，imgtype）VALUES
（null，''" 。 base64_encode（$ filecontents）。 ''，''" 。 $ imgtype。 "''）" ;;
$ result = mysql_query（$ insert，$ db）;
if（$ result）{
echo"< p> Result！< / p>" ;;
} else {
echo"< p> No，no，no ..它没有用< / p>" ;;
}
}
}其他{
echo" ;< p>请上传图片 - 错误的文件类型（。
$ tmpfiletype。）< / p>" ;;
}
}其他{
echo"< p>文件未上传< / p>" ;;
}
}
}
？>
< form enctype =" multipart / form-data"名称= QUOT;" method =" post"
action =" simple.php？action = add">
< input type =" hidden"名称= QUOT; MAX_FILE_SIZE" value =" 5000000"><！ - about。
5MB - >
< input name =" imgfile" type =" file">< br />
< input name =" submit"类型= [提交" value =" Submit">
< / form>
<？php
$ select =" SELECT * FROM imglib";
$ result = mysql_query（$ select，$ db）;
if（mysql_num_rows（$ result）> 0）{
echo"< ul> \ n";
while（$ row = mysql_fetch_array（$ result））{
$ imgid = $ row [ ''imgid''];
echo < li>< a href = \" simpleimageviewer.php？imgid =" 。 $ imgid。
" \">图片" 。 $ imgid。 "< / a>< / li> \ n";
}
echo"< / ul>" ;;
}
？>< / body>
< / html>
抱歉关于代码的质量，但我想我应该提供它
all。我也在某些时候需要调整上传图像的大小，所以我似乎最好创建一个临时文件，我可以在写作之前用
来播放。
问题是，当我使用以下
脚本（simpleimageviewer.php）查看图像时：
include（'' functions.php''）;
//图片输出
$ select =" SELECT imgdata，imgtype FROM imglib WHERE imgid =" ; 。
$ _GET [''imgid''];
$ result = mysql_query（$ select，$ db）;
while（$ row = mysql_fetch_array（$ result））{
$ imgdata = $ row [''imgdata''];
$ imgtype = $ row [''imgtype''] ;
}
// header
header（" Content-type："。$ imagetype）;
echo base64_decode（$ imgdata）;
flush（）;
它不会显示所有图像。请参阅
http：// www .martynbissett.co.uk / test / ... er.php？imgid = 5
看看我的意思。当我突出显示图像时，它确实显示了很多白色
区域，这些区域在上传的图像中创建初始
图像时似乎一直存在问题。
任何人都可以告诉我的代码可能有什么问题，干杯
Burnsy
_GET [''action' ]）==" add"）{
tmpfilesize =
Hi,
I am trying to upload an image, create a new file based on that image
and then store the base64 encoded image data in a database.
I dont really know where my code is going wrong so I will just display
it below:
The following code contains the script to process the uploaded file,
the form to upload the file and then a few links to view the file from
a php script called simpleimageviewer.php:
<?php
//this file includes the function, ''dbconn()'', that allows you to
//connect to the database
include(''functions.php'');
?>
<html>
<body>
<?php
//
if (isset($_GET[''action''])) {
if (isset($_GET[''action'']) == "add") {
$tmpfilesize = $_FILES[''imgfile''][''size''];
$tmpfilename = $_FILES[''imgfile''][''tmp_name''];
$tmpfiletype = $_FILES[''imgfile''][''type''];
if ($tmpfilesize > 0) {
if (substr($tmpfiletype, 0, 6) == ''image/'') {
//create image from uploaded image
switch ($tmpfiletype)
{
case "image/jpeg":
case "image/pjpeg":
$img = imagecreatefromjpeg($tmpfilename);
break;
case "image/gif":
$img = imagecreatefromgif($tmpfilename);
break;
case "image/png":
$img = imagecreatefrompng($tmpfilename);
break;
}
//create new file
$newfile = tempnam("/tmp", "img_");
//create jpeg and save as newfile
imagejpeg($img, $newfile, 80);
//open new file
$nf = fopen($newfile, "r");
//get contents of opened file and store in var
$filecontents = fread($nf, filesize($newfile));
//destroy file
fclose($nf);
unlink($newfile);
if ($filecontents) {
$insert = "INSERT INTO imglib (imgid, imgdata, imgtype) VALUES
(null, ''" . base64_encode($filecontents) . "'', ''" . $imgtype . "'')";
$result = mysql_query($insert, $db);
if ($result) {
echo "<p>Result!</p>";
}else{
echo "<p>No, no, no ... it didnt work</p>";
}
}
} else {
echo "<p>Please upload an image - wrong file type (" .
$tmpfiletype . ")</p>";
}
}else{
echo "<p>File not uploaded</p>";
}
}
}
?>
<form enctype="multipart/form-data" name="" method="post"
action="simple.php?action=add">
<input type="hidden" name="MAX_FILE_SIZE" value="5000000"><!-- approx.
5MB -->
<input name="imgfile" type="file"><br />
<input name="submit" type="submit" value="Submit">
</form>
<?php
$select = "SELECT * FROM imglib";
$result = mysql_query($select, $db);
if (mysql_num_rows($result) > 0) {
echo "<ul>\n";
while ($row = mysql_fetch_array($result)) {
$imgid = $row[''imgid''];
echo " <li><a href=\"simpleimageviewer.php?imgid=" . $imgid .
"\">Image " . $imgid . "</a></li>\n";
}
echo "</ul>";
}
?></body>
</html>
Sorry about the mass of code but I thought I should just provide it
all. I also at some point need to resize the uploaded image so I seemed
the best idea to create a temporary file that I could play around with
before writing.
The problem is that when I go to view the image using the following
script (simpleimageviewer.php):
include(''functions.php'');
//IMAGE OUTPUT
$select = "SELECT imgdata, imgtype FROM imglib WHERE imgid = " .
$_GET[''imgid''];
$result = mysql_query($select, $db);
while ($row = mysql_fetch_array($result)) {
$imgdata = $row[''imgdata''];
$imgtype = $row[''imgtype''];
}
//header
header("Content-type: " . $imagetype);
echo base64_decode($imgdata);
flush();
It doesnt display all the image. See
http://www.martynbissett.co.uk/test/...er.php?imgid=5 to
see what I mean. When I highlight the image it does show a lot of white
area that appears to have been a problem when creating the initial
image from the uploaded one.
Can anyone tell from my code what might be the problem, cheers
Burnsy
_GET[''action''])) {
if (isset(
_GET[''action'']) == "add") {
tmpfilesize =
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