javascript – 我可以使用jquery ajax和php上传文件,需要一些解释和修改
作者:互联网
我读过这个.
https://developer.mozilla.org/en/docs/Web/API/FormData
仍然无法理解为什么formdata(frm)在代码中是空的.的console.log(FRM);返回“FormData {}”为空.为什么?
file_form.php =>
<title>Upload File</title>
<div id="targetLayer">No Image</div>
<img id="my_image1" src="" />
<img id="my_image2" src="" />
<form id="uploadForm1" enctype="multipart/form-data">
<input name="image1" type="file" />
<input type="submit" value="Submit" />
</form>
<form id="uploadForm2" enctype="multipart/form-data">
<input name="image2" type="file" />
<input type="submit" value="Submit" />
</form>
<script type="text/javascript" src="jquery-1.11.3.js"></script>
<script type="text/javascript">
$(document).ready(function (e) {
$("#uploadForm1").on('submit', (function (e) {
frm = new FormData($(this)[0]);
console.log(frm);
e.preventDefault();
$.ajax({
url: "upload_script1.php",
type: "POST",
data: frm,
contentType: false,
cache: false,
processData: false,
success: function (response) {
var JsonObject = JSON.parse(response);
$("#targetLayer").html(response);
var fl1 = "uploads/" + JsonObject.image1;
$("#my_image1").attr("src", fl1).height(100).width(100);
}
});
}));
$("#uploadForm2").on('submit', (function (e) {
frm = new FormData($(this)[0]);
e.preventDefault();
$.ajax({
url: "upload_script2.php",
type: "POST",
data: frm,
contentType: false,
cache: false,
processData: false,
success: function (response) {
var JsonObject = JSON.parse(response);
$("#targetLayer").html(response);
var fl2 = "uploads/" + JsonObject.image2;
$("#my_image2").attr("src", fl2).height(100).width(100);
}
});
}));
});
</script>
upload_script1.php =>
<?php
$uploaddir = '/var/www/html/file-upload/uploads/';
$uploadfile = $uploaddir . basename($_FILES['image1']['name']);
move_uploaded_file($_FILES['image1']['tmp_name'], $uploadfile);
$image1 = $_FILES['image1']['name'];
$images["image1"] = $image1;
echo json_encode($images);
?>
upload_script2.php =>
<?php
$uploaddir = '/var/www/html/file-upload/uploads/';
$uploadfile = $uploaddir . basename($_FILES['image2']['name']);
move_uploaded_file($_FILES['image2']['tmp_name'], $uploadfile);
$image2 = $_FILES['image2']['name'];
$images["image2"] = $image2;
echo json_encode($images);
?>
>任何人都可以将两个$.ajax()代码重写为一个$.ajax()代码,使代码更短吗?如有必要,重写html和php代码.
任何帮助将受到高度赞赏.
提前致谢.
解决方法:
我只修改了javascript部分.
我的例子是……
<title>Upload File</title>
<div id="targetLayer">No Image</div>
<img id="my_image1" src=""/>
<img id="my_image2" src=""/>
<form id="uploadForm1" enctype="multipart/form-data">
<input name="image1" type="file" />
<input type="submit" value="Submit" />
</form>
<form id="uploadForm2" enctype="multipart/form-data">
<input name="image2" type="file" />
<input type="submit" value="Submit" />
</form>
<script type="text/javascript" src="jquery-1.11.3.js"></script>
<script type="text/javascript">
function fetch( type ) {
var typeMap = {
form1: { formId: 'uploadForm1', url: 'upload_script1.php', flKey: 'image1', imgId: 'my_image1' },
form2: { formId: 'uploadForm2', url: 'upload_script2.php', flKey: 'image2', imgId: 'my_image2' }
};
var info = typeMap[ type ];
if( ! info ) return console.error('no type: ' + type );
$("#"+info.formId).on('submit',(function(e) {
frm = new FormData($(this)[0]);
console.log(frm);
e.preventDefault();
$.ajax({
url: info.url,
type: "POST",
data: frm,
contentType: false,
cache: false,
processData:false,
success: function(response){
var JsonObject = JSON.parse(response);
$("#targetLayer").html(response);
var fl = "uploads/" + JsonObject[ info.flKey ];
$("#"+info.imgId).attr("src",fl).height(100).width(100);
}
});
}));
}
$(document).ready(function (e) {
fetch( 'form1' );
fetch( 'form2' );
});
</script>
标签:javascript,php,jquery,ajax,form-data 来源: https://codeday.me/bug/20190727/1556767.html