dongmeixian9665 2017-09-30 11:34
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将文件上传到服务器并返回文件名

I'm trying to upload a file to my php server, then return the name of the file to display in the html document. But I get the following 

`error: Objektet wasn't found! The requested address was not found on this server. The link on the previous page appears to be incorrect or out of date Error 404

localhost
Apache/2.4.27 (Win32) OpenSSL/1.0.2l PHP/7.1.8`

My html Doc 

<html>
    <body>
        <form method="post" enctype="multipart/form-data" action="server.php">
        <input type="file" name="fileToUpload" id="fileToUpload" size="35">
        <br>
        <br>
        <input type="submit" value="Upload" name="submit">
    </body>
</html>

My php doc

<?php
header('Content-type: text/plain');
    if(isset($_POST["fileToUpload"])){
        $file = $_FILES["fileToUpload"];
        echo("File: ".$file);
    }
?>
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2条回答 默认 最新

  • doulan8330 2017-09-30 11:38
    关注

    You have many errors in PHP 

    <?php
 if(isset($_FILES["fileToUpload"])){
        $file = $_FILES["fileToUpload"]["name"];
        echo "File: ".$file;
    }
?>

    HTML

    <html>
    <body>
        <form method="post" enctype="multipart/form-data" action="server.php">
        <input type="file" name="fileToUpload" id="fileToUpload" size="35">
        <br>
        <br>
        <input type="submit" value="Upload" name="submit">
    </body>
</html>

    Errors

    1.if(isset($_POST["file"])){ its not post it should be $_FILES["fileToUpload"]) since its a file upload

    1. $file = $_FILES["file"]; and in your html you have defined file name as fileToUpload but your accessign unknown name so it should be $file = $_FILES["fileToUpload"]["name"];
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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