How to limit data in a single cell of DataTable
I have DataTable version 1.10.12 and I want to display only images in one cell of particular row. There are many images in database but I only want to show 2 images.
How can I achieve this?
How to set limit in forEach loop?
This is my code:
{
data: 'plays', name: 'plays', "defaultContent": "",
render: function (data, type, row, meta) {
var plays = '';
data.forEach(function (item) {
plays += '<img class="img" src="/images/theme/image_placeholder.jpg" alt="">'
})
return plays;
}
}
jquery datatable
edited Nov 23 '18 at 13:09
Pedro Gaspar
549321
asked Nov 23 '18 at 13:01
unknownunknown
207
add a comment |
I have DataTable version 1.10.12 and I want to display only images in one cell of particular row. There are many images in database but I only want to show 2 images.
How can I achieve this?
How to set limit in forEach loop?
This is my code:
{
data: 'plays', name: 'plays', "defaultContent": "",
render: function (data, type, row, meta) {
var plays = '';
data.forEach(function (item) {
plays += '<img class="img" src="/images/theme/image_placeholder.jpg" alt="">'
})
return plays;
}
}
jquery datatable
edited Nov 23 '18 at 13:09
Pedro Gaspar
549321
asked Nov 23 '18 at 13:01
unknownunknown
207
add a comment |
I have DataTable version 1.10.12 and I want to display only images in one cell of particular row. There are many images in database but I only want to show 2 images.
How can I achieve this?
How to set limit in forEach loop?
This is my code:
{
data: 'plays', name: 'plays', "defaultContent": "",
render: function (data, type, row, meta) {
var plays = '';
data.forEach(function (item) {
plays += '<img class="img" src="/images/theme/image_placeholder.jpg" alt="">'
})
return plays;
}
}
jquery datatable
edited Nov 23 '18 at 13:09
Pedro Gaspar
549321
asked Nov 23 '18 at 13:01
unknownunknown
207
I have DataTable version 1.10.12 and I want to display only images in one cell of particular row. There are many images in database but I only want to show 2 images.
How can I achieve this?
How to set limit in forEach loop?
This is my code:
{
data: 'plays', name: 'plays', "defaultContent": "",
render: function (data, type, row, meta) {
var plays = '';
data.forEach(function (item) {
plays += '<img class="img" src="/images/theme/image_placeholder.jpg" alt="">'
})
return plays;
}
}
jquery datatable
jquery datatable
edited Nov 23 '18 at 13:09
Pedro Gaspar
549321
asked Nov 23 '18 at 13:01
unknownunknown
207
edited Nov 23 '18 at 13:09
Pedro Gaspar
549321
asked Nov 23 '18 at 13:01
unknownunknown
207
edited Nov 23 '18 at 13:09
Pedro Gaspar
549321
edited Nov 23 '18 at 13:09
Pedro Gaspar
549321
edited Nov 23 '18 at 13:09
Pedro Gaspar
549321
549321
asked Nov 23 '18 at 13:01
unknownunknown
207
asked Nov 23 '18 at 13:01
unknownunknown
207
asked Nov 23 '18 at 13:01
unknownunknown
207
207
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The forEach method has an aditional parameter that gives you the iteration index:
arr.forEach(function callback(currentValue, index, array) {
// Your code
}[, thisArg]);
So you can simply check if index is greater than 1 to have only 2 iterations:
data.forEach(function (item, index) {
// Add this
if (index > 1) {
return;
}
plays += '<img class="img" src="/images/theme/image_placeholder.jpg" alt="">'
})
answered Nov 23 '18 at 13:26
Iker VázquezIker Vázquez
355215
Thank you very much for reply, solved my problem.
– unknown
Nov 24 '18 at 4:08
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
The forEach method has an aditional parameter that gives you the iteration index:
arr.forEach(function callback(currentValue, index, array) {
// Your code
}[, thisArg]);
So you can simply check if index is greater than 1 to have only 2 iterations:
data.forEach(function (item, index) {
// Add this
if (index > 1) {
return;
}
plays += '<img class="img" src="/images/theme/image_placeholder.jpg" alt="">'
})
answered Nov 23 '18 at 13:26
Iker VázquezIker Vázquez
355215
Thank you very much for reply, solved my problem.
– unknown
Nov 24 '18 at 4:08
add a comment |
The forEach method has an aditional parameter that gives you the iteration index:
arr.forEach(function callback(currentValue, index, array) {
// Your code
}[, thisArg]);
So you can simply check if index is greater than 1 to have only 2 iterations:
data.forEach(function (item, index) {
// Add this
if (index > 1) {
return;
}
plays += '<img class="img" src="/images/theme/image_placeholder.jpg" alt="">'
})
answered Nov 23 '18 at 13:26
Iker VázquezIker Vázquez
355215
Thank you very much for reply, solved my problem.
– unknown
Nov 24 '18 at 4:08
add a comment |
The forEach method has an aditional parameter that gives you the iteration index:
arr.forEach(function callback(currentValue, index, array) {
// Your code
}[, thisArg]);
So you can simply check if index is greater than 1 to have only 2 iterations:
data.forEach(function (item, index) {
// Add this
if (index > 1) {
return;
}
plays += '<img class="img" src="/images/theme/image_placeholder.jpg" alt="">'
})
answered Nov 23 '18 at 13:26
Iker VázquezIker Vázquez
355215
The forEach method has an aditional parameter that gives you the iteration index:
arr.forEach(function callback(currentValue, index, array) {
// Your code
}[, thisArg]);
So you can simply check if index is greater than 1 to have only 2 iterations:
data.forEach(function (item, index) {
// Add this
if (index > 1) {
return;
}
plays += '<img class="img" src="/images/theme/image_placeholder.jpg" alt="">'
})
answered Nov 23 '18 at 13:26
Iker VázquezIker Vázquez
355215
answered Nov 23 '18 at 13:26
Iker VázquezIker Vázquez
355215
answered Nov 23 '18 at 13:26
Iker VázquezIker Vázquez
355215
answered Nov 23 '18 at 13:26
Iker VázquezIker Vázquez
355215
355215
Thank you very much for reply, solved my problem.
– unknown
Nov 24 '18 at 4:08
add a comment |
Thank you very much for reply, solved my problem.
– unknown
Nov 24 '18 at 4:08
Thank you very much for reply, solved my problem.
– unknown
Nov 24 '18 at 4:08
Thank you very much for reply, solved my problem.
– unknown
Nov 24 '18 at 4:08
add a comment |
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