Want to fetch data from input and construct in JSON
0
I am trying to fetch the data which is stored in a table in a JSON format, but while it is using the .each to fetch data, then the latest value is getting repeated all the time.
I have shown the actual output where it should have different data from the table rows
function getPackTableDetails() {
var packDetailsRow = ;
var tableRowCount = $('#addPackTable tr').length;
var returnJSONFormat = "";
$('#addPackTable tr').each(function(index) {
if(index != 0) {
var packDetailsObj = {};
($(this).find("td")).each(function(idx) {
switch(idx) {
case 0:
packDetailsObj.table_pack_name = $('#table_pack_name').val();
break;
case 1:
packDetailsObj.table_pack_type = $('#table_pack_type').val();
break;
case 2:
packDetailsObj.base_amount = $('#base_amount').val();
break;
case 3:
packDetailsObj.tax_amount = $('#tax_amount').val();
break;
case 4:
packDetailsObj.total_amount = $('#total_amount').val();
break;
}
});
packDetailsRow.push(packDetailsObj);
}
});
// console.log(JSON.stringify(packDetailsRow));
return packDetailsRow;
}
As i am using a input type text where i use value to store my data and then make it retrieve to the following jquery code
Output:
{
"pack_details": [
{
"table_pack_name": "Demo Pack Name5",
"table_pack_type": "Demo Pack Name5",
"base_amount": "200.00",
"tax_amount": "0.00",
"total_amount": "200.00"
},
{
"table_pack_name": "Demo Pack Name5",
"table_pack_type": "Demo Pack Name5",
"base_amount": "200.00",
"tax_amount": "0.00",
"total_amount": "200.00"
},
{
"table_pack_name": "Demo Pack Name5",
"table_pack_type": "Demo Pack Name5",
"base_amount": "200.00",
"tax_amount": "0.00",
"total_amount": "200.00"
}
]
}
Actual Output:
{
"pack_details": [
{
"table_pack_name": "Demo Pack Name5",
"table_pack_type": "Demo Pack Name5",
"base_amount": "20.00",
"tax_amount": "10.00",
"total_amount": "30.00"
},
{
"table_pack_name": "Demo Pack Name6",
"table_pack_type": "Demo Pack Name6",
"base_amount": "200.00",
"tax_amount": "0.00",
"total_amount": "200.00"
},
{
"table_pack_name": "Demo Pack Name7",
"table_pack_type": "Demo Pack Name7",
"base_amount": "400.00",
"tax_amount": "40.00",
"total_amount": "440.00"
}
]
}
javascript jquery
asked Dec 26 at 13:34
Kiran Bhagannavar
135
add a comment |
0
I am trying to fetch the data which is stored in a table in a JSON format, but while it is using the .each to fetch data, then the latest value is getting repeated all the time.
I have shown the actual output where it should have different data from the table rows
function getPackTableDetails() {
var packDetailsRow = ;
var tableRowCount = $('#addPackTable tr').length;
var returnJSONFormat = "";
$('#addPackTable tr').each(function(index) {
if(index != 0) {
var packDetailsObj = {};
($(this).find("td")).each(function(idx) {
switch(idx) {
case 0:
packDetailsObj.table_pack_name = $('#table_pack_name').val();
break;
case 1:
packDetailsObj.table_pack_type = $('#table_pack_type').val();
break;
case 2:
packDetailsObj.base_amount = $('#base_amount').val();
break;
case 3:
packDetailsObj.tax_amount = $('#tax_amount').val();
break;
case 4:
packDetailsObj.total_amount = $('#total_amount').val();
break;
}
});
packDetailsRow.push(packDetailsObj);
}
});
// console.log(JSON.stringify(packDetailsRow));
return packDetailsRow;
}
As i am using a input type text where i use value to store my data and then make it retrieve to the following jquery code
Output:
{
"pack_details": [
{
"table_pack_name": "Demo Pack Name5",
"table_pack_type": "Demo Pack Name5",
"base_amount": "200.00",
"tax_amount": "0.00",
"total_amount": "200.00"
},
{
"table_pack_name": "Demo Pack Name5",
"table_pack_type": "Demo Pack Name5",
"base_amount": "200.00",
"tax_amount": "0.00",
"total_amount": "200.00"
},
{
"table_pack_name": "Demo Pack Name5",
"table_pack_type": "Demo Pack Name5",
"base_amount": "200.00",
"tax_amount": "0.00",
"total_amount": "200.00"
}
]
}
Actual Output:
{
"pack_details": [
{
"table_pack_name": "Demo Pack Name5",
"table_pack_type": "Demo Pack Name5",
"base_amount": "20.00",
"tax_amount": "10.00",
"total_amount": "30.00"
},
{
"table_pack_name": "Demo Pack Name6",
"table_pack_type": "Demo Pack Name6",
"base_amount": "200.00",
"tax_amount": "0.00",
"total_amount": "200.00"
},
{
"table_pack_name": "Demo Pack Name7",
"table_pack_type": "Demo Pack Name7",
"base_amount": "400.00",
"tax_amount": "40.00",
"total_amount": "440.00"
}
]
}
javascript jquery
asked Dec 26 at 13:34
Kiran Bhagannavar
135
gj for not providing the html, makes it that challenging
– WASD
Dec 26 at 13:41
If I understand your code right, code like$('#table_pack_name').val()will get the same value for every iteration.
