这段代码，我已经完成了注册号码的搜索。所有代码都正确且运行正常但未找到匹配数据。没有显示消息。所以，我想在未找到匹配数据时显示消息＆＃34;未找到此类数据＆＃34;。 我怎样才能做到这一点？请帮忙解决这个问题， 我的代码如下。
的index.php
<!DOCTYPE html>
<html>
<head>
<title>How to return JSON Data from PHP Script using Ajax Jquery</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" />
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
<style>
#result {
position: absolute;
width: 100%;
max-width:870px;
cursor: pointer;
overflow-y: auto;
max-height: 400px;
box-sizing: border-box;
z-index: 1001;
}
.link-class:hover{
background-color:#f1f1f1;
}
</style>
</head>
<body>
<br /><br />
<div class="container" style="width:900px;">
<h2 align="center">How to return JSON Data from PHP Script using Ajax Jquery</h2>
<h3 align="center">Search Employee Data</h3><br />
<div class="row">
<div class="col-md-4">
<input type="text" name="employee_list" id="employee_list" class="form-control">
</div>
<div class="col-md-4">
<button type="button" name="search" id="search" class="btn btn-info">Search</button>
</div>
</div>
<br />
<div class="table-responsive" id="employee_details" style="display:none">
<table class="table table-bordered">
<tr>
<td width="10%" align="right"><b>Name</b></td>
<td width="90%"><span id="name"></span></td>
</tr>
<tr>
<td width="10%" align="right"><b>Address</b></td>
<td width="90%"><span id="address"></span></td>
</tr>
<tr>
<td width="10%" align="right"><b>Gender</b></td>
<td width="90%"><span id="total_marks"></span></td>
</tr>
<tr>
<td width="10%" align="right"><b>Designation</b></td>
<td width="90%"><span id="email"></span></td>
</tr>
<tr>
<td width="10%" align="right"><b>Age</b></td>
<td width="90%"><span id="ph"></span></td>
</tr>
</table>
</div>
</div>
</body>
</html>
<script>
$(document).ready(function(){
$('#search').click(function(){
var enroll= $('#employee_list').val();
if(enroll != '')
{
$.ajax({
url:"fetch.php",
method:"POST",
data:{enroll:enroll},
dataType:"JSON",
success:function(data)
{
if(data.length != 0){
$('#employee_details').css("display", "block");
$('#name').text(data.name);
$('#address').text(data.address);
$('#total_marks').text(data.total_marks);
$('#ph').text(data.ph);
$('#email').text(data.email);
}
else { alert("Please Select Employee"); }
}
})
}
else
{
alert("Please Select Employee");
$('#employee_details').css("display", "none");
}
});
});
</script>
fetch.php
<?php
//fetch.php
if(isset($_POST["enroll"]))
{
$connect = mysqli_connect("localhost", "root", "", "aviation");
$query = "SELECT * FROM student WHERE enroll = '".$_POST["enroll"]."'";
$result = mysqli_query($connect, $query);
if(mysqli_num_rows($result)>0)
{
while($row = mysqli_fetch_array($result))
{
$data["name"] = $row["name"];
$data["address"] = $row["address"];
$data["total_marks"] = $row["total_marks"];
$data["email"] = $row["email"];
$data["ph"] = $row["ph"];
}
echo json_encode($data);
}
}
?>
答案 0 :(得分：1)
你可以添加
error: function(){
alert('No Such Data Found!');
}
在success:function(data)
所以你的ajax调用就像：
$.ajax({
url:"fetch.php",
method:"POST",
data:{enroll:enroll},
dataType:"JSON",
success:function(data)
{
if ($.trim(data)){
$('#employee_details').css("display", "block");
$('#name').text(data.name);
$('#address').text(data.address);
$('#total_marks').text(data.total_marks);
$('#ph').text(data.ph);
$('#email').text(data.email);}
else{alert('No Such Data Found!')}
},
error: function(){
alert('error !');
}
或在php代码中添加else，如果结果为空则返回false：
if(mysql_num_rows($result)>0)
{
while($row = mysqli_fetch_array($result))
{
//code ......
}
echo json_encode($data);
}else {
return false;
}
然后你的ajax调用就像：
$.ajax({
url:"fetch.php",
method:"POST",
data:{enroll:enroll},
dataType:"JSON",
success:function(data)
{
if(data != false){
$('#employee_details').css("display", "block");
$('#name').text(data.name);
$('#address').text(data.address);
$('#total_marks').text(data.total_marks);
$('#ph').text(data.ph);
$('#email').text(data.email);}else{alert('No Such Data Found!');}
}
答案 1 :(得分：1)
在PHP代码中，您将使用循环来获取数据库行。但是你的PHP代码就像你只收到一行数据一样。
除非您另有说明，否则我会考虑您只获得1行数据。
您已经检查过mysql是否返回1个或更多结果。 因此，我们所需要的只是一个变量，告诉您是否找到了结果。
示例：
<?php
$output = array(
'result' => 0
);
if (isset($_POST["enroll"])) {
$connect = mysqli_connect("localhost", "root", "", "aviation");
$query = "SELECT name,address,total_marks,email,ph FROM student WHERE enroll = '" . $_POST["enroll"] . "'";
$result = mysqli_query($connect, $query);
if (mysql_num_rows($result) == 1) {
$output['row'] = mysqli_fetch_array($result);
$output['result'] = 1;
}
}
echo json_encode($output);
然后，如果你的javascript代码，你只需要测试data.result是否等于1或0并相应地采取行动。示例：
<script>
$(document).ready(function () {
$('#search').click(function () {
var enroll = $('#employee_list').val();
if (enroll != '')
{
$.ajax({
url: "fetch.php",
method: "POST",
data: {enroll: enroll},
dataType: "JSON",
success: function (data)
{
if(data.result == 1){
$('#employee_details').css("display", "block");
$('#name').text(data.row.name);
$('#address').text(data.row.address);
$('#total_marks').text(data.row.total_marks);
$('#ph').text(data.row.ph);
$('#email').text(data.row.email);
}
else{
alert("No Such Data Found");
}
}
})
} else
{
alert("Please Select Employee");
$('#employee_details').css("display", "none");
}
});
});
</script>
编辑：小心，你还应该在你的mysql查询中使用它之前清理POST变量