我有以下文本字段：
<input type="text" name="empid" id="empid" tabindex="1" onblur="getname()">
<input type="text" name="name" id="name" tabindex="2"/>
<input type="text" name="city" id="name" tabindex="3"/>
<input type="text" name="state" id="name" tabindex="4"/>
数据库表为：
empid name city state
EMP471 BBB bbbbb cccccc
EMP444 AAA xxxx yyyyyy
我是 php 的新手。我在互联网上找到了一些代码来检索数据。但它不起作用。
Ajax 代码是：
function getname() {
var id=$("#id").val();
$.ajax({
type:"post",
dataType:"text",
data:"id="+id,
url:"getinsdata.php",
success:function(response)
{
$("#name").val(response.name);
$("#city").val(response.city);
$("#state").val(response.state);
}
});
}
而 PHP 代码是
<?php
include 'connection.php';
$id=$_POST['id'];
$id=$_POST['id'];
$query=mysql_query("select name,city,state from ins_master where id=$id");
$result=mysql_fetch_row($query);
echo $result[0];
exit;
?>
当我们选择 empid 时，当使用 AJAX 在 PHP 中触发 Onblur 事件时，相应的名称、城市、状态应该显示在文本框中。
你想实现什么？发送数据并根据查询获得响应？获取一些数据？
我会去
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/4.0.0-beta/css/bootstrap.min.css" integrity="sha384-/Y6pD6FV/Vv2HJnA6t+vslU6fwYXjCFtcEpHbNJ0lyAFsXTsjBbfaDjzALeQsN6M" crossorigin="anonymous">
<script src="https://cdnjs.cloudflare.com/ajax/libs/popper.js/1.11.0/umd/popper.min.js" integrity="sha384-b/U6ypiBEHpOf/4+1nzFpr53nxSS+GLCkfwBdFNTxtclqqenISfwAzpKaMNFNmj4" crossorigin="anonymous"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/4.0.0-beta/js/bootstrap.min.js" integrity="sha384-h0AbiXch4ZDo7tp9hKZ4TsHbi047NrKGLO3SEJAg45jXxnGIfYzk4Si90RDIqNm1" crossorigin="anonymous"></script>
</head>
<body>
<form id="test" method="POST">
<input type="text" id="name" required minlength="5" name="name"/>
<input type="password" id="pw" required name="pw"/>
<input id ="sub" type="submit"/>
</form>
<div id="answer"></div>
</body>
<script>
$("#sub").click(function(event){
event.preventDefault();
query = $.post({
url : 'check_ajax.php',
data : {'name': $('input[name=name]').val(), 'pw': $('#pw').val()},
});
query.done(function(response){
$('#answer').html(response);
});
});
</script>
这是check_ajax.php：
<?php
var_dump($_POST);
?>
在第二个文件中，但这是您应该执行查询并插入/选择的位置
正如人们所说，我们不写代码，而是提供线索，因为它是基础知识/基础知识，我无法帮助更多，因为您必须理解。复制粘贴不是一个好主意
试试这个html
<!DOCTYPE html>
<html>
<head>
<title></title>
<link rel="stylesheet" type="text/css" href="https://cdnjs.cloudflare.com/ajax/libs/jqueryui/1.12.1/jquery-ui.min.css">
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>
<script type="text/javascript" src="https://cdnjs.cloudflare.com/ajax/libs/jqueryui/1.12.1/jquery-ui.min.js"></script>
<script type="text/javascript">
function getname(val) {
$.ajax({
url: 'getinsdata.php',
type: 'POST',
data: 'state_id='+val,
dataType: 'json',
success:function(data){
var len = data.length;
if(len > 0){
var id = data[0]['id'];
var name = data[0]['name'];
var city = data[0]['city'];
var state = data[0]['state'];
document.getElementById('name').value = name;
document.getElementById('city').value = city;
document.getElementById('state').value = state;
}
}
});
}
</script>
</head>
<body>
<form method="post">
<input type="text" name="empid" id="empid" tabindex="1" onblur="getname(this.value);">
<input type="text" name="name" id="name" tabindex="2"/>
<input type="text" name="city" id="city" tabindex="3"/>
<input type="text" name="state" id="state" tabindex="4"/>
</form>
</body>
</html>
而getinsdata.php是
<?php
include('connection.php');
$id = $_POST['state_id'];
$sql = "SELECT * FROM ins_master WHERE id='$id'";
$result = mysqli_query($conn,$sql);
$users_arr = array();
while( $row = mysqli_fetch_array($result) ){
$id = $row['id'];
$name = $row['name'];
$city = $row['city'];
$state = $row['state'];
$users_arr[] = array("id" => $id, "name" => $name, "city" => $city, "state" => $state);
}
// encoding array to json format
echo json_encode($users_arr);
exit;
?>
还有你的连接.php
<?php
$username = "";
$password = "";
$dbname = "";
$conn = mysqli_connect("localhost",$username,$password,$dbname);
if(!$conn){
die("Error in Connecation");
}
?>
放置$dbname= 数据库名称