– obfish
Dec 26 at 13:41
yes @obfish you are right. my iteration is getting repeated
– Kiran Bhagannavar
Dec 26 at 13:47
add a comment |
0
0
0
I am trying to fetch the data which is stored in a table in a JSON format, but while it is using the .each to fetch data, then the latest value is getting repeated all the time.
I have shown the actual output where it should have different data from the table rows
function getPackTableDetails() {
var packDetailsRow = ;
var tableRowCount = $('#addPackTable tr').length;
var returnJSONFormat = "";
$('#addPackTable tr').each(function(index) {
if(index != 0) {
var packDetailsObj = {};
($(this).find("td")).each(function(idx) {
switch(idx) {
case 0:
packDetailsObj.table_pack_name = $('#table_pack_name').val();
break;
case 1:
packDetailsObj.table_pack_type = $('#table_pack_type').val();
break;
case 2:
packDetailsObj.base_amount = $('#base_amount').val();
break;
case 3:
packDetailsObj.tax_amount = $('#tax_amount').val();
break;
case 4:
packDetailsObj.total_amount = $('#total_amount').val();
break;
}
});
packDetailsRow.push(packDetailsObj);
}
});
// console.log(JSON.stringify(packDetailsRow));
return packDetailsRow;
}
As i am using a input type text where i use value to store my data and then make it retrieve to the following jquery code
Output:
{
"pack_details": [
{
"table_pack_name": "Demo Pack Name5",
"table_pack_type": "Demo Pack Name5",
"base_amount": "200.00",
"tax_amount": "0.00",
"total_amount": "200.00"
},
{
"table_pack_name": "Demo Pack Name5",
"table_pack_type": "Demo Pack Name5",
"base_amount": "200.00",
"tax_amount": "0.00",
"total_amount": "200.00"
},
{
"table_pack_name": "Demo Pack Name5",
"table_pack_type": "Demo Pack Name5",
"base_amount": "200.00",
"tax_amount": "0.00",
"total_amount": "200.00"
}
]
}
Actual Output:
{
"pack_details": [
{
"table_pack_name": "Demo Pack Name5",
"table_pack_type": "Demo Pack Name5",
"base_amount": "20.00",
"tax_amount": "10.00",
"total_amount": "30.00"
},
{
"table_pack_name": "Demo Pack Name6",
"table_pack_type": "Demo Pack Name6",
"base_amount": "200.00",
"tax_amount": "0.00",
"total_amount": "200.00"
},
{
"table_pack_name": "Demo Pack Name7",
"table_pack_type": "Demo Pack Name7",
"base_amount": "400.00",
"tax_amount": "40.00",
"total_amount": "440.00"
}
]
}
javascript jquery
asked Dec 26 at 13:34
Kiran Bhagannavar
135
I am trying to fetch the data which is stored in a table in a JSON format, but while it is using the .each to fetch data, then the latest value is getting repeated all the time.
