以PHP JQUERY AJAX的形式搜索和显示
我试图在mysql数据库中搜索表单输入值,如果数据库中存在该输入值,那么我希望在同一表单中显示匹配的输入行详细信息。请帮忙,我想我的php代码有问题 谢谢 index.php以PHP JQUERY AJAX的形式搜索和显示,php,jquery,mysql,ajax,search,Php,Jquery,Mysql,Ajax,Search,我试图在mysql数据库中搜索表单输入值,如果数据库中存在该输入值,那么我希望在同一表单中显示匹配的输入行详细信息。请帮忙,我想我的php代码有问题 谢谢 index.php <div class="col-sm-6"> <div class="form-group"> <label for="campaignname">Link</label> <input type="text" class="form
<div class="col-sm-6">
<div class="form-group">
<label for="campaignname">Link</label>
<input type="text" class="form-control" id="link" name="link" placeholder="Link" required>
</div>
</div>
<div class="col-sm-4">
<div class="form-group">
<label for="campaignname">First Name</label>
<input type="text" class="form-control" id="suppfirstname" name="suppfirstname" placeholder="First Name" required>
</div>
</div>
<div class="col-sm-4">
<div class="form-group">
<label for="campaignname">Last Name</label>
<input type="text" class="form-control" id="supplastname" name="supplastname" placeholder="Last Name" required>
</div>
<?php
$connect = mysqli_connect("localhost", "root", "", "test");
$output = '';
if(isset($_POST["link"]))
{
if($_POST["link"] != '')
{
$sql = "SELECT * FROM customertable WHERE link = '".$_POST["link"]."'";
}
else
{
$sql = "SELECT * FROM customertable WHERE link = 'jesusischrist'";
// I dont want to load any data here so used wrong code
}
$result = mysqli_query($connect, $sql);
while($row = mysqli_fetch_array($result))
{
$output = $result;
}
echo $output;
}
链接
名字
姓
Jquery调用Ajax
<script>
$(document).ready(function(){
$('#link').change(function(){
var link = $(this).val();
$.ajax({
url:"php_action/addnewlead/getlinkdata.php",
method:"POST",
data:{link:link},
success:function(response){
$("#suppfirstname").val(response.firstname);
$("#supplastname").val(response.lastname);
}
});
});
});
</script>
$(文档).ready(函数(){
$('#link').change(function(){
var link=$(this.val();
$.ajax({
url:“php_action/addnewlead/getlinkdata.php”,
方法:“张贴”,
数据:{link:link},
成功:功能(响应){
$(“#suppfirstname”).val(response.firstname);
$(“#supplastname”).val(response.lastname);
}
});
});
});
getlinkdata.php
<div class="col-sm-6">
<div class="form-group">
<label for="campaignname">Link</label>
<input type="text" class="form-control" id="link" name="link" placeholder="Link" required>
</div>
</div>
<div class="col-sm-4">
<div class="form-group">
<label for="campaignname">First Name</label>
<input type="text" class="form-control" id="suppfirstname" name="suppfirstname" placeholder="First Name" required>
</div>
</div>
<div class="col-sm-4">
<div class="form-group">
<label for="campaignname">Last Name</label>
<input type="text" class="form-control" id="supplastname" name="supplastname" placeholder="Last Name" required>
</div>
<?php
$connect = mysqli_connect("localhost", "root", "", "test");
$output = '';
if(isset($_POST["link"]))
{
if($_POST["link"] != '')
{
$sql = "SELECT * FROM customertable WHERE link = '".$_POST["link"]."'";
}
else
{
$sql = "SELECT * FROM customertable WHERE link = 'jesusischrist'";
// I dont want to load any data here so used wrong code
}
$result = mysqli_query($connect, $sql);
while($row = mysqli_fetch_array($result))
{
$output = $result;
}
echo $output;
}
因此,首先,正如有人提到的,您应该真正清理您的输入,准备好的语句也是一个很好的实践。我在这里稍微修改了您的原始代码,因为当您知道缺少链接时,没有理由查询数据库。我认为您的问题的一部分是您在最后回显了您的输出,您不能回显数组元素(使用print\u r或var\u dump)。我将响应部分包装在JSON编码中,以将其转换为Javascript可解析字符串,从而使键/值结构保持完整
<?php
try {
if (isset($_POST['link'] && !empty($_POST['link']) {
$mysqli = connect();
// Sanitize your inputs. I'm guessing link is a string?
$link = filter_var($_POST['link'], FILTER_SANITIZE_STRING);
// create query string for prepared statement
$sql = "SELECT firstname, lastname FROM customertable WHERE link = ? LIMIT 1";
// prepare statement and bind variables
$stmt = $mysqli->prepare($sql);
$stmt->bind_param('s', $link);
$stmt->execute();
$result = $stmt->get_result();
$row = $result->fetch_assoc();
$stmt->close();
sendResponse(200, $row);
}
sendResponse(400, ['status' => 'Link not supplied']);
} catch (\Exception $e) {
sendResponse(500, ['status' => $e->getMessage()]);
}
/**
* Sends JSON encoded response to client side with response http code.
*/
function sendResponse($code, $response)
{
http_response_code($code);
echo json_encode($response);
exit();
}
/**
* Handles connecting to mysql.
*
* @return object $connect instance of MySQLi class
*/
function connect()
{
try {
$connect = new \mysqli("localhost", "root", "", "test");
return $connect;
} catch (mysqli_sql_exception $e) {
throw $e;
}
}
$output=$result代码>这行是错误的。应该是$output=$row['FIELDNAME']代码>此外,如果只有一个结果,则不需要循环。用户将如何输入数据库中的确切名称?您可以给出下拉列表或jquery自动完成用户的数据,直接在这样的查询中进行操作是非常危险的!你应该考虑使用。这将有助于感谢老学徒