我可以使用以下jQuery代码使用ajax请求的POST方法执行文件上载吗?
$.ajax({
type: "POST",
timeout: 50000,
url: url,
data: dataString,
success: function (data) {
alert('success');
return false;
}
});
如果可能,我需要填写数据部分吗?这是正确的方式吗?我只将文件POST到服务器端。
我一直在搜索,但我发现的是一个插件,而在我的计划中,我不想使用它。至少目前是这样。
使用FormData。它工作得很好:-)。。。
var jform = new FormData();
jform.append('user',$('#user').val());
jform.append('image',$('#image').get(0).files[0]); // Here's the important bit
$.ajax({
url: '/your-form-processing-page-url-here',
type: 'POST',
data: jform,
dataType: 'json',
mimeType: 'multipart/form-data', // this too
contentType: false,
cache: false,
processData: false,
success: function(data, status, jqXHR){
alert('Hooray! All is well.');
console.log(data);
console.log(status);
console.log(jqXHR);
},
error: function(jqXHR,status,error){
// Hopefully we should never reach here
console.log(jqXHR);
console.log(status);
console.log(error);
}
});
我已经很晚了,但我正在寻找一个基于ajax的图像上传解决方案,我正在寻找的答案在这篇文章中有点分散。我确定的解决方案涉及FormData对象。我组装了一个基本形式的代码。您可以看到它演示了如何使用fd.append()向表单添加自定义字段,以及如何在完成ajax请求时处理响应数据。
上传html:
<!DOCTYPE html>
<html>
<head>
<title>Image Upload Form</title>
<script src="//code.jquery.com/jquery-1.9.1.js"></script>
<script type="text/javascript">
function submitForm() {
console.log("submit event");
var fd = new FormData(document.getElementById("fileinfo"));
fd.append("label", "WEBUPLOAD");
$.ajax({
url: "upload.php",
type: "POST",
data: fd,
processData: false, // tell jQuery not to process the data
contentType: false // tell jQuery not to set contentType
}).done(function( data ) {
console.log("PHP Output:");
console.log( data );
});
return false;
}
</script>
</head>
<body>
<form method="post" id="fileinfo" name="fileinfo" onsubmit="return submitForm();">
<label>Select a file:</label><br>
<input type="file" name="file" required />
<input type="submit" value="Upload" />
</form>
<div id="output"></div>
</body>
</html>
如果您使用php,这里有一种处理上传的方法,包括使用上面html中演示的两个自定义字段。
上传.php
<?php
if ($_POST["label"]) {
$label = $_POST["label"];
}
$allowedExts = array("gif", "jpeg", "jpg", "png");
$temp = explode(".", $_FILES["file"]["name"]);
$extension = end($temp);
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/jpg")
|| ($_FILES["file"]["type"] == "image/pjpeg")
|| ($_FILES["file"]["type"] == "image/x-png")
|| ($_FILES["file"]["type"] == "image/png"))
&& ($_FILES["file"]["size"] < 200000)
&& in_array($extension, $allowedExts)) {
if ($_FILES["file"]["error"] > 0) {
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
} else {
$filename = $label.$_FILES["file"]["name"];
echo "Upload: " . $_FILES["file"]["name"] . "<br>";
echo "Type: " . $_FILES["file"]["type"] . "<br>";
echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br>";
if (file_exists("uploads/" . $filename)) {
echo $filename . " already exists. ";
} else {
move_uploaded_file($_FILES["file"]["tmp_name"],
"uploads/" . $filename);
echo "Stored in: " . "uploads/" . $filename;
}
}
} else {
echo "Invalid file";
}
?>
<html>
<head>
<title>Ajax file upload</title>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script>
$(document).ready(function (e) {
$("#uploadimage").on('submit', (function(e) {
e.preventDefault();
$.ajax({
url: "upload.php", // Url to which the request is send
type: "POST", // Type of request to be send, called as method
data: new FormData(this), // Data sent to server, a set of key/value pairs (i.e. form fields and values)
contentType: false, // The content type used when sending data to the server.
cache: false, // To unable request pages to be cached
processData:false, // To send DOMDocument or non processed data file it is set to false
success: function(data) // A function to be called if request succeeds
{
alert(data);
}
});
}));
</script>
</head>
<body>
<div class="main">
<h1>Ajax Image Upload</h1><br/>
<hr>
<form id="uploadimage" action="" method="post" enctype="multipart/form-data">
<div id="image_preview"><img id="previewing" src="noimage.png" /></div>
<hr id="line">
<div id="selectImage">
<label>Select Your Image</label><br/>
<input type="file" name="file" id="file" required />
<input type="submit" value="Upload" class="submit" />
</div>
</form>
</div>
</body>
</html>
您可以使用ajaxSubmit方法,如下所示:)当您选择需要上传到服务器的文件时,表单将提交到服务器:)
$(document).ready(function () {
var options = {
target: '#output', // target element(s) to be updated with server response
timeout: 30000,
error: function (jqXHR, textStatus) {
$('#output').html('have any error');
return false;
}
},
success: afterSuccess, // post-submit callback
resetForm: true
// reset the form after successful submit
};
$('#idOfInputFile').on('change', function () {
$('#idOfForm').ajaxSubmit(options);
// always return false to prevent standard browser submit and page navigation
return false;
});
});
我用简单的代码处理了这些问题。您可以从这里下载工作演示
就你的情况而言,这些都是可能的。我将逐步介绍如何使用AJAX jquery将文件上传到服务器。
首先让我们创建一个HTML文件,添加如下表单文件元素,如下所示。
<form action="" id="formContent" method="post" enctype="multipart/form-data" >
<input type="file" name="file" required id="upload">
<button class="submitI" >Upload Image</button>
</form>
其次,创建一个jquery.js文件,并添加以下代码来处理向服务器提交的文件
$("#formContent").submit(function(e){
e.preventDefault();
var formdata = new FormData(this);
$.ajax({
url: "ajax_upload_image.php",
type: "POST",
data: formdata,
mimeTypes:"multipart/form-data",
contentType: false,
cache: false,
processData: false,
success: function(){
alert("file successfully submitted");
},error: function(){
alert("okey");
}
});
});
});
你完成了。查看更多信息