I successfully created an upload field in create mode in jQuery jTable as follows:
upload: {
title: 'Upload Image',
input: function (data) {
return '<input type="file" name="file">';
'<input type="submit" value="Submit" id="submitBtn" />';
},
},
I am able to successfully display the browse button and select files. I am unable to submit the form:
Error: newcourse.php' not found or unable to stat.
with the same file name in which the code is.
I am at a dead end. Where will the file be uploaded to? In the same directory?
How to do it in jTable ? Can it be done using ajax ? If so, how to proceed? jTable documentation is very poor.
I finally found a soluton to upload files on jTable
Now with the version of jtable 2.4.0 (the most recent at the moment writing this post)
on your file.js add the following methods:
// Read a page's GET URL variables and return them as an associative array.
function getVars(url)
{
var formData = new FormData();
var split;
$.each(url.split("&"), function(key, value) {
split = value.split("=");
formData.append(split[0], decodeURIComponent(split[1].replace(/\+/g, " ")));
});
return formData;
}
// Variable to store your files
var files;
$( document ).delegate('#input-image','change', prepareUpload);
// Grab the files and set them to our variable
function prepareUpload(event)
{
files = event.target.files;
}
//The actions for your table:
$('#table-container').jtable({
actions: {
listAction: 'backoffice/catalogs/display',
deleteAction: 'backoffice/catalogs/delete',
createAction: function (postData) {
var formData = getVars(postData);
if($('#input-image').val() !== ""){
formData.append("userfile", $('#input-image').get(0).files[0]);
}
return $.Deferred(function ($dfd) {
$.ajax({
url: 'backoffice/catalogs/update',
type: 'POST',
dataType: 'json',
data: formData,
processData: false, // Don't process the files
contentType: false, // Set content type to false as jQuery will tell the server its a query string request
success: function (data) {
$dfd.resolve(data);
$('#table-container').jtable('load');
},
error: function () {
$dfd.reject();
}
});
});
},
updateAction: function (postData) {
var formData = getVars(postData);
if($('#input-image').val() !== ""){
formData.append("userfile", $('#input-image').get(0).files[0]);
}
return $.Deferred(function ($dfd) {
$.ajax({
url: 'backoffice/catalogs/update',
type: 'POST',
dataType: 'json',
data: formData,
processData: false, // Don't process the files
contentType: false, // Set content type to false as jQuery will tell the server its a query string request
success: function (data) {
$dfd.resolve(data);
$('#table-container').jtable('load');
},
error: function () {
$dfd.reject();
}
});
});
},
// Now for the fields:
fields: {
id_catalog: {
key: true,
create: false,
edit: false,
list: false
},
thumb_url: {
title: 'Image',
type: 'file',
create: false,
edit: true,
list: true,
display: function(data){
return '<img src="' + data.record.thumb_url + '" width="150" height="207" class="preview">';
},
input: function(data){
return '<img src="' + data.record.thumb_url + '" width="150" height="207" class="preview">';
},
width: "150",
listClass: "class-row-center"
},
image: {
title: 'Select File',
list: false,
create: true,
input: function(data) {
html = '<input type ="file" id="input-image" name="userfile" accept="image/*" />';
return html;
}
}
}
});
Now, you are able to process the files on your server side.
That's all folks.
Hello shels
I lost a lot of time to searching of a solution of this issue. I read a many articles and found that this solution working for me fine:
actions: {
listAction: 'modules_data.php?action=list',
deleteAction: 'modules_data.php?action=delete',
updateAction: 'modules_data.php?action=update',
createAction: 'modules_data.php?action=create'
},
...
image_file: {
title: 'Album cover',
list: true,
create: true,
edit: true,
input: function(data) {
return '<form target="iframeTarget" class="formUploadFile" action="file_upload.php" method="post" enctype="multipart/form-data"> <input type="file" onchange="this.form.submit()" name="myFile"/> </form> <iframe class="upload-iframe" style="display: none;" src="#" name="iframeTarget"></iframe>';
}
},
image: {
title: 'Album cover',
list: true,
create: true,
edit: true,
input: function(data) {
html = '<input type ="hidden" id="image" name="image" />';
return html;
}
},
....
How to capture submit response in a form created dynamically?
So you can capture submit event using:
...
formSubmitting: function(event, data) {
filename = $('input[type=file]').val().split('\\').pop();
($("#" + data.form.attr("id")).find('input[name="image"]').val(filename));
},
And save data to the server side script.
I think, it's a good solution and hope will be helpful.
Best Regards
Kameliya
console.log not working for FormData. Use instead
for (var data of formData) {
console.log(data);
}
