我正试图开发一个web应用程序,在那里你可以创建帖子,并有一个(只是一个)图像,但你不需要有一个,它也可以只是一个文本帖子。这只是一个个人项目,试图使php和ajax变得更好,所以我可能做错了.我想要做的是将表单数据提交给一个php函数,这个函数将决定是否上传它,那么我想要做的是,如果它成功地将图像名发送到另一个函数,将post插入到数据库中,但是如果图像上传没有成功地显示错误。此外,如果它不是成功的,但错误是‘没有图片上传’,继续继续并创建文章无论如何。
这是我到目前为止所拥有的
The jQuery
$(".new-post").submit(function (e) {
e.preventDefault();
$.ajax({
type: "post",
url: "includes/image-upload.php",
data: new FormData(this),
processData: false,
contentType: false,
error: function (response) {
console.log(response);
},
success: function (response) {
var body = $("#post-body").val();
$.ajax({
type: "post",
url: "create-post-feed.php",
data: {
body: body,
image: response
},
error: function (response) {
console.log(response);
},
success: function (response) {
// console.log(response);
$('section.feed').prepend(response);
$('article.post p').each(function () {
$(this).html(linkHashtags($(this).html()));
});
$('article.post p').each(function () {
$(this).html(linkatsymbols($(this).html()));
});
revealPosts();
}
});
}
});
});在进行了一些研究之后,我发现如果这是一个错误,我必须将它作为JSON发送回来,然后在jQuery中检查结果。但是我还没有开始工作..。
<?php
require_once('../dbconnect.php');
include_once( INCLUDES_PATH .'functions.php');
function uploadFile ($file_field = null, $check_image = false, $random_name = false) {
//Config Section
$user_id = $_SESSION['user_id'];
//check if directory exist if not create it
if (!file_exists(HOME_PATH ."users/user_".$user_id)) {
mkdir(HOME_PATH ."users/user_".$user_id, 0777, true);
}
if (!file_exists(HOME_PATH ."users/user_".$user_id."/posts")) {
mkdir(HOME_PATH ."users/user_".$user_id."/posts", 0777, true);
}
//Set file upload path
$path = "../users/user_".$user_id."/posts/"; //with trailing slash
//Set max file size in bytes
$max_size = 1000000;
//Set default file extension whitelist
$whitelist_ext = array('jpeg','jpg','png','gif');
//Set default file type whitelist
$whitelist_type = array('image/jpeg', 'image/jpg', 'image/png','image/gif');
//The Validation
// Create an array to hold any output
$out = array('error'=>null);
if (!$file_field) {
$out['error'][] = "Please specify a valid form field name";
}
if (!$path) {
$out['error'][] = "Please specify a valid upload path";
}
if (count($out['error'])>0) {
return $out;
}
//Make sure that there is a file
if((!empty($_FILES[$file_field])) && ($_FILES[$file_field]['error'] == 0)) {
// Get filename
$file_info = pathinfo($_FILES[$file_field]['name']);
$name = $file_info['filename'];
$ext = $file_info['extension'];
//Check file has the right extension
if (!in_array($ext, $whitelist_ext)) {
$out['error'][] = "Invalid file Extension";
}
//Check that the file is of the right type
if (!in_array($_FILES[$file_field]["type"], $whitelist_type)) {
$out['error'][] = "Invalid file Type";
}
//Check that the file is not too big
if ($_FILES[$file_field]["size"] > $max_size) {
$out['error'][] = "File is too big";
}
//If $check image is set as true
if ($check_image) {
if (!getimagesize($_FILES[$file_field]['tmp_name'])) {
$out['error'][] = "Uploaded file is not a valid image";
}
}
//Create full filename including path
if ($random_name) {
// Generate random filename
$tmp = str_replace(array('.',' '), array('',''), microtime());
if (!$tmp || $tmp == '') {
$out['error'][] = "File must have a name";
}
$newname = $tmp.'.'.$ext;
} else {
$newname = $name.'.'.$ext;
}
//Check if file already exists on server
if (file_exists($path.$newname)) {
$out['error'][] = "A file with this name already exists";
}
if (count($out['error'])>0) {
//The file has not correctly validated
return $out;
}
if (move_uploaded_file($_FILES[$file_field]['tmp_name'], $path.$newname)) {
echo $newname;
} else {
$out['error'][] = "Server Error!";
}
} else {
$out['error'][] = "No file uploaded";
return $out;
}
}
$file = uploadFile('file', true, true);
if (is_array($file['error'])) {
echo json_encode($file['error']);
}
die();真正感谢任何帮助或评论,如何使它更好的BTW,这目前的工作,唯一的问题是,帖子总是创建,无论图片上传与否。
通常我至少会在同一天得到答案..。这个问题有那么糟吗?我听说编辑的XD会帮你解决这个问题
发布于 2016-12-19 02:45:26
你说你想做的和你实际做的不一样。在您的代码中,您正在尝试两个AJAX调用:(1)上传图像,(2)保存post。
为什么不在一个电话里同时做两件事呢?这样,您就不必等待上传图片(或无)的响应才能创建新的帖子。
$(".new-post").submit(function (e) {
e.preventDefault();
$.ajax({
type: "post",
url: "create-post-feed.php",
data: new FormData(this), // send post information including selected file
processData: false,
contentType: false,
error: function (response) {
console.log(response);
},
success: function (response) {
// etc
}
});
});然后在create-post-feed.php中,将image-upload.php中的逻辑与当前代码合并。我通常是这样做的:
// validate post
// validate image (if any)
// if no errors:
// upload image (if any) and retrieve filename
// if no errors:
// save post (with filename if any); if it fails, delete image (if any)
// send response https://stackoverflow.com/questions/41203004
复制相似问题