帮我修改此代码以写入数据库
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所以我必须写一个表格将图像上传到网站...我找到了下面的代码,将其添加到我的网站,并对其进行了测试。似乎工作很棒。问题是写入数据库的代码中没有任何地方。。。它只是将图像上传到文件夹,并且没有数据库可以对其进行跟踪……我知道SQL足够好,可以完成编写工作代码...。但是我不太了解实际的sql代码在哪里.....如果我单击GO按钮上传图像,则用户将从此页面移走...因此sql也会运行?...我如何修改它以添加sql代码
有任何想法吗?
继承人的代码...任何帮助将是巨大的!
<?php
//define a maxim size for the uploaded images in Kb
define (\"MAX_SIZE\",\"500\");
//This function reads the extension of the file. It is used to determine if the file is an image by checking the extension.
function getExtension($str) {
$i = strrpos($str,\".\");
if (!$i) { return \"\"; }
$l = strlen($str) - $i;
$ext = substr($str,$i+1,$l);
return $ext;
}
//This variable is used as a flag. The value is initialized with 0 (meaning no error found)
//and it will be changed to 1 if an errro occures.
//If the error occures the file will not be uploaded.
$errors=0;
//checks if the form has been submitted
if(isset($_POST[\'Submit\']))
{
//reads the name of the file the user submitted for uploading
$image=$_FILES[\'image\'][\'name\'];
//if it is not empty
if ($image)
{
//get the original name of the file from the clients machine
$filename = stripslashes($_FILES[\'image\'][\'name\']);
//get the extension of the file in a lower case format
$extension = getExtension($filename);
$extension = strtolower($extension);
//if it is not a known extension, we will suppose it is an error and will not upload the file,
//otherwise we will do more tests
if (($extension != \"jpg\") && ($extension != \"jpeg\") && ($extension != \"png\") && ($extension != \"gif\"))
{
//print error message
echo \'<h1>Unknown extension!</h1>\';
$errors=1;
}
else
{
//get the size of the image in bytes
//$_FILES[\'image\'][\'tmp_name\'] is the temporary filename of the file
//in which the uploaded file was stored on the server
$size=filesize($_FILES[\'image\'][\'tmp_name\']);
//compare the size with the maxim size we defined and print error if bigger
if ($size > MAX_SIZE*1024)
{
echo \'<h1>You have exceeded the size limit!</h1>\';
$errors=1;
}
//we will give an unique name, for example the time in unix time format
$image_name=time().\'.\'.$extension;
//the new name will be containing the full path where will be stored (images folder)
$newname=\"images/\".$image_name;
//we verify if the image has been uploaded, and print error instead
$copied = copy($_FILES[\'image\'][\'tmp_name\'], $newname);
if (!$copied)
{
echo \'<h1>Copy unsuccessfull!</h1>\';
$errors=1;
}}}}
//If no errors registred, print the success message
if(isset($_POST[\'Submit\']) && !$errors)
{
echo \"<h1>File Uploaded Successfully! Try again!</h1>\";
}
?>
<!--next comes the form, you must set the enctype to \"multipart/frm-data\" and use an input type \"file\" -->
<form name=\"newad\" method=\"post\" enctype=\"multipart/form-data\" action=\"\">
<table>
<tr><td><input type=\"file\" name=\"image\"></td></tr>
<tr><td><input name=\"Submit\" type=\"submit\" value=\"Upload image\"></td></tr>
</table>
</form>
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1 个回复
琶竞捆栓
至此,您知道上传已成功。您没有提供表架构或SQL风格。我将假设MySQL。在这种情况下,请检查mysql_query()。 PHP文档为连接和运行查询提供了一个很好的示例。