[英]passing data from javascript using ajax and saving it to a database
嗨,大家好,我目前有一个网站,允许用户搜索网站,我想将这些搜索词以及每次搜索的次数保存在数据库中。 为此,我创建了一个具有两列搜索的表,该表是varchar(30),而将值int(30)搜索设置为唯一,因此实例设计的术语不会填充两行,但是当我运行程序时却什么也没有正在保存到数据库中,我想知道是否有人可以帮助您。
javascript
function performSearch() {
var searchURL = "http://api.lmiforall.org.uk/api/v1/soc/search?q=";
var searchTerms = $('#searchterm').val();
var searchReady = searchTerms.toLowerCase();
$.ajax({
type: "POST",
url: 'findfavourites.php',
data: {searchReady : searchReady},
success: function(data)
{
alert("success!");
}
});
$('#soctable tbody').empty();
$.getJSON(searchURL + searchReady, function(results) {
results.forEach(function (result) {
var row = $("<tr></tr>");
var codeCell = $("<td></td>");
var titleCell = $("<td></td>");
var descriptionCell = $("<td></td>");
var qualificationsCell = $("<td></td>");
var tasksCell = $("<td></td>");
codeCell.html(result.soc);
titleCell.html(result.title);
descriptionCell.html(result.description);
qualificationsCell.html(result.qualifications);
tasksCell.html(result.tasks);
row.append(codeCell);
row.append(titleCell);
row.append(descriptionCell);
row.append(qualificationsCell);
row.append(tasksCell);
$('#soctable tbody').append(row);
});
});
}
$(function() {
// when the page is loaded
$('#dosearch').on('click', performSearch);
});
的PHP
<?php
include 'includes/dbConnection.php';
if(isset($_POST['searchReady']))
{
$uid = $_POST['searchReady'];
$value='1';
$query=$conn->prepare("INSERT INTO favