无法在php中使用上载文件逻辑
我有一个php代码片段无法在php中使用上载文件逻辑,php,Php,我有一个php代码片段 <?php $target_dir = "uploads/"; $target_file_name = $target_dir .basename($_FILES["file"]["name"]); $response = array(); // Check if image file is a actual image or fake image if (isset($_FILES["file"])) { if (move_uploaded_file($
<?php
$target_dir = "uploads/";
$target_file_name = $target_dir .basename($_FILES["file"]["name"]);
$response = array();
// Check if image file is a actual image or fake image
if (isset($_FILES["file"]))
{
if (move_uploaded_file($_FILES["file"]["tmp_name"], $target_file_name))
{
$success = true;
$message = "Successfully Uploaded";
}
else
{
$success = false;
$message = "Error while uploading";
}
}
else
{
$success = false;
$message = "Required Field Missing";
}
$response["success"] = $success;
$response["message"] = $message;
echo json_encode($response);
?>
如何解决此问题尝试此
$target\u file\u name=$\u SERVER['DOCUMENT\u ROOT']./'。.basename($\u FILES[“file”][“name”])
似乎$target\u dir
不存在
在移动文件之前,请检查目标目录
if (!is_dir($target_dir)) {
if (!mkdir($target_dir, 0777, true)) {
exit(json_encode(['success'=>false, 'message'=>'failed to create target directory']));
}
}
请检查文件权限。在哪里?在服务器上?是的,请在您上载文件的服务器或本地计算机上。它是远程主机,如何检查?好的,这会给我一条消息,表示上载成功,但我必须上载到
$target_dir=“uploads/”代码>目录$target_file_name=$_SERVER['DOCUMENT_ROOT']./uploads/'.basename($_FILES[“file”][“name”);文件夹“uploads”必须存在或$target_dir=$\u SERVER['DOCUMENT_ROOT']./uploads/;$target_file_name=$target_dir.basename($_FILES[“file”][“name”]);
if (!is_dir($target_dir)) {
if (!mkdir($target_dir, 0777, true)) {
exit(json_encode(['success'=>false, 'message'=>'failed to create target directory']));
}
}