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Ajax with JSON GET request undefined
Detecting an undefined object propertyHow do I format a Microsoft JSON date?Can comments be used in JSON?How can I pretty-print JSON in a shell script?Abort Ajax requests using jQueryWhat is the correct JSON content type?Why does Google prepend while(1); to their JSON responses?How to check for “undefined” in JavaScript?How can I pretty-print JSON using JavaScript?How to send FormData objects with Ajax-requests in jQuery?
I am trying to get the front end to display the contents of my dynamo db table although it displays the information as undefined.
$(document).ready(function()
$.ajax(
type: 'GET',
url:someURL,
success: function(data)
$('#incidentid').html('');
data.forEach(function(IncidentNotesItem)
$('#incidentid').append ('<p>'+ IncidentNotesItem.incidentid + '</p>');
)
);
);
javascript html json ajax amazon-dynamodb
asked Mar 8 at 20:37
ZootopiaZootopia
105
|
show 2 more comments
I am trying to get the front end to display the contents of my dynamo db table although it displays the information as undefined.
$(document).ready(function()
$.ajax(
type: 'GET',
url:someURL,
success: function(data)
$('#incidentid').html('');
data.forEach(function(IncidentNotesItem)
$('#incidentid').append ('<p>'+ IncidentNotesItem.incidentid + '</p>');
)
);
);
javascript html json ajax amazon-dynamodb
asked Mar 8 at 20:37
ZootopiaZootopia
105
Are you sure the function is working? Add anerror: function(response) console.log(response.responsetext);function to your AJAX so we can see what the error is if one exists.
– torogadude
Mar 8 at 20:41
Is the AJAX request to the same domain?
– circusdei
Mar 8 at 20:45
I added that line and refreshed my webpage but the console is giving me a status 200 and yes, the AJAX request is to the same domain
– Zootopia
Mar 8 at 20:59
But I still see undefined: undefined undefined undefined undefined undefined
– Zootopia
Mar 8 at 21:02
Are you getting the response you are expecting? What's insidedata?
– EternalHour
Mar 8 at 21:05
|
show 2 more comments
I am trying to get the front end to display the contents of my dynamo db table although it displays the information as undefined.
$(document).ready(function()
$.ajax(
type: 'GET',
url:someURL,
success: function(data)
$('#incidentid').html('');
data.forEach(function(IncidentNotesItem)
$('#incidentid').append ('<p>'+ IncidentNotesItem.incidentid + '</p>');
)
);
);
javascript html json ajax amazon-dynamodb
asked Mar 8 at 20:37
ZootopiaZootopia
105
I am trying to get the front end to display the contents of my dynamo db table although it displays the information as undefined.
$(document).ready(function()
$.ajax(
type: 'GET',
url:someURL,
success: function(data)
$('#incidentid').html('');
data.forEach(function(IncidentNotesItem)
$('#incidentid').append ('<p>'+ IncidentNotesItem.incidentid + '</p>');
)
);
);
javascript html json ajax amazon-dynamodb
javascript html json ajax amazon-dynamodb
asked Mar 8 at 20:37
ZootopiaZootopia
105
asked Mar 8 at 20:37
ZootopiaZootopia
105
asked Mar 8 at 20:37
ZootopiaZootopia
105
asked Mar 8 at 20:37
ZootopiaZootopia
105
asked Mar 8 at 20:37
ZootopiaZootopia
105
105
Are you sure the function is working? Add anerror: function(response) console.log(response.responsetext);function to your AJAX so we can see what the error is if one exists.
– torogadude
Mar 8 at 20:41
Is the AJAX request to the same domain?
– circusdei
Mar 8 at 20:45
I added that line and refreshed my webpage but the console is giving me a status 200 and yes, the AJAX request is to the same domain
– Zootopia
Mar 8 at 20:59
But I still see undefined: undefined undefined undefined undefined undefined
– Zootopia
Mar 8 at 21:02
Are you getting the response you are expecting? What's insidedata?
– EternalHour
Mar 8 at 21:05
|
show 2 more comments
Are you sure the function is working? Add anerror: function(response) console.log(response.responsetext);function to your AJAX so we can see what the error is if one exists.
– torogadude
Mar 8 at 20:41
Is the AJAX request to the same domain?
– circusdei
Mar 8 at 20:45
I added that line and refreshed my webpage but the console is giving me a status 200 and yes, the AJAX request is to the same domain
– Zootopia
Mar 8 at 20:59
But I still see undefined: undefined undefined undefined undefined undefined
– Zootopia
Mar 8 at 21:02
Are you getting the response you are expecting? What's insidedata?
