Question

164

Shift is not a function in javascript

rated 0 times [ 166] [ 2] / answers: 1 / hits: 23499 / 7 Years ago, tue, march 21, 2017, 12:00:00

I have multiple tables(with same structure)in my html. They are in tabs. I am creating tabs dynamically and i want to send those table data to the mysql database. So i wanted to get those data using javascript. I am correctly created TableData1,TableData2.... arrays using for loop. Problem is I cannot increment TableData here ('TableData'+i).shift(); . I am getting an error. I want to create TableData1.shift(),TableData2.shift().....

function myDataSendFunction(){

var i;



     for(i = 1; i <= array_size; i++){

         eval(var TableData+i+=[];);



      $('#mytable'+i+ ' tr').each(function(row, tr){

          ('TableData'+i)[row]={

              colum1 : $(tr).find('td:eq(1)').text()

              , colum2 :$(tr).find('td:eq(2)').text()

              , colum3 : $(tr).find('td:eq(3)').text()

              , colum4 : $(tr).find('td:eq(4)').text()

              , colum5 : $(tr).find('td:eq(5)').text()

              , colum6 : $(tr).find('td:eq(6)').text()

              , colum7 : $(tr).find('td:eq(7)').text()

              , colum8 : $(tr).find('td:eq(8)').text()

          }

      }); 



        ('TableData'+i).shift(); 



    }



}

I am getting this error.

Uncaught TypeError: (TableData + i).shift is not a function

    at myDataSendFunction (<anonymous>:25:25)

    at HTMLInputElement.onclick (create.php:1)

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leslijessalyng

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answered 7 Years ago meliashelbief · Accepted Answer

It's clear that you are very new to this, so I'm going to show you first and then explain second.

function myDataSendFunction(){

    var i,

        TableData = [];



    for(i = 1; i <= array_size; i++){

        TableData[i] = [];



        $('#mytable' + i + ' tr').each(function(row, tr){

            TableData[i][row]={

                  colum1 : $(tr).find('td:eq(1)').text()

                , colum2 : $(tr).find('td:eq(2)').text()

                , colum3 : $(tr).find('td:eq(3)').text()

                , colum4 : $(tr).find('td:eq(4)').text()

                , colum5 : $(tr).find('td:eq(5)').text()

                , colum6 : $(tr).find('td:eq(6)').text()

                , colum7 : $(tr).find('td:eq(7)').text()

                , colum8 : $(tr).find('td:eq(8)').text()

            };

        }); 



        TableData[i].shift();

    }

}

Base on your comment:

what should i do then?? I tried with TableData array, then problem occurs in here TableData[i][row] with Uncaught TypeError: Cannot set property '0' of undefined

Your troubles revolve around setting properties on variables that haven't been defined. You must define TableData before attempting to assign a property.

Once TableData is initialized to an array (var TableData = []), you can set properties on TableData. But you cannot immediately set properties on properties of TableData. For example, you cannot jump directly to TableData[i][row]. You must first set TableData[i] to an array (TableData[i] = []), and then you can set TableData[i][row] to some value.

By trying to solve that problem with eval, you ran into a whole new world of problems. Try to avoid eval... it's a very complicated beast that tends to cause a lot of confusion and pain.

It may be helpful to review MDN's Working with objects documentation to better understand what's going on with JavaScript's array and objects.