[英]PHP Database ajax not updating
A while ago i made a search function with ajax and php. 前一阵子我用ajax和php进行了搜索。 You could fill in a textbox with text and it would try to find a match among all countries stored in the database.
您可以在文本框中填写文本,然后尝试在数据库中存储的所有国家/地区中找到匹配项。 Now i am refining the code and making it PDO, but i broke something and i cant find out what.
现在,我正在完善代码并使其成为PDO,但是我弄坏了一些东西,却找不到。
this is my plain HTML 这是我的纯HTML
<head>
<title>Ajax</title>
<link href="style/style.css" rel="stylesheet" type="text/css" />
<link rel="stylesheet" type="text/css" />
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
<script type="text/javascript" src="scripts/Javascript.js"></script>
</head>
<body>
<div id="main">
<h1 class="title">Enter your country please</h1>
<input type="text" id="search" autocomplete="off" onchange="">
<h4 id="results-text">Showing results for: <b id="search-string">Array</b></h4>
<ul id="results"></ul>
</div>
</body>
here is my Jquery and javascript. 这是我的Jquery和javascript。 note i have not changed anything to the HTML nor javascript so it can not by a type error.
请注意,我没有更改HTML或javascript的任何内容,因此它不会出现类型错误。
$(document).ready(function() {
alert('asdf');
function search() {
var query_value = $('input#search').val();
$('b#search-string').html(query_value);
if(query_value !== ''){
$.ajax({
type: "POST",
url: "search.php",
data: { query: query_value },
cache: false,
success: function(html){
$("ul#results").html(html);
}
});
}
return false;
}
$("input#search").live("keyup", function(e) {
clearTimeout($.data(this, 'timer'));
var search_string = $(this).val();
if (search_string == '') {
$("ul#results").fadeOut();
$('h4#results-text').fadeOut();
}
else {
$("ul#results").fadeIn();
$('h4#results-text').fadeIn();
$(this).data('timer', setTimeout(search, 100));
};
});
});
And here is my Search.PHP 这是我的Search.PHP
<?php
class SearchEngine{
private $html;
public function __construct($conn){
$this->html = '<li class="result">
<h3>NameReplace</h3>
<a target="_blank" href="ULRReplace"></a>
</li>';
if (isset($_POST["query"])) {
$search_string = mysql_real_escape_string($search_string);
}
else{
$search_string = 'b';
}
if (strlen($search_string) >= 1 && $search_string !== ' ') {
$query = 'SELECT * FROM country WHERE name LIKE "%' . $search_string . '%"';
$result = $conn->prepare($query);
$result->execute();
$result_array = $result->fetchAll();
foreach ($result_array as $result) {
$display_name = preg_replace("/" . $search_string . "/i", "<b>" . $search_string . "</b>", $result['name']);
$display_url = 'sadf';
$output = str_replace('NameReplace', $display_name, $this->html);
$output = str_replace('ULRReplace', $display_url, $output);
echo($output);
}
}
}
}
?>
The problems: 问题:
the Post query is never created, for this i made a isset so for now when there is no Post Query created. 永远不会创建Post查询,为此,我进行了一个isset设置,因此现在没有创建Post Query时。 It will create a Post Query with value "B".
它将创建一个值为“ B”的后查询。
I think the page never gets updated, but i cant be 100% sure since the post never gets created so there is never a update to the query. 我认为该页面永远不会更新,但是我不能100%确定,因为该帖子永远不会被创建,所以查询永远不会更新。 And for some reason the results are not placed in there correct spot.
由于某种原因,结果没有放在正确的位置。
Any help will be much appreciated. 任何帮助都感激不尽。 Please be gentle i am new to Ajax and i rather want to understand than have the solution.
请保持谦虚,我是Ajax的新手,我想了解而不是找到解决方案。 Thank you
谢谢
Are you really posting your search string to a PHP class? 您是否真的将搜索字符串发布到PHP类?
That's not how it works, you need to create an instance of that class and use it. 这不是它的工作方式,您需要创建该类的实例并使用它。
Create a separate php file for that class, and included it in search.php
为该类创建一个单独的php文件,并将其包含在
search.php
You can have a simple class looking like the following: 您可以拥有一个类似于以下内容的简单类:
search_engine.php search_engine.php
<?php
class SearchEngine{
private $conn;
function __construct($conn){
$this->conn = $conn;
}
function get_search_results($search_string){
$query = 'SELECT * FROM country WHERE name LIKE :search';
$result = $this->conn->prepare($query);
$result->execute(array(':search'=>'%'.$search_string.'%'));
$result_array = $result->fetchAll();
return $result_array;
}
}
?>
search.php search.php中
if (isset($_POST["query"])) {
$search_string = $_POST["query"];
if(strlen($search_string) >= 1 && $search_string !== ' ') {
include 'search_engine.php'
$engine = new SearchEngine($conn);
$results = $engine->get_search_results($search_string);
foreach($results as $result){
//do something
}
}
}
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