使用AJAX从Laravel中的数据库检索数据
好吧，我是拉威尔的新手。如果有人能帮我解决这个问题，它可以为我解释很多。我想做的事情是将这个PHP ajax请求转换为laravel ajax。这是一个动态选择框 PHP和ajax代码是：使用AJAX从Laravel中的数据库检索数据,ajax,dynamic,drop-down-menu,laravel,Ajax,Dynamic,Drop Down Menu,Laravel,好吧，我是拉威尔的新手。如果有人能帮我解决这个问题，它可以为我解释很多。我想做的事情是将这个PHP ajax请求转换为laravel ajax。这是一个动态选择框 PHP和ajax代码是： <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Sections Demo</title>
<script type="text/javascript" src="http://ajax.googleapis.com/
ajax/libs/jquery/1.4.2/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function()
{
$(".country").change(function()
{
var id=$(this).val();
var dataString = 'id='+ id;
$.ajax
({
type: "POST",
url: "ajax_city.php",
data: dataString,
cache: false,
success: function(html)
{
$(".city").html(html);
}
});
});
});
</script>
<style>
label
{
font-weight:bold;
padding:10px;
}
</style>
</head>
<body>
<div style="margin:80px">
<label>Country :</label> <select name="country" class="country">
<option selected="selected">--Select Country--</option>
<?php
include('db.php');
$sql=mysql_query("select id,data from data where weight='1'");
while($row=mysql_fetch_array($sql))
{
$id=$row['id'];
$data=$row['data'];
echo '<option value="'.$id.'">'.$data.'</option>';
} ?>
</select> <br/><br/>
<label>City :</label> <select name="city" class="city">
<option selected="selected">--Select City--</option>
</select>
</div>
</body>
</html>
public function city()
{
$city= City::all();
return View::make('country', ['city'=>$city]);
}
Route::post('/contry/city', 'RegistrationController@city');
$(".country").change(function()
{
var id=$(this).val();
var dataString = 'id='+ id;
$.ajax
({
type: "POST",
url: "/country/city",
data: dataString,
cache: false,
dataType : "html",
success: function(html)
{
$(".stage").html(html);
} ,
// code to run if the request fails; the raw request and
// status codes are passed to the function
error: function( xhr, status, errorThrown ) {
alert( "Sorry, there was a problem!" );
console.log( "Error: " + errorThrown );
console.log( "Status: " + status );
console.dir( xhr );
},
// code to run regardless of success or failure
complete: function( xhr, status ) {
alert( "The request is complete!" );
}
});
});
那是,，我不知道，但是，我的意见是，如果你使用的是laravel，为什么要使用mysql\u query？我没有使用mysql query，我正在尝试将mysql查询转换为laravel thingsIs
Route:：post（'/contry/city'，'RegistrationController@city');输入错误？您键入了“contry”。您还可以检查从app/storage/logs/laravel.log
中得到的错误消息吗RegistrationController@city'); 这是一个打字错误。我得到了错误500。如果在您看来，您将如何使用MVC将上面的php代码转换为laravel？
$(".country").change(function()
{
var id=$(this).val();
var dataString = 'id='+ id;
$.ajax
({
type: "POST",
url: "/country/city",
data: dataString,
cache: false,
dataType : "html",
success: function(html)
{
$(".stage").html(html);
} ,
// code to run if the request fails; the raw request and
// status codes are passed to the function
error: function( xhr, status, errorThrown ) {
alert( "Sorry, there was a problem!" );
console.log( "Error: " + errorThrown );
console.log( "Status: " + status );
console.dir( xhr );
},
// code to run regardless of success or failure
complete: function( xhr, status ) {
alert( "The request is complete!" );
}
});
});