Question

storedObjective = "";
storedSignal_type = "";
$.each(data, function(index, key) {

    if (key.objective == storedObjective)
    {
        print_1 = "";
    }
    else
    {
        print_1 = key.objective;
        storedObjective = key.objective;
    }
    if (key.signal_type == storedSignal_type)
    {
        print_2 = "";
    }
    else
    {
        print_2 = key.signal_type;
        storedSignal_type = key.signal_type;
    }

    $('#myTable').append('<tr><td>'+print_1+'</td><td>'+print_2+'</td><td>'+key.signal_name+'</td></tr>');
    console.log(print_1);
    console.log(print_2);


    if (print_1==="" && print_2==="")
    {
        $('tr').addClass('yes')
    }
    else{
        $('tr').addClass('no')
    }
});

代码的所有第一部分都可以正常工作。但最后，我的if始终评估为true，即使我console.log这些变量肯定不是。python manage.py test。我在这里缺少什么？

Answer 1

您正在为每个现有行分配类（是/否），因此如果if匹配一次，则所有现有表行将获得该类 - 如果else匹配则相同。您可以简化append并在那里添加课程。

var className = print_1==="" && print_2==="" ? 'yes' : 'no';
$('#myTable').append('<tr class="' + className + '"><td>'+print_1+'</td><td>'+print_2+'</td><td>'+key.signal_name+'</td></tr>');

Answer 2

您需要将yes / no类仅添加到for each中的当前元素中，否则如果任何其他元素满足条件，则每个tr将具有yes / no。

这里，只给当前元素一个类：

let $nextRow = $('<tr><td>'+print_1+'</td><td>'+print_2+'</td> <td>'+key.signal_name+'</td></tr>');

if (print_1==="" && print_2==="")  
    $nextRow.addClass('yes');
else 
    $nextRow.addClass('no');

$('#myTable').append($nextRow);

var data = [
    {
        "objective": "Awareness",
        "signal_type": "Efficiency",
        "signal_name": "CPM"
    },
    {
        "objective": "Awareness",
        "signal_type": "Quality",
        "signal_name": "Click-thru Rate"
    },
    {
        "objective": "Awareness",
        "signal_type": "Quality",
        "signal_name": "Frequency"
    },
    {
        "objective": "Awareness",
        "signal_type": "Quality",
        "signal_name": "Interaction Rate"
    },
    {
        "objective": "Awareness",
        "signal_type": "Quality",
        "signal_name": "Viewability"
    },
    {
        "objective": "Awareness",
        "signal_type": "Volume",
        "signal_name": "Display Time"
    },
    {
        "objective": "Awareness",
        "signal_type": "Volume",
        "signal_name": "Impression"
    },
    {
        "objective": "Awareness",
        "signal_type": "Volume",
        "signal_name": "Reach"
    },
    {
        "objective": "Conversion",
        "signal_type": "Efficiency",
        "signal_name": "Cost per Action"
    },
    {
        "objective": "Conversion",
        "signal_type": "Efficiency",
        "signal_name": "ROI/ROAS"
    },
    {
        "objective": "Conversion",
        "signal_type": "Quality",
        "signal_name": "Avg Order Value"
    },
    {
        "objective": "Conversion",
        "signal_type": "Volume",
        "signal_name": "Conversion"
    },
    {
        "objective": "Conversion",
        "signal_type": "Volume",
        "signal_name": "Revenue"
    },
    {
        "objective": "Engagement",
        "signal_type": "Efficiency",
        "signal_name": "Cost per Click"
    },
    {
        "objective": "Engagement",
        "signal_type": "Quality",
        "signal_name": "Avg Interaction Time"
    },
    {
        "objective": "Engagement",
        "signal_type": "Quality",
        "signal_name": "Conversion Rate"
    },
    {
        "objective": "Engagement",
        "signal_type": "Volume",
        "signal_name": "Click"
    },
    {
        "objective": "Engagement",
        "signal_type": "Volume",
        "signal_name": "Interaction"
    },
    {
        "objective": "Engagement",
        "signal_type": "Volume",
        "signal_name": "Interaction Time"
    }
];

storedObjective = "";
storedSignal_type = "";
$.each(data, function(index, key) {

    if (key.objective == storedObjective)
    {
        print_1 = "";
    }
    else
    {
        print_1 = key.objective;
        storedObjective = key.objective;
    }
    if (key.signal_type == storedSignal_type)
    {
        print_2 = "";
    }
    else
    {
        print_2 = key.signal_type;
        storedSignal_type = key.signal_type;
    } 

   let $nextRow = $('<tr><td>'+print_1+'</td><td>'+print_2+'</td> <td>'+key.signal_name+'</td></tr>');


    if (print_1==="" && print_2==="")
    {   
        $nextRow.addClass('yes');
    } 
    else{
        $nextRow.addClass('no');
    }
  $('#myTable').append($nextRow);
});

.yes {
  background-color: green;
}

.no {
  background-color: red;
  color: white;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table id="myTable">
</table>

检查空字符串的IF语句始终返回true

2 个答案: