Trying to upload image using jqueryIs there an “exists” function for jQuery?How can I upload files asynchronously?Add table row in jQueryHow do I check if an element is hidden in jQuery?Setting “checked” for a checkbox with jQuery?How to check whether a checkbox is checked in jQuery?Disable/enable an input with jQuery?Preview an image before it is uploadedjQuery scroll to element“Thinking in AngularJS” if I have a jQuery background?
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Trying to upload image using jquery
Is there an “exists” function for jQuery?How can I upload files asynchronously?Add table row in jQueryHow do I check if an element is hidden in jQuery?Setting “checked” for a checkbox with jQuery?How to check whether a checkbox is checked in jQuery?Disable/enable an input with jQuery?Preview an image before it is uploadedjQuery scroll to element“Thinking in AngularJS” if I have a jQuery background?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
0
can anybody help me? I' m trying to upload an image using jquery and laravel but don't know how to pass the data
Ajax Function
var form_data = new FormData();
form_data.append('file', file_data);
form_data.append('link', link);
form_data.append('comment', comment);
form_data.append('reportid', reportid);
$.ajax(
url:" route('report.update') ",
type:"GET",
data:form_data,
contentType:false,
cache:false,
processData:false,
success:function(data)
alert(data);
);
Route
Route::get('report-update', ['uses'=>'Executive@reportUpdate', 'as'=>'report.update']);
Controller:
public function reportUpdate(Request $request)
$file = $request->file('file');
echo $file->getClientOriginalName();
jquery laravel
edited Mar 22 at 10:30
bhavinjr
981213
asked Mar 22 at 10:17
Mr. XMr. X
12
add a comment |
0
can anybody help me? I' m trying to upload an image using jquery and laravel but don't know how to pass the data
Ajax Function
var form_data = new FormData();
form_data.append('file', file_data);
form_data.append('link', link);
form_data.append('comment', comment);
form_data.append('reportid', reportid);
$.ajax(
url:" route('report.update') ",
type:"GET",
data:form_data,
contentType:false,
cache:false,
processData:false,
success:function(data)
alert(data);
);
Route
Route::get('report-update', ['uses'=>'Executive@reportUpdate', 'as'=>'report.update']);
Controller:
public function reportUpdate(Request $request)
$file = $request->file('file');
echo $file->getClientOriginalName();
jquery laravel
edited Mar 22 at 10:30
bhavinjr
981213
asked Mar 22 at 10:17
Mr. XMr. X
12
1
Change typeGETtoPOST, and change your Route too
– Saromase
Mar 22 at 10:19
1
Firstly,your route is wrong, it should be POST. And you cannot directly send a file via AJAX like this. What you can do is upload the file using some other methods and pass the name or path of file in your form data.. You can use libraries like blueimp (github.com/blueimp/jQuery-File-Upload )to upload the file , and then pass the return file data in AJAX
– Sauav
Mar 22 at 10:30
add a comment |
0
0
0
can anybody help me? I' m trying to upload an image using jquery and laravel but don't know how to pass the data
Ajax Function
var form_data = new FormData();
form_data.append('file', file_data);
form_data.append('link', link);
form_data.append('comment', comment);
form_data.append('reportid', reportid);
$.ajax(
url:" route('report.update') ",
type:"GET",
data:form_data,
contentType:false,
cache:false,
processData:false,
success:function(data)
alert(data);
);
Route
Route::get('report-update', ['uses'=>'Executive@reportUpdate', 'as'=>'report.update']);
Controller:
public function reportUpdate(Request $request)
$file = $request->file('file');
echo $file->getClientOriginalName();
jquery laravel
edited Mar 22 at 10:30
bhavinjr
981213
asked Mar 22 at 10:17
Mr. XMr. X
12
can anybody help me? I' m trying to upload an image using jquery and laravel but don't know how to pass the data
Ajax Function
var form_data = new FormData();
form_data.append('file', file_data);
form_data.append('link', link);
form_data.append('comment', comment);
form_data.append('reportid', reportid);
$.ajax(
url:" route('report.update') ",
type:"GET",
data:form_data,
contentType:false,
cache:false,
processData:false,
success:function(data)
alert(data);
);
Route
Route::get('report-update', ['uses'=>'Executive@reportUpdate', 'as'=>'report.update']);
Controller:
public function reportUpdate(Request $request)
$file = $request->file('file');
echo $file->getClientOriginalName();
jquery laravel
jquery laravel
edited Mar 22 at 10:30
bhavinjr
981213
asked Mar 22 at 10:17
Mr. XMr. X
12
edited Mar 22 at 10:30
bhavinjr
981213
asked Mar 22 at 10:17
Mr. XMr. X
12
edited Mar 22 at 10:30
bhavinjr
981213
edited Mar 22 at 10:30
bhavinjr
981213
edited Mar 22 at 10:30
bhavinjr
981213
981213
asked Mar 22 at 10:17
Mr. XMr. X
12
asked Mar 22 at 10:17
Mr. XMr. X
12
asked Mar 22 at 10:17
Mr. XMr. X
12
12
1
Change typeGETtoPOST, and change your Route too
– Saromase
Mar 22 at 10:19
1
Firstly,your route is wrong, it should be POST. And you cannot directly send a file via AJAX like this. What you can do is upload the file using some other methods and pass the name or path of file in your form data.. You can use libraries like blueimp (github.com/blueimp/jQuery-File-Upload )to upload the file , and then pass the return file data in AJAX
– Sauav
Mar 22 at 10:30
add a comment |
1
Change typeGETtoPOST, and change your Route too
– Saromase
Mar 22 at 10:19
1
Firstly,your route is wrong, it should be POST. And you cannot directly send a file via AJAX like this. What you can do is upload the file using some other methods and pass the name or path of file in your form data.. You can use libraries like blueimp (github.com/blueimp/jQuery-File-Upload )to upload the file , and then pass the return file data in AJAX
– Sauav
Mar 22 at 10:30
1
1
Change type
GET to POST, and change your Route too– Saromase
Mar 22 at 10:19
Change type
GET to POST, and change your Route too– Saromase
Mar 22 at 10:19
1
1
Firstly,your route is wrong, it should be POST. And you cannot directly send a file via AJAX like this. What you can do is upload the file using some other methods and pass the name or path of file in your form data.. You can use libraries like blueimp (github.com/blueimp/jQuery-File-Upload )to upload the file , and then pass the return file data in AJAX
– Sauav
Mar 22 at 10:30
Firstly,your route is wrong, it should be POST. And you cannot directly send a file via AJAX like this. What you can do is upload the file using some other methods and pass the name or path of file in your form data.. You can use libraries like blueimp (github.com/blueimp/jQuery-File-Upload )to upload the file , and then pass the return file data in AJAX
– Sauav
Mar 22 at 10:30
add a comment |
3 Answers
3
active
oldest
votes
1
Ajax
$("#FormId").on('submit', (function(e)
e.preventDefault();
$.ajax(
url: " route('report.update') ",
type: "POST",
data: new FormData(this),
contentType: false,
cache: false,
processData: false,
success: function(data)
)
));
Route
Route::POST('report-update', ['uses'=>'Executive@reportUpdate', 'as'=>'report.update']);
Function
public function reportUpdate(Request $request)
$file = $request->file('file');
$fileName = time().$request->file('file')->getClientOriginalName();
$path=$file->move('FolderPath',$fileName);
answered Mar 23 at 4:32
JIJOMON K.AJIJOMON K.A
1,2481724
add a comment |
0
add enctype="multipart/form-data" in your form
enctype="multipart/form-data"
$.ajax(
url:" route('report.update') ",
type:"POST",
data:form_data,
contentType:false,
processData:false,
success:function(data)
alert(data);
);
Route::POST('report-update', ['uses'=>'Executive@reportUpdate', 'as'=>'report.update']);
answered Mar 22 at 10:24
Akash Kumar VermaAkash Kumar Verma
830414
add a comment |
0
just try this
var formData = new FormData($("#formid")[0]);
$.ajax(
url: " route('report.update') ",
type: 'POST',
dataType: 'JSON',
data: formData,
processData: false,
contentType: false,
enctype: 'multipart/form-data',
mimeType: 'multipart/form-data',
cache: false,
success: function(data, textStatus, jqXHR)
console.log(data)
);
and use form method is POST
answered Mar 23 at 4:21
Paras RaiyaniParas Raiyani
793
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Ajax
$("#FormId").on('submit', (function(e)
e.preventDefault();
$.ajax(
url: " route('report.update') ",
type: "POST",
data: new FormData(this),
contentType: false,
cache: false,
processData: false,
success: function(data)
)
));
Route
Route::POST('report-update', ['uses'=>'Executive@reportUpdate', 'as'=>'report.update']);
Function
public function reportUpdate(Request $request)
$file = $request->file('file');
$fileName = time().$request->file('file')->getClientOriginalName();
$path=$file->move('FolderPath',$fileName);
answered Mar 23 at 4:32
JIJOMON K.AJIJOMON K.A
1,2481724
add a comment |
1
Ajax
$("#FormId").on('submit', (function(e)
e.preventDefault();
$.ajax(
url: " route('report.update') ",
type: "POST",
data: new FormData(this),
contentType: false,
cache: false,
processData: false,
success: function(data)
)
));
Route
Route::POST('report-update', ['uses'=>'Executive@reportUpdate', 'as'=>'report.update']);
Function
public function reportUpdate(Request $request)
$file = $request->file('file');
$fileName = time().$request->file('file')->getClientOriginalName();
$path=$file->move('FolderPath',$fileName);
answered Mar 23 at 4:32
JIJOMON K.AJIJOMON K.A
1,2481724
add a comment |
1
1
1
Ajax
$("#FormId").on('submit', (function(e)
e.preventDefault();
$.ajax(
url: " route('report.update') ",
type: "POST",
data: new FormData(this),
contentType: false,
cache: false,
processData: false,
success: function(data)
)
));
Route
Route::POST('report-update', ['uses'=>'Executive@reportUpdate', 'as'=>'report.update']);
Function
public function reportUpdate(Request $request)
$file = $request->file('file');
$fileName = time().$request->file('file')->getClientOriginalName();
$path=$file->move('FolderPath',$fileName);
answered Mar 23 at 4:32
JIJOMON K.AJIJOMON K.A
1,2481724
Ajax
$("#FormId").on('submit', (function(e)
e.preventDefault();
$.ajax(
url: " route('report.update') ",
type: "POST",
data: new FormData(this),
contentType: false,
cache: false,
processData: false,
success: function(data)
)
));
Route
Route::POST('report-update', ['uses'=>'Executive@reportUpdate', 'as'=>'report.update']);
Function
public function reportUpdate(Request $request)
$file = $request->file('file');
$fileName = time().$request->file('file')->getClientOriginalName();
$path=$file->move('FolderPath',$fileName);
answered Mar 23 at 4:32
JIJOMON K.AJIJOMON K.A
1,2481724
answered Mar 23 at 4:32
JIJOMON K.AJIJOMON K.A
1,2481724
answered Mar 23 at 4:32
JIJOMON K.AJIJOMON K.A
1,2481724
answered Mar 23 at 4:32
JIJOMON K.AJIJOMON K.A
1,2481724
1,2481724
add a comment |
add a comment |
0
add enctype="multipart/form-data" in your form
enctype="multipart/form-data"
$.ajax(
url:" route('report.update') ",
type:"POST",
data:form_data,
contentType:false,
processData:false,
success:function(data)
alert(data);
);
Route::POST('report-update', ['uses'=>'Executive@reportUpdate', 'as'=>'report.update']);
answered Mar 22 at 10:24
Akash Kumar VermaAkash Kumar Verma
830414
add a comment |
0
add enctype="multipart/form-data" in your form
enctype="multipart/form-data"
$.ajax(
url:" route('report.update') ",
type:"POST",
data:form_data,
contentType:false,
processData:false,
success:function(data)
alert(data);
);
Route::POST('report-update', ['uses'=>'Executive@reportUpdate', 'as'=>'report.update']);
answered Mar 22 at 10:24
Akash Kumar VermaAkash Kumar Verma
830414
add a comment |
0
0
0
add enctype="multipart/form-data" in your form
enctype="multipart/form-data"
$.ajax(
url:" route('report.update') ",
type:"POST",
data:form_data,
contentType:false,
processData:false,
success:function(data)
alert(data);
);
Route::POST('report-update', ['uses'=>'Executive@reportUpdate', 'as'=>'report.update']);
answered Mar 22 at 10:24
Akash Kumar VermaAkash Kumar Verma
830414
add enctype="multipart/form-data" in your form
enctype="multipart/form-data"
$.ajax(
url:" route('report.update') ",
type:"POST",
data:form_data,
contentType:false,
processData:false,
success:function(data)
alert(data);
);
Route::POST('report-update', ['uses'=>'Executive@reportUpdate', 'as'=>'report.update']);
answered Mar 22 at 10:24
Akash Kumar VermaAkash Kumar Verma
830414
answered Mar 22 at 10:24
Akash Kumar VermaAkash Kumar Verma
830414
answered Mar 22 at 10:24
Akash Kumar VermaAkash Kumar Verma
830414
answered Mar 22 at 10:24
Akash Kumar VermaAkash Kumar Verma
830414
830414
add a comment |
add a comment |
0
just try this
var formData = new FormData($("#formid")[0]);
$.ajax(
url: " route('report.update') ",
type: 'POST',
dataType: 'JSON',
data: formData,
processData: false,
contentType: false,
enctype: 'multipart/form-data',
mimeType: 'multipart/form-data',
cache: false,
success: function(data, textStatus, jqXHR)
console.log(data)
);
and use form method is POST
answered Mar 23 at 4:21
Paras RaiyaniParas Raiyani
793
add a comment |
0
just try this
var formData = new FormData($("#formid")[0]);
$.ajax(
url: " route('report.update') ",
type: 'POST',
dataType: 'JSON',
data: formData,
processData: false,
contentType: false,
enctype: 'multipart/form-data',
mimeType: 'multipart/form-data',
cache: false,
success: function(data, textStatus, jqXHR)
console.log(data)
);
and use form method is POST
answered Mar 23 at 4:21
Paras RaiyaniParas Raiyani
793
add a comment |
0
0
0
just try this
var formData = new FormData($("#formid")[0]);
$.ajax(
url: " route('report.update') ",
type: 'POST',
dataType: 'JSON',
data: formData,
processData: false,
contentType: false,
enctype: 'multipart/form-data',
mimeType: 'multipart/form-data',
cache: false,
success: function(data, textStatus, jqXHR)
console.log(data)
);
and use form method is POST
answered Mar 23 at 4:21
Paras RaiyaniParas Raiyani
793
just try this
var formData = new FormData($("#formid")[0]);
$.ajax(
url: " route('report.update') ",
type: 'POST',
dataType: 'JSON',
data: formData,
processData: false,
contentType: false,
enctype: 'multipart/form-data',
mimeType: 'multipart/form-data',
cache: false,
success: function(data, textStatus, jqXHR)
console.log(data)
);
and use form method is POST
answered Mar 23 at 4:21
Paras RaiyaniParas Raiyani
793
answered Mar 23 at 4:21
Paras RaiyaniParas Raiyani
793
answered Mar 23 at 4:21
Paras RaiyaniParas Raiyani
793
answered Mar 23 at 4:21
Paras RaiyaniParas Raiyani
793
793
add a comment |
add a comment |
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1
Change type
GETtoPOST, and change your Route too– Saromase
Mar 22 at 10:19
1
Firstly,your route is wrong, it should be POST. And you cannot directly send a file via AJAX like this. What you can do is upload the file using some other methods and pass the name or path of file in your form data.. You can use libraries like blueimp (github.com/blueimp/jQuery-File-Upload )to upload the file , and then pass the return file data in AJAX
– Sauav
Mar 22 at 10:30