我混合了“ajax select”和“滚动加载数据时”脚本，并且这是有效的，但我不知道如何在 div.status 中打印“未找到数据”输出变量(在 animals.php 上)为空。
index.html
<!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>
动物.php
<?php
$connect = mysqli_connect("localhost", "root", "", "petsdata");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_set_charset($connect,"utf8");
$output = '';
$animal_start = $connect->real_escape_string($_POST["animal_start"]);
$animal_limit = $connect->real_escape_string($_POST["animal_limit"]);
$category = $connect->real_escape_string($_POST["category"]);
$chipnumber = $connect->real_escape_string($_POST["chipnumber"]);
if (isset($animal_start, $animal_limit, $category, $chipnumber)) {
if (!empty($category) && !empty($chipnumber)) {
$query = mysqli_query($connect, "SELECT * FROM animals WHERE chipnumber LIKE '%".$chipnumber."%' AND category = '".$category."' ORDER BY id LIMIT ".$animal_start.", ".$animal_limit."");
}
else if (!empty($category)) {
$query = mysqli_query($connect, "SELECT * FROM animals WHERE category = '".$category."' ORDER BY id LIMIT ".$animal_start.", ".$animal_limit."");
}
else if (!empty($chipnumber)) {
$query = mysqli_query($connect, "SELECT * FROM animals WHERE chipnumber LIKE '%".$chipnumber."%' AND status = '1' ORDER BY id DESC LIMIT ".$animal_start.", ".$animal_limit."");
}
else {
$query = mysqli_query($connect, "SELECT * FROM animals ORDER BY id DESC LIMIT ".$animal_start.", ".$animal_limit."");
}
while ($row = mysqli_fetch_array($query)) {
$output .= '<div class="animal">';
$output .= '<span>Category: ' . $row["category"] . '</span>';
$output .= '<span>Chipnumber: ' . $row["chipnumber"] . '</span>';
$output .= '</div>';
}
}
echo $output;
?>
最佳答案
if($query->num_rows > 0){
//Proceed as normally
}else{
$output = 'No data Found';
}
关于javascript - 当代码输出为空时如何打印 "not found"？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/47024579/