Question

我有一个问题是在getJson之后在变量数据中携带数据..下面是我的代码

function format ( d ) {
  var id = d[1],$base = '<?php echo Routed::url('/api/course_fees_setup/getSponsorList?'); ?>';
   var rows = ''; 
    $.getJSON($base+'id=' + id,function(data){
        var index =0;
        var newOptions = data.data;
        var SponsorbyFee = newOptions.Sponsorship[0].SponsorbyFee;
        $.each(SponsorbyFee, function(value,key) {
          // console.log(key);
          switch(key['fee_type_id']){
            case '1' :
                  // console.log('program');
                  break;
            case '2' :
                  // console.log(key['OthersFee']);
                  rows[index] = '<tr><td>' + key['OthersFee']['payment_description'] + '</td><td>' + key['OthersFee']['price'] + '</td><td>' + key['OthersFee']['remarks'] + '</td><tr>';
                  index++;
                  break;
            case '3' : 
                  // console.log(key['DiscountFee']);
                  rows[index] = '<tr><td>' + key['DiscountFee']['payment_description'] + '</td><td>' + key['DiscountFee']['price'] + '</td><td>' + key['DiscountFee']['remarks'] + '</td><tr>';
                  index++;
                  break;
          }
        });
    });
    return '<br /><br /><table id="sponsorlist-tbl" class="hover cell-border" width="100%">'+
                    '<thead>'+
                          '<tr>'+
                              '<th width="30%">Payment Description</th>'+
                              '<th width="10%">Price</th>'+
                              '<th width="30%">Remarks</th>'+
                          '</tr>'+
                    '</thead>'+
                    '<tbody>'+
                      rows +
                    '</tbody>'+
                '<table><br /><br /><br />';
}

我希望在第36行完成循环我的行后，在第36行返回变量数据。但好像它返回null。有人帮助我纠正我的代码，因为我只研究生成jquery中的表。任何帮助都会非常感激。谢谢你们

Answer 1

$.getJSON()以异步方式返回结果。您可以在html之后的data来电中退回$.getJSON()字符串和$.each();使用.then()来处理结果

function format ( d ) {
  var id = d[1],$base = '<?php echo Routed::url('/api/course_fees_setup/getSponsorList?'); ?>';
   var rows = ''; 
    return $.getJSON($base+'id=' + id)
    .then(function(data){
        var index =0;
        var newOptions = data.data;
        var SponsorbyFee = newOptions.Sponsorship[0].SponsorbyFee;
        $.each(SponsorbyFee, function(value,key) {
          // console.log(key);
          switch(key['fee_type_id']){
            case '1' :
                  // console.log('program');
                  break;
            case '2' :
                  // console.log(key['OthersFee']);
                  rows[index] = '<tr><td>' + key['OthersFee']['payment_description'] + '</td><td>' + key['OthersFee']['price'] + '</td><td>' + key['OthersFee']['remarks'] + '</td><tr>';
                  index++;
                  break;
            case '3' : 
                  // console.log(key['DiscountFee']);
                  rows[index] = '<tr><td>' + key['DiscountFee']['payment_description'] + '</td><td>' + key['DiscountFee']['price'] + '</td><td>' + key['DiscountFee']['remarks'] + '</td><tr>';
                  index++;
                  break;
          }
        });
        return ['<br /><br /><table id="sponsorlist-tbl" class="hover cell-border" width="100%">'+
                    '<thead>'+
                          '<tr>'+
                              '<th width="30%">Payment Description</th>'+
                              '<th width="10%">Price</th>'+
                              '<th width="30%">Remarks</th>'+
                          '</tr>'+
                    '</thead>'+
                    '<tbody>'+
                      rows +
                    '</tbody>'+


         '<table><br /><br /><br />'
         , data];
    });

}

format(/* parameter */).then(function(results) {
  console.log(results)
})

如何在getJson之后传输数据

1 个答案: