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文件上传无效| 腓

[英]File upload not working | Php

 原文  2016-08-22 06:39:13  5   7    php/ forms/ file

问题描述

我正在尝试上传php中的文件而无法执行此操作。 我正在尝试上传的文件是一个csv文件，但它不应该是一个问题。 我正在使用php上传我的文件。 我也试图在同一页面处理表单。 下面是我的文件上传代码，它无法正常工作...... 

<!DOCTYPE html>
<html>
<head>
<title>File Upload</title>
</head>
<body>

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
    <input type="file" name="csv_file">
    <input type="submit" name="submit">
</form>


<?php   
    if(isset($_POST['submit'])) {
        if(isset($_POST['csv_file'])) {             
            echo "<p>".$_POST['csv_file']." => file input successfull</p>";
            fileUpload();
        }
    }   

function fileUpload () {
    $target_dir = "var/import/";
    $file_name = $_FILES['csv_file']['name'];
    $file_tmp = $_FILES['csv_file']['tmp_name'];

    if (move_uploaded_file($file_tmp, $target_dir.$file_name)) {
        echo "<h1>File Upload Success</h1>";            
    }
    else {
        echo "<h1>File Upload not successfull</h1>";
    }
}

?>

7 个解决方案

解决方案1
 3  2016-08-22 06:41:09

使用enctype =“multipart / form-data” 

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST" enctype="multipart/form-data">
    <input type="file" name="csv_file"> 
    <input type="submit" name="submit">
</form>

解决方案2
 3  2016-08-22 06:41:48

使用enctype属性更新您的表单代码 

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST" enctype = "multipart/form-data">
    <input type="file" name="csv_file">
    <input type="submit" name="submit">
</form>

解决方案3
 2 已采纳  2016-08-22 06:50:58

我尝试下面的代码，它的工作完美。 

<!DOCTYPE html>
<html>
    <head>
        <title>File Upload</title>
    </head>
    <body>

        <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST" enctype="multipart/form-data">
            <input type="file" name="csv_file">
            <input type="submit" name="submit">
        </form>


        <?php
        if (isset($_POST['submit'])) {
            echo "<p>" . $_POST['csv_file'] . " => file input successfull</p>";
            $target_dir = "images ";
            $file_name = $_FILES['csv_file']['name'];
            $file_tmp = $_FILES['csv_file']['tmp_name'];

            if (move_uploaded_file($file_tmp, $target_dir . $file_name)) {
                echo "<h1>File Upload Success</h1>";
            } else {
                echo "<h1>File Upload not successfull</h1>";
            }
        }
        ?>
    </body>
</html>

解决方案4
 1  2016-08-22 06:43:16

用PHP上传代码[不检查其扩展名] 

<?php
if(isset($_POST['save'])){
$path="var/import/";
$name = $_FILES['csv_file']['name'];//Name of the File
$temp = $_FILES['csv_file']['tmp_name'];
if(move_uploaded_file($temp, $path . $name)){
    echo "success";
}else{
    echo "failed";
}
}
?>
<form method="post" action="" enctype="multipart/form-data">
<input type="file" name="csv_file">
<input type="submit" name="save" value="submit">
</form>

解决方案5
 1  2016-08-22 06:56:16

它应该是这样的 

    <!DOCTYPE html>
    <html>
    <head>
    <title>File Upload</title>
    </head>
    <body>

    <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST" enctype="multipart/form-data">
        <input type="file" name="csv_file">
        <input type="submit" name="submit">
    </form>


    <?php
        if(isset($_POST['submit'])) {

            if ($_FILES['csv_file']['size'] > 0)
            {
                echo "<p>".$_FILES['csv_file']['name']." => file input successfull</p>";

                fileUpload();
            }
        }

    function fileUpload () {
        $target_dir = "var/import/";
        $file_name = $_FILES['csv_file']['name'];


        $file_tmp = $_FILES['csv_file']['tmp_name'];

        if (move_uploaded_file($file_tmp, $target_dir.$file_name)) {
            echo "<h1>File Upload Success</h1>";
        }
        else {
            echo "<h1>File Upload not successfull</h1>";
        }
    }

    ?>
</body>

以下是您需要进行的更改

  • 在表单标记中添加enctype = "multipart/form-data"作为新属性
  • 从这个改变if(isset($_POST['csv_file'])) {到这个if($_FILES['csv_file']['size'] > 0) {因为你需要检查大小是否上传
  • 更改此echo "<p>".$_POST['csv_file']." => file input successfull</p>"; 回到echo "<p>".$_FILES['csv_file']['name']." => file input successfull</p>"; 因为你需要使用$_FILES来获取文件名而不是$_POST
  • 最后但并非最不重要的，完整的</body>标签，如果你还没有。

解决方案6
 1  2016-08-22 07:54:44

First Give permission to folder where you going to upload Ex: "var/import/" folder.

<!DOCTYPE html>
<html>
<head>
<title>File Upload</title>
</head>
<body>

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST" enctype= "multipart/form-data">
    <input type="file" name="csv_file">
    <input type="submit" name="submit">
</form>


<?php   
    if(isset($_POST['submit'])) {
        if(isset($_FILES['csv_file']) && $_FILES['csv_file']['size'] > 0) {             
            echo "<p>".$_FILES['csv_file']['name']." => file input successfull</p>";
            fileUpload();
        }
    }   

function fileUpload () {
    $target_dir = "var/import/";
    $file_name = $_FILES['csv_file']['name'];
    $file_tmp = $_FILES['csv_file']['tmp_name'];

    if (move_uploaded_file($file_tmp, $target_dir.$file_name)) {
        echo "<h1>File Upload Success</h1>";            
    }
    else {
        echo "<h1>File Upload not successfull</h1>";
    }
}

解决方案7
 0  2018-07-14 10:56:49

对于上传文件，技巧是enctype =“multipart / form-data”使用此表单定义

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