Reputation: 5250
I have some HTML, JSON, and Javascript. I want to render a table consisting of 3 rows (as of JSON). Problem is that all my data comes out in TD's in one tr
<table>
<tbody>
<tr>
<td>car</td>
<td>1000</td>
<td>2000</td>
<td>1500</td>
<td>1000</td>
<td>house</td>
<td>500</td>
<td>600</td>
<td>700</td>
<td>800</td>
<td>internet</td>
<td>100</td>
<td>200</td>
<td>2350</td>
<td>1940</td>
</tr>
</tbody>
</table>
I want
<table>
<tbody>
<tr>
<td>car</td>
<td>1000</td>
<td>2000</td>
<td>1500</td>
<td>1000</td>
</tr>
<tr>
<td>house</td>
<td>500</td>
<td>600</td>
<td>700</td>
<td>800</td>
</tr>
<tr>
<td>internet</td>
<td>100</td>
<td>200</td>
<td>2350</td>
<td>1940</td>
</tr>
</tbody></table>
Here is my code HTML
<div class="calc"></div>
JSON
{
"item": [
{
"name": "car",
"cost": [1000, 2000, 1500, 1000]
},
{
"name": "house",
"cost": [500, 600, 700, 800]
},
{
"name": "internet",
"cost": [100, 200, 2350, 1940]
}
]
}
Javascript
$.getJSON("calc.json", function(data) {
var $table = $('<table/>'),
$row = $('<tr/>');
for(var i = 0; i < data.item.length; i ++){
$row.append( '<td>' + data.item[i].name + '</td>' );
for (var cost = 0; cost < data.item[i].cost.length; cost ++) {
$row.append( '<td>' + data.item[i].cost[cost] + '</td>' );
}
}
$('.calc').append($table.append($row));
});
EDIT a less code intensive version
$.getJSON("calc.json", function(data) {
var $table = $("<table/>");
$.each(data.item, function(k,v) {
var $row = $("<tr/>");
$row.append("<td>" + v.name + "</td>");
$.each(v.cost, function(k,v) {
$row.append("<td>" + v + "</td>");
});
$table.append($row);
});
$(".calc").append($table);
});
Upvotes: 0
Views: 81
Reputation: 25882
Problem is this because you have declared only one row for all Items.
Create new Row for Each item.
change your function like bellow.
$.getJSON("calc.json", function(data) {
var $table = $('<table/>');
for(var i = 0; i < data.item.length; i ++){
var $row = $('<tr/>'); //new Row for current ITEM
$row.append( '<td>' + data.item[i].name + '</td>' );
for (var cost = 0; cost < data.item[i].cost.length; cost ++) {
$row.append( '<td>' + data.item[i].cost[cost] + '</td>' );
}
$table.append($row); // append current row
}
$('.calc').append($table);
});
Upvotes: 1
Reputation: 423
var $table = $('<table/>');
for(var i = 0; i < data.item.length; i ++){
var $row = $('<tr/>');
$row.append( '<td>' + data.item[i].name + '</td>' );
for (var cost = 0; cost < data.item[i].cost.length; cost ++) {
$row.append( '<td>' + data.item[i].cost[cost] + '</td>' );
}
$table.append($row)
}
$('.calc').append($table);
Upvotes: 3
Reputation: 1934
Hi i have changed you function a little but now you can use below given function:-
$.getJSON("calc.json", function(data) {
var $table = $('<table/>');
for(var i = 0; i < data.item.length; i ++){
var $row = $('<tr/>');
$row.append( '<td>' + data.item[i].name + '</td>' );
for (var cost = 0; cost < data.item[i].cost.length; cost ++) {
$row.append( '<td>' + data.item[i].cost[cost] + '</td>' );
}
$table.append($row);
}
$('.calc').append($table );
});
Upvotes: 3