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所以我必须编写一个表格来将图像上传到网站......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。似乎工作得很好。问题是在写入数据库的代码中没有任何地方......它只是将图像上传到一个文件夹并且没有数据库来跟踪它......我知道 SQL 足以完成写作代码....但我不太了解将实际的 sql 代码放在哪里......如果我点击 GO 按钮上传图片，用户就会被带离这个页面...... sql 也会如此运行？...如何修改它以添加 sql 代码

有任何想法吗？

继承人的代码......任何帮助都会很棒！

<?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 回答 1

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将您的代码放在以下块中：

//If no errors registred, print the success message
if(isset($_POST['Submit']) && !$errors) {
   // do database stuff here...
   echo "<h1>File Uploaded Successfully! Try again!</h1>";
}

至此你就知道上传成功了。您没有提供表架构或 SQL 风格。我将假设 MySQL。在这种情况下，请查看mysql_query()。PHP 文档有一个很好的例子来连接和运行查询。

于 2011-05-16T01:38:17.743 回答