table displaying [object Object] in every cell

Mohan Srinivas

based on drop down i m displaying table using ajax jquery call.while i m selecting drop down list table is displaying but every cells showing [object Object] instead of data's

This my jquery code:

function generateTable(json) {

var $table = $('#tblDynamic');

$table.find('thead').empty()

$table.find('tbody').empty()

if (json && json.length > 0) {

var header = json[0];

var columns = [];

for (var col in header) {

columns.push('' + col + '');

}

$table.find('thead').append('' + columns.join('') + '');

var rows = [];

for (var i = 0; i < json.length; i++) {

var row = json[i];

var tds = [];

for (var col in row) {

tds.push('' + row + '');

}

rows.push('' + tds.join() + '');

}

$table.find('tbody').append(rows.join(''));

}

7 Replies

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Mohan Srinivas9y ago
One table means we will set,but table names randomly changing from drop down, so property also changing right,,,so how we set
Manas Mohapatra9y ago
Then try like:

for (var col in row) {
tds.push('' + row.City + '');
}

If it helps please accept the answer...
Mohan Srinivas9y ago
for json[0] displaying row like this above
Mohan Srinivas9y ago
Object
City:"chennai"
ContactName:"Raja"
Country:"india"
Salary:"25000"
__proto__:Object
Manas Mohapatra9y ago
I don't know what is structure of JSON object.. Please do console.log(json).
Press F12 in browser and see output on console tab. And then provide the data..
Mohan Srinivas9y ago
what have to write can you say it?
Manas Mohapatra9y ago
I think you need change yellow marked code. It is showing the JSON object directly in table cell. So as per JSON format, you need to assign properties like row.Id(if Id is one property there..)
function generateTable(json) {
    var $table = $('#tblDynamic');
    $table.find('thead').empty()
    $table.find('tbody').empty()
    if (json && json.length > 0) {
        var header = json[0];
        var columns = [];
        for (var col in header) {
            columns.push('' + col + '');
        }
        $table.find('thead').append('' + columns.join('') + '');
        var rows = [];
        for (var i = 0; i < json.length; i++) {
            var row = json[i];
            var tds = [];
            for (var col in row) {
                tds.push('' + row + '');
            }
            rows.push('' + tds.join() + '');
        }
        $table.find('tbody').append(rows.join(''));
    }
}