Question

我混合了一个＆＃34; ajax select＆＃34;和＃34;滚动加载数据＆＃34;脚本，这是有效的，但我不知道如何打印＆＃34;未找到数据＆＃34;当输出变量（在base_index.html上）为空时，在div.status中。

的index.html

animals.php

animals.php

<!DOCTYPE html>
<html>
<head>
    <meta charset="utf-8">
    <title>Animals</title>
    <script src="https://code.jquery.com/jquery-3.2.1.min.js"></script>
</head>
<body>

<div class="search">

<div class="filter category">
    <select name="category" id="category">
        <option value="">All</option>
        <option value="free">Free</option>
        <option value="lost">Lost</option>
        <option value="found">Found</option>
    </select>
</div>

<div class="filter chipnumber">
    <input type="text" name="chipnumber" id="chipnumber"></div>
</div>

<div class="send">

<button type="submit" id="submit">Search</button>

</div>

</div>

<script>

$(document).ready(function() {

    var animal_limit = 6;
    var animal_start = 0;
    var animal_action = 'inactive';

    function load_animal_data() {

        var category = $('#category').val();
        var chipnumber = $('#chipnumber').val();

        $.ajax({

            url: "animals.php",
            method: "POST",
            data: {animal_limit:animal_limit, animal_start:animal_start, animal_action:animal_action, category:category, chipnumber:chipnumber},
            success:function(data) {

                $('div.animals').append(data);

                if (data == '') {

                    animal_action = 'active';

                } else {

                    animal_action = 'inactive';

                }

            }

        });

    }

    load_animal_data();

    function search() {

        var category = $('#category').val();
        var chipnumber = $('#chipnumber').val();

        animal_start = 0;

        load_animal_data();

    }

    $('#search').on('click', function() {

        search();

    });

    $(window).scroll(function () {

        if ($(window).scrollTop() + $(window).height() > $('div.animals').height() && animal_action == 'inactive') {

            animal_action = 'active';

            animal_start = animal_start + animal_limit;

            setTimeout(function() {

                load_animal_data();

            }, 1000);

        }

    });

});

</script>

<div class="animals"></div>
<div class="status"></div>

</body>
</html>

Answer 1

if($query->num_rows > 0){
   //Proceed as normally 
}else{
   $output = 'No data Found';
}

如何打印＆＃34;未找到＆＃34;当代码输出为空时？

1 个答案: