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我正忙于PHP上传和邮件功能，基本上它需要上传的图像并将其作为附件放入电子邮件中，然后还将其上传到目录，我将如何根据日期重命名图像但保留扩展名，人们可以上传具有相同文件名的图像...

这是上传功能：

$newdirectory = "applicants";
$count = 0;
foreach ($_FILES['myfile']['name'] as $filename)
{
    $temp = $_FILES['myfile']['tmp_name'][$count];
    move_uploaded_file($temp, $newdirectory . '/' . $filename);
    $count++;
}

然后是整个 php 函数：

<?php
    function printMember($member) {
        foreach($member as $key=>$value) {
            //Fill the aux string first
            $str.= "$key : $value <br />";
        }
        //string that will be added to $msg variable inside the loop
        return $str;
    }

    $json = $_POST['parameters'];
    $json_string = stripslashes($json);
    $data = json_decode($json_string, true);
    $depCount = count($data["dependants"]);

    $msg .= "<h2>Main member data:</h2>";
    $msg .= printMember($data["mainmember"]);
    $msg .= "<h2>There are $depCount Dependants</h2>";

    foreach ($data["dependants"] as $index => $dependant) {
       $msg .= "<h2>Dependant $index</h2>";
       $msg .= printMember($dependant);
    }

    $strTo = "chante@jamfactory.co.za";
    $strSubject = "Image Testing";
    $strMessage = nl2br($msg);

    //*** Uniqid Session ***//
    $strSid = md5(uniqid(time()));

    $strHeader = "";
    $strHeader .= "From: Dawid<xxx@xxx.co.za>\nReply-To:xxxx@xxx.co.za";

    $strHeader .= "MIME-Version: 1.0\n";
    $strHeader .= "Content-Type: multipart/mixed; boundary=\"".$strSid."\"\n\n";
    $strHeader .= "This is a multi-part message in MIME format.\n";

    $strHeader .= "--".$strSid."\n";
    $strHeader .= "Content-type: text/html; charset=utf-8\n";
    $strHeader .= "Content-Transfer-Encoding: 7bit\n\n";
    $strHeader .= $strMessage."\n\n";

    //*** Attachment ***//
    $count = 0;
    foreach($_FILES['myfile']['name'] as $filename)
    {
        $temp = $_FILES['myfile']['tmp_name'][$count];
        $strFilesName = $filename;
        $strContent = chunk_split(base64_encode(file_get_contents($temp))); 
        $strHeader .= "--".$strSid."\n";
        $strHeader .= "Content-Type: application/octet-stream; name=\"".$strFilesName."\"\n"; 
        $strHeader .= "Content-Transfer-Encoding: base64\n";
        $strHeader .= "Content-Disposition: attachment; filename=\"".$strFilesName."\"\n\n";
        $strHeader .= $strContent."\n\n";
        $count++;
    }


    $flgSend = @mail($strTo,$strSubject,null,$strHeader);  // @ = No Show Error //

    if($flgSend)
    {
        echo "Mail send completed.";
        $newdirectory = "applicants";
        $count = 0;
        foreach ($_FILES['myfile']['name'] as $filename)
        {
            $temp = $_FILES['myfile']['tmp_name'][$count];
            move_uploaded_file($temp, $newdirectory . '/' . $filename);
            $count++;
        }
    }
    else
    {
        echo "Cannot send mail.";
    }
?>
4

3 回答 3

0

用move_uploaded_file()..你可以选择文件名..

该函数检查以确保 filename 指定的文件是有效的上传文件（意味着它是通过 PHP 的 HTTP POST 上传机制上传的）。如果文件有效，它将被移动到目的地给定的文件名。

所以你可以做

 $newfilename='yourFileName';
 move_uploaded_file($temp, $newdirectory . '/' . $newfilename);// this will save your filename with newfilename from temp to newdirectory
于 2013-02-04T13:17:22.903 回答
0

试试这个方法来获取扩展名并重命名新文件名：

foreach ($_FILES['myfile']['name'] as $filename)
{ 
     $temp = $_FILES['myfile']['tmp_name'][$count];

     $ImageExt = substr($filename, strrpos($filename, '.'));
     $ImageExt = str_replace('.','',$ImageExt);

     $filename = "YOUR_NEW_NAME_HERE".$ImageExt;
     move_uploaded_file($temp, $newdirectory . '/' . $filename);
     count++;
}

如果文件名包含多个点（。）那么你可以使用pathinfosubstr(strrchr($file_name,'.'),1);也会帮助你。

于 2013-02-04T13:18:50.483 回答
0

我将pathinfo用于排除文件扩展名，并添加额外的timestamp内容以防止文件名冲突并替换现有文件。

$newdirectory = "applicants";
$count = 0;
foreach ($_FILES['myfile']['name'] as $filename)
{

    $temp = $_FILES['myfile']['tmp_name'][$count];
    $file_info = pathinfo($filename);

    // Don't you forget to have the $user available here ...
    $filename = $user . '_' . time() . '.' . $file_info['extension'];

    move_uploaded_file($temp, $newdirectory . '/' . $filename);
    $count++;
}
于 2013-02-04T13:26:00.167 回答