– EternalHour
Mar 8 at 21:05
Are you sure the function is working? Add an
error: function(response) console.log(response.responsetext); function to your AJAX so we can see what the error is if one exists.– torogadude
Mar 8 at 20:41
Are you sure the function is working? Add an
error: function(response) console.log(response.responsetext); function to your AJAX so we can see what the error is if one exists.– torogadude
Mar 8 at 20:41
Is the AJAX request to the same domain?
– circusdei
Mar 8 at 20:45
Is the AJAX request to the same domain?
– circusdei
Mar 8 at 20:45
I added that line and refreshed my webpage but the console is giving me a status 200 and yes, the AJAX request is to the same domain
– Zootopia
Mar 8 at 20:59
I added that line and refreshed my webpage but the console is giving me a status 200 and yes, the AJAX request is to the same domain
– Zootopia
Mar 8 at 20:59
But I still see undefined: undefined undefined undefined undefined undefined
– Zootopia
Mar 8 at 21:02
But I still see undefined: undefined undefined undefined undefined undefined
– Zootopia
Mar 8 at 21:02
Are you getting the response you are expecting? What's inside
data?– EternalHour
Mar 8 at 21:05
Are you getting the response you are expecting? What's inside
data?– EternalHour
Mar 8 at 21:05
|
show 2 more comments
1 Answer
1
active
oldest
votes
I solved it! The issue was that I wasn't calling the column name properly. I did console.log(IncidentNotesItem) to see what is stored inside the IncidentNotesItem Function and saw on the console that it is called id instead of incidentid.
$(document).ready(function()
$.ajax(
type: 'GET',
url:someURL,
success: function(data)
$('#incidentid').html('');
data.forEach(function(IncidentNotesItem)
$('#incidentid').append ('<p>'+ IncidentNotesItem.id + '</p>');
)
);
);
answered Mar 12 at 19:01
ZootopiaZootopia
105
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I solved it! The issue was that I wasn't calling the column name properly. I did console.log(IncidentNotesItem) to see what is stored inside the IncidentNotesItem Function and saw on the console that it is called id instead of incidentid.
$(document).ready(function()
$.ajax(
type: 'GET',
url:someURL,
success: function(data)
$('#incidentid').html('');
data.forEach(function(IncidentNotesItem)
$('#incidentid').append ('<p>'+ IncidentNotesItem.id + '</p>');
)
);
);
answered Mar 12 at 19:01
ZootopiaZootopia
105
add a comment |
I solved it! The issue was that I wasn't calling the column name properly. I did console.log(IncidentNotesItem) to see what is stored inside the IncidentNotesItem Function and saw on the console that it is called id instead of incidentid.
$(document).ready(function()
$.ajax(
type: 'GET',
url:someURL,
success: function(data)
$('#incidentid').html('');
data.forEach(function(IncidentNotesItem)
$('#incidentid').append ('<p>'+ IncidentNotesItem.id + '</p>');
)
);
);
answered Mar 12 at 19:01
ZootopiaZootopia
105
add a comment |
I solved it! The issue was that I wasn't calling the column name properly. I did console.log(IncidentNotesItem) to see what is stored inside the IncidentNotesItem Function and saw on the console that it is called id instead of incidentid.
$(document).ready(function()
$.ajax(
type: 'GET',
url:someURL,
success: function(data)
$('#incidentid').html('');
data.forEach(function(IncidentNotesItem)
$('#incidentid').append ('<p>'+ IncidentNotesItem.id + '</p>');
)
);
);
answered Mar 12 at 19:01
ZootopiaZootopia
105
I solved it! The issue was that I wasn't calling the column name properly. I did console.log(IncidentNotesItem) to see what is stored inside the IncidentNotesItem Function and saw on the console that it is called id instead of incidentid.
$(document).ready(function()
$.ajax(
type: 'GET',
url:someURL,
success: function(data)
$('#incidentid').html('');
data.forEach(function(IncidentNotesItem)
$('#incidentid').append ('<p>'+ IncidentNotesItem.id + '</p>');
)
);
);
answered Mar 12 at 19:01
ZootopiaZootopia
105
answered Mar 12 at 19:01
ZootopiaZootopia
105
answered Mar 12 at 19:01
ZootopiaZootopia
105
answered Mar 12 at 19:01
ZootopiaZootopia
105
105
add a comment |
add a comment |
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Are you sure the function is working? Add an
error: function(response) console.log(response.responsetext);function to your AJAX so we can see what the error is if one exists.– torogadude
Mar 8 at 20:41
Is the AJAX request to the same domain?
– circusdei
Mar 8 at 20:45
I added that line and refreshed my webpage but the console is giving me a status 200 and yes, the AJAX request is to the same domain
– Zootopia
Mar 8 at 20:59
But I still see undefined: undefined undefined undefined undefined undefined
– Zootopia
Mar 8 at 21:02
Are you getting the response you are expecting? What's inside
data?– EternalHour
Mar 8 at 21:05