发送AJAX获取请求
[英]Send AJAX get request
问题描述
I just started to study AJAX and I already have some trouble. 
我刚开始学习AJAX，已经遇到了一些麻烦。
I created database and table users with id, name, email fields. 我创建了具有id, name, email字段的数据库和表users 。
Then I created php file db.php for connection with the database 然后我创建了php文件db.php来与数据库连接
<?php
$dblocation = "localhost";
$dbname = "test5";
$dbuser = "root";
$dbpasswd = "";
$dbcnx = @mysql_connect($dblocation, $dbuser, $dbpasswd);
if (!$dbcnx) {
exit ( "<p>Error.</p>" );
}
if ( ! @mysql_select_db($dbname, $dbcnx)) {
exit ( "<p>Error.</p>" );
}
?>
,created file get-user.php to return an answer for ajax-request ，创建文件get-user.php返回ajax请求的答案
<?php
include ("db.php");
$result = mysql_query("SELECT name, email FROM users WHERE id='$id'", $dbcnx);
return $result;
?>
And this is my index.php file 这是我的index.php文件
<!DOCTYPE html>
<html>
<head>
<title>index</title>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script>
$(document).ready(function () {
$('#button-1').on('click', function () {
var id = $('#user-id').val();
$.ajax({
type: 'GET',
url: 'get-user.php',
datatype: "html",
data: id,
success: function (data) {
alert(data);
}
});
});
});
</script>
</head>
<body>
<h2>Title</h2>
<input name="id" id="user-id" type="text" size="15" />
<button id="button-1">Get data</button>
</html>
I would like to input the number, click button and get information about user from database, but I get only empty alert. 我想输入数字，单击按钮并从数据库中获取有关用户的信息，但是我只会收到空警报。 How to make it work? 如何使其运作？
Sorry for asking, I'm just newbie. 抱歉问，我只是新手。
5 个解决方案
解决方案1
3 已采纳 2014-01-16 18:29:28
These changes are require: 这些更改是必需的：
Ajax: data: id, to data: "{id : id}, Ajax： data: id,至data: "{id : id},
PHP: return $result; PHP： return $result; to echo $result; echo $result;
PHP & MySQL: PHP和MySQL：
$result = mysql_query("SELECT name, email FROM users WHERE id='$id'", $dbcnx);
$row = mysql_fetch_array($result);
return $result[0] . ";" . $row[1];
解决方案2
1 2014-01-16 18:29:44
The correct way to pass the parameters is this 传递参数的正确方法是这样
$.ajax({
type: 'GET',
url: 'get-user.php',
datatype: "html",
data: { id: id },
success: function (data) {
alert(data);
}
});
In the php return the value like this 在php返回这样的值
<?php
include ("db.php");
$result = mysql_query("SELECT name, email FROM users WHERE id='$id'", $dbcnx);
$fetch=mysql_fetch_array($result);
return $fetch["email"];
?>
解决方案3
1 2014-01-16 18:33:39
Try this, 尝试这个，
Javascript: Javascript：
$.ajax({
type: 'GET',
url: 'get-user.php',
datatype: "html",
data: {id:id},
success: function (data) {
alert(data);
}
});
in get-user.php: 在get-user.php中：
<?php
if(isset($_GET['id'])){
include ("db.php");
$id = $_GET['id'];
....
}
?>
Note: Use mysqli_* functions or PDO instead of mysql_* functions(deprecated) 注意：使用mysqli_ *函数或PDO代替mysql_ *函数（已弃用）
解决方案4
1 2014-01-16 18:35:15
You are very close, but there are a few things you need to change. 您非常亲密，但是需要更改一些内容。
first inside of your ajax call: 首先在您的ajax调用中：
$.ajax({
type: 'GET',
url: 'get-user.php',
datatype: "html",
data: {'id': id},
success: function (data) {
alert(data);
}
});
notice the change in the data option. 注意data选项中的更改。
Second, inside your get-user.php you refer to the value using $_GET['id'] . 其次，在您的get-user.php中，使用$_GET['id']引用值。
Finally, instead of return you want to echo the response. 最后，您要echo显响应而不是return 。
解决方案5
1 2014-01-16 18:36:22
Ajax code Ajax代码
data: {id: id};
php code PHP代码
use pdo instead of mysql_query, the code you use on mysql_query its not prevent sql injection. 使用pdo代替mysql_query，在mysql_query使用的代码不会阻止sql注入。
include ("db.php");
$id = $_GET['id'];
$stmt = $pdo->prepare('SELECT name, email FROM users WHERE id = :id');
$stmt->execute(array('id' => $id));
$result = $stmt->get_result();
echo $result;
$stmt = $dbConnection->prepare('SELECT name, email FROM users WHERE id = ?');
$stmt->bind_param('id', $id);
$stmt->execute();
$result = $stmt->get_result();
echo $result;
声明:本站的技术帖子网页，遵循CC BY-SA 4.0协议，如果您需要转载，请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.