我正试图立即对输入变化的关键字进行数据库查询。现在,我得到了一个成功的查询,存储了所有的结果,但是在GET
上显示它们,我的ajax
函数返回false。为什么会这样呢?
:
POST http://example.com/functions/ajax.php
response: the data i need to display
后接
GET http://example.com/functions/ajax.php
response: false
这是我的JS:
<script id="source" language="javascript" type="text/javascript">
$(function () {
$('#url').bind('input', function () {
$(this).val() // get value
$.ajax({
type: 'POST',
url: 'functions/ajax.php',
data: {
url: $('#url').val()
},
success: function (data) //on receive of reply
{
$(function () {
$.ajax({
url: 'functions/ajax.php', //the script to call to get data
data: "", //you can insert url arguments here to pass to api.php
//for example "id=5&parent=6"
dataType: 'json', //data format
success: function (data) //on receive of reply
{
var namePHP = data[1];
var categoryPHP = data[2];
//--------------------------------------------------------------------
// 3) Update html content
//--------------------------------------------------------------------
$('#name').html(namePHP); // # name and #category are input fields I want autofilled
$('#category').html(categoryPHP);
}
});
});
}
});
});
});
</script>
ajax.php
<?php
require_once ('DBconnect.php');
$url = $_POST['url'];
$url = mysqli_real_escape_string($con, $url);
$query = "SELECT * FROM `inserted_posts` WHERE `search_name` = '$url'";
$result = mysql_query($query);
$array = mysql_fetch_array($result);
echo json_encode($array);
?>
您的ajax.php
文件依赖于$_POST
变量。这意味着你必须使用POST
方法发送数据。否则,PHP就看不到它了
将$url的变量赋值改为从$_REQUEST而不是$_POST中提取
$url = $_POST['url'];
$url = $_REQUEST['url'];