One Pagination controls for Two different tables side by side

One Pagination controls for Two different tables side by side

NagarajanCNagarajanC Posts: 5Questions: 1Answers: 0

Hi Team,
I have two different tables displayed side by side (Table1 and Table2)and two pagination are present for each datatable.
If user selects page 3 of table 1 then data of Table 2 must also display page 3 data.

Thanks,
Nagarajan

Answers

  • NagarajanCNagarajanC Posts: 5Questions: 1Answers: 0

    Can somebody please let me know how it can be done in datatable?

  • colincolin Posts: 12,490Questions: 0Answers: 2,118
    edited May 8

    You would use page - and when changed on table 1, call page() for the second table. You can get the paging detail for the first table with page() or page.info(),

    Colin

  • NagarajanCNagarajanC Posts: 5Questions: 1Answers: 0
    edited May 10

    Hi Colin,

    Thanks a lot for the quick info about the API's. I have implemented and is working fine

    Thanks,
    Nagarajan

    Please find below the code.

    $(document).ready(function(){
    
       $('#empTable').DataTable({
    
    
            'processing':true,
            'serverSide':true,
            'serverMethod':'post',
            'ajax':{
            url:'/ajaxfile'},
    
            'scrollY':200,
            'scrollCollapse': true,
    
            'lengthMenu':[[5,50],[5,50]],
            searching:true,
            sort:false,
            "serverSide":true,
            'columns':[
                {data:'name'},
                {data:'position'},
                {data:'age'},
                {data:'salary'},
                {data:'office'}
            ]
    
    
    
        })
    
    $('#empTable1').DataTable({
    
    
            'processing':true,
            'serverSide':true,
            'serverMethod':'post',
            'ajax':{
            url:'/ajaxfile'},
    
            'scrollY':200,
            'scrollCollapse': true,
    
            'lengthMenu':[[5,50],[5,50]],
            searching:true,
            sort:false,
            "serverSide":true,
            'columns':[
                {data:'name'},
                {data:'position'},
                {data:'age'},
                {data:'salary'},
                {data:'office'}
            ]
    
    
    
        })
    
    var table = $('#empTable').DataTable();
    
    $('#empTable').on( 'page.dt', function () {
        var info = table.page.info();
       // $('#pageInfo').html( 'Showing page: '+info.page+' of '+info.pages );
        console.log(info);
    
    //select second table
    test(info.page);
    
    } );
    
    
    function test(page_number){
    var table = $('#empTable1').DataTable();
    table.page(page_number).draw( false );
    }
    
    
    
    
    } );
    
  • colincolin Posts: 12,490Questions: 0Answers: 2,118

    Nice, glad to hear you got it working,

    Colin

  • NagarajanCNagarajanC Posts: 5Questions: 1Answers: 0

    Hi Colin,
    I have now the two tables loaded successfully and i have also implemented the callback functions for the ajax data load from the server.
    I want to highlight the cells as red if table1 data does not matches with the Table 2.
    I am not able to compare the table 1 with the table 2 on the call back function as either one of the table data is not loaded yet.
    Any thoughts on this?
    Do i need to start the compare only when both the ajax data is loaded?or this can be done on the callback function?
    do let me know all the views if possible
    Thanks,
    Nagarajan

  • colincolin Posts: 12,490Questions: 0Answers: 2,118

    The problem, as you say, is ensuring both the tables are settled after the loads. You could use a combinations of global variables and the xhr (which triggers when the data is received) and draw (which triggers after the draw). So something like (and excuse bugs/typos as untested)

    table1Drawn = false;
    table2Drawn = false;
    $('#table1').on('xhr', function() {
      $('#table1').one(`draw`, function() {
        if (table2Drawn) {
           // both are drawn so do your code
           // reset both table1Drawn and table2Drawn to false
        }
        else {
          table1Drawn = true;
        }
      ​});
    });
    

    Colin

  • NagarajanCNagarajanC Posts: 5Questions: 1Answers: 0

    Hi Colin,
    Thanks a lot for the quick info. I will give it a try and confirm.
    Thanks,
    Nagarajan

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