I have shown the actual output where it should have different data from the table rows
function getPackTableDetails() {
var packDetailsRow = ;
var tableRowCount = $('#addPackTable tr').length;
var returnJSONFormat = "";
$('#addPackTable tr').each(function(index) {
if(index != 0) {
var packDetailsObj = {};
($(this).find("td")).each(function(idx) {
switch(idx) {
case 0:
packDetailsObj.table_pack_name = $('#table_pack_name').val();
break;
case 1:
packDetailsObj.table_pack_type = $('#table_pack_type').val();
break;
case 2:
packDetailsObj.base_amount = $('#base_amount').val();
break;
case 3:
packDetailsObj.tax_amount = $('#tax_amount').val();
break;
case 4:
packDetailsObj.total_amount = $('#total_amount').val();
break;
}
});
packDetailsRow.push(packDetailsObj);
}
});
// console.log(JSON.stringify(packDetailsRow));
return packDetailsRow;
}
As i am using a input type text where i use value to store my data and then make it retrieve to the following jquery code
Output:
{
"pack_details": [
{
"table_pack_name": "Demo Pack Name5",
"table_pack_type": "Demo Pack Name5",
"base_amount": "200.00",
"tax_amount": "0.00",
"total_amount": "200.00"
},
{
"table_pack_name": "Demo Pack Name5",
"table_pack_type": "Demo Pack Name5",
"base_amount": "200.00",
"tax_amount": "0.00",
"total_amount": "200.00"
},
{
"table_pack_name": "Demo Pack Name5",
"table_pack_type": "Demo Pack Name5",
"base_amount": "200.00",
"tax_amount": "0.00",
"total_amount": "200.00"
}
]
}
Actual Output:
{
"pack_details": [
{
"table_pack_name": "Demo Pack Name5",
"table_pack_type": "Demo Pack Name5",
"base_amount": "20.00",
"tax_amount": "10.00",
"total_amount": "30.00"
},
{
"table_pack_name": "Demo Pack Name6",
"table_pack_type": "Demo Pack Name6",
"base_amount": "200.00",
"tax_amount": "0.00",
"total_amount": "200.00"
},
{
"table_pack_name": "Demo Pack Name7",
"table_pack_type": "Demo Pack Name7",
"base_amount": "400.00",
"tax_amount": "40.00",
"total_amount": "440.00"
}
]
}
javascript jquery
javascript jquery
asked Dec 26 at 13:34
Kiran Bhagannavar
135
asked Dec 26 at 13:34
Kiran Bhagannavar
135
asked Dec 26 at 13:34
Kiran Bhagannavar
135
asked Dec 26 at 13:34
Kiran Bhagannavar
135
asked Dec 26 at 13:34
Kiran Bhagannavar
135
135
gj for not providing the html, makes it that challenging
– WASD
Dec 26 at 13:41
If I understand your code right, code like$('#table_pack_name').val()will get the same value for every iteration.
– obfish
Dec 26 at 13:41
yes @obfish you are right. my iteration is getting repeated
– Kiran Bhagannavar
Dec 26 at 13:47
add a comment |
gj for not providing the html, makes it that challenging
– WASD
Dec 26 at 13:41
If I understand your code right, code like$('#table_pack_name').val()will get the same value for every iteration.
– obfish
Dec 26 at 13:41
yes @obfish you are right. my iteration is getting repeated
– Kiran Bhagannavar
Dec 26 at 13:47
gj for not providing the html, makes it that challenging
– WASD
Dec 26 at 13:41
gj for not providing the html, makes it that challenging
– WASD
Dec 26 at 13:41
If I understand your code right, code like
$('#table_pack_name').val() will get the same value for every iteration.– obfish
Dec 26 at 13:41
If I understand your code right, code like
$('#table_pack_name').val() will get the same value for every iteration.– obfish
Dec 26 at 13:41
yes @obfish you are right. my iteration is getting repeated
– Kiran Bhagannavar
Dec 26 at 13:47
yes @obfish you are right. my iteration is getting repeated
– Kiran Bhagannavar
Dec 26 at 13:47
add a comment |
2 Answers
2
active
oldest
votes
0
because you're assigning the same values every iteration
switch(idx) {
case 0:
packDetailsObj.table_pack_name = $('#table_pack_name').val();
break;
case 1:
packDetailsObj.table_pack_type = $('#table_pack_type').val();
break;
case 2:
packDetailsObj.base_amount = $('#base_amount').val();
break;
case 3:
packDetailsObj.tax_amount = $('#tax_amount').val();
break;
case 4:
packDetailsObj.total_amount = $('#total_amount').val();
break;
}
answered Dec 26 at 13:40
WASD
1646
No it didnt work @WASD
– Kiran Bhagannavar
Dec 26 at 13:44
I trolled by accident
– WASD
Dec 26 at 13:46
add a comment |
0
A few things to should follow is id should be unique to an element and no two elements should have same id. Check this for further reading. You can instead use name attribute to read values as below.
