我正在尝试上传php中的文件但无法执行此操作。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>";
}
}
?>
答案 0 :(得分:3)
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>
答案 1 :(得分:3)
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>
答案 2 :(得分:2)
我尝试过以下代码并且它的工作非常完美。
<!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 . $file_name)) {
echo "<h1>File Upload Success</h1>";
} else {
echo "<h1>File Upload not successfull</h1>";
}
}
?>
</body>
</html>
答案 3 :(得分:1)
在PHP中上传代码[不检查其扩展名]
<?php
if(isset($_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>
答案 4 :(得分:1)
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
enctype = "multipart/form-data"
作为新属性if(isset($_POST['csv_file'])) {
更改为此if($_FILES['csv_file']['size'] > 0) {
是否上传echo "<p>".$_POST['csv_file']." => file input
successfull</p>";
更改为此echo "<p>".$_FILES['csv_file']['name']."
=> file input successfull</p>";
,因为您需要使用$_FILES
来获取文件名而不是$_POST
</body>
标记。答案 5 :(得分:1)
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>";
}
}
答案 6 :(得分:0)
对于上传文件,窍门是enctype =“ multipart / form-data” 使用此表单定义