$('#addPackTable tr').each(function (index) {
if (index != 0) {
var packDetailsObj = {};
($(this).find("td")).each(function (idx) {
switch (idx) {
case 0:
packDetailsObj.table_pack_name = $(this).find('[name="table_pack_name"]').val();
break;
case 1:
packDetailsObj.table_pack_type = $(this).find('[name="table_pack_type"]').val();
break;
case 2:
packDetailsObj.base_amount = $(this).find('[name="base_amount"]').val();
break;
case 3:
packDetailsObj.tax_amount = $(this).find('[name="tax_amount"]').val();
break;
case 4:
packDetailsObj.total_amount = $(this).find('[name="total_amount"]').val();
break;
}
});
packDetailsRow.push(packDetailsObj);
}
answered Dec 26 at 13:51
Avi
588215
I guess that selectors like 'td #table_pack_name' makes no sense.
– obfish
Dec 26 at 14:01
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
because you're assigning the same values every iteration
switch(idx) {
case 0:
packDetailsObj.table_pack_name = $('#table_pack_name').val();
break;
case 1:
packDetailsObj.table_pack_type = $('#table_pack_type').val();
break;
case 2:
packDetailsObj.base_amount = $('#base_amount').val();
break;
case 3:
packDetailsObj.tax_amount = $('#tax_amount').val();
break;
case 4:
packDetailsObj.total_amount = $('#total_amount').val();
break;
}
answered Dec 26 at 13:40
WASD
1646
No it didnt work @WASD
– Kiran Bhagannavar
Dec 26 at 13:44
I trolled by accident
– WASD
Dec 26 at 13:46
add a comment |
0
because you're assigning the same values every iteration
switch(idx) {
case 0:
packDetailsObj.table_pack_name = $('#table_pack_name').val();
break;
case 1:
packDetailsObj.table_pack_type = $('#table_pack_type').val();
break;
case 2:
packDetailsObj.base_amount = $('#base_amount').val();
break;
case 3:
packDetailsObj.tax_amount = $('#tax_amount').val();
break;
case 4:
packDetailsObj.total_amount = $('#total_amount').val();
break;
}
answered Dec 26 at 13:40
WASD
1646
No it didnt work @WASD
– Kiran Bhagannavar
Dec 26 at 13:44
I trolled by accident
– WASD
Dec 26 at 13:46
add a comment |
0
0
0
because you're assigning the same values every iteration
switch(idx) {
case 0:
packDetailsObj.table_pack_name = $('#table_pack_name').val();
break;
case 1:
packDetailsObj.table_pack_type = $('#table_pack_type').val();
break;
case 2:
packDetailsObj.base_amount = $('#base_amount').val();
break;
case 3:
packDetailsObj.tax_amount = $('#tax_amount').val();
break;
case 4:
packDetailsObj.total_amount = $('#total_amount').val();
break;
}
answered Dec 26 at 13:40
WASD
1646
because you're assigning the same values every iteration
switch(idx) {
case 0:
packDetailsObj.table_pack_name = $('#table_pack_name').val();
break;
case 1:
packDetailsObj.table_pack_type = $('#table_pack_type').val();
break;
case 2:
packDetailsObj.base_amount = $('#base_amount').val();
break;
case 3:
packDetailsObj.tax_amount = $('#tax_amount').val();
break;
case 4:
packDetailsObj.total_amount = $('#total_amount').val();
break;
}
answered Dec 26 at 13:40
WASD
1646
edited Dec 26 at 13:46
answered Dec 26 at 13:40
WASD
1646
answered Dec 26 at 13:40
WASD
1646
answered Dec 26 at 13:40
WASD
1646
1646
No it didnt work @WASD
– Kiran Bhagannavar
Dec 26 at 13:44
I trolled by accident
– WASD
Dec 26 at 13:46
add a comment |
No it didnt work @WASD
– Kiran Bhagannavar
Dec 26 at 13:44
I trolled by accident
– WASD
Dec 26 at 13:46
No it didnt work @WASD
– Kiran Bhagannavar
Dec 26 at 13:44
No it didnt work @WASD
– Kiran Bhagannavar
Dec 26 at 13:44
I trolled by accident
– WASD
Dec 26 at 13:46
I trolled by accident
– WASD
Dec 26 at 13:46
add a comment |
0
A few things to should follow is id should be unique to an element and no two elements should have same id. Check this for further reading. You can instead use name attribute to read values as below.
$('#addPackTable tr').each(function (index) {
if (index != 0) {
var packDetailsObj = {};
($(this).find("td")).each(function (idx) {
switch (idx) {
case 0:
packDetailsObj.table_pack_name = $(this).find('[name="table_pack_name"]').val();
break;
case 1:
packDetailsObj.table_pack_type = $(this).find('[name="table_pack_type"]').val();
break;
case 2:
packDetailsObj.base_amount = $(this).find('[name="base_amount"]').val();
break;
case 3:
packDetailsObj.tax_amount = $(this).find('[name="tax_amount"]').val();
break;
case 4:
packDetailsObj.total_amount = $(this).find('[name="total_amount"]').val();
break;
}
});
packDetailsRow.push(packDetailsObj);
}
answered Dec 26 at 13:51
Avi
588215
I guess that selectors like 'td #table_pack_name' makes no sense.
– obfish
Dec 26 at 14:01
add a comment |
0
A few things to should follow is id should be unique to an element and no two elements should have same id. Check this for further reading. You can instead use name attribute to read values as below.
$('#addPackTable tr').each(function (index) {
if (index != 0) {
var packDetailsObj = {};
($(this).find("td")).each(function (idx) {
switch (idx) {
case 0:
packDetailsObj.table_pack_name = $(this).find('[name="table_pack_name"]').val();
break;
case 1:
packDetailsObj.table_pack_type = $(this).find('[name="table_pack_type"]').val();
break;
case 2:
packDetailsObj.base_amount = $(this).find('[name="base_amount"]').val();
break;
case 3:
packDetailsObj.tax_amount = $(this).find('[name="tax_amount"]').val();
break;
case 4:
packDetailsObj.total_amount = $(this).find('[name="total_amount"]').val();
break;
}
});
packDetailsRow.push(packDetailsObj);
}
answered Dec 26 at 13:51
Avi
588215
I guess that selectors like 'td #table_pack_name' makes no sense.
– obfish
Dec 26 at 14:01
add a comment |
0
0
0
A few things to should follow is id should be unique to an element and no two elements should have same id. Check this for further reading. You can instead use name attribute to read values as below.
$('#addPackTable tr').each(function (index) {
if (index != 0) {
var packDetailsObj = {};
($(this).find("td")).each(function (idx) {
switch (idx) {
case 0:
packDetailsObj.table_pack_name = $(this).find('[name="table_pack_name"]').val();
break;
case 1:
packDetailsObj.table_pack_type = $(this).find('[name="table_pack_type"]').val();
break;
case 2:
packDetailsObj.base_amount = $(this).find('[name="base_amount"]').val();
break;
case 3:
packDetailsObj.tax_amount = $(this).find('[name="tax_amount"]').val();
break;
case 4:
packDetailsObj.total_amount = $(this).find('[name="total_amount"]').val();
break;
}
});
packDetailsRow.push(packDetailsObj);
}
answered Dec 26 at 13:51
Avi
588215
A few things to should follow is id should be unique to an element and no two elements should have same id. Check this for further reading. You can instead use name attribute to read values as below.
$('#addPackTable tr').each(function (index) {
if (index != 0) {
var packDetailsObj = {};
($(this).find("td")).each(function (idx) {
switch (idx) {
case 0:
packDetailsObj.table_pack_name = $(this).find('[name="table_pack_name"]').val();
break;
case 1:
packDetailsObj.table_pack_type = $(this).find('[name="table_pack_type"]').val();
break;
case 2:
packDetailsObj.base_amount = $(this).find('[name="base_amount"]').val();
break;
case 3:
packDetailsObj.tax_amount = $(this).find('[name="tax_amount"]').val();
break;
case 4:
packDetailsObj.total_amount = $(this).find('[name="total_amount"]').val();
break;
}
});
packDetailsRow.push(packDetailsObj);
}
answered Dec 26 at 13:51
Avi
588215
edited Dec 27 at 13:35
answered Dec 26 at 13:51
Avi
588215
answered Dec 26 at 13:51
Avi
588215
answered Dec 26 at 13:51
Avi
588215
588215
I guess that selectors like 'td #table_pack_name' makes no sense.
– obfish
Dec 26 at 14:01
add a comment |
I guess that selectors like 'td #table_pack_name' makes no sense.
– obfish
Dec 26 at 14:01
I guess that selectors like 'td #table_pack_name' makes no sense.
– obfish
Dec 26 at 14:01
I guess that selectors like 'td #table_pack_name' makes no sense.
– obfish
Dec 26 at 14:01
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gj for not providing the html, makes it that challenging
– WASD
Dec 26 at 13:41
If I understand your code right, code like
$('#table_pack_name').val()will get the same value for every iteration.– obfish
Dec 26 at 13:41
yes @obfish you are right. my iteration is getting repeated
– Kiran Bhagannavar
Dec 26 at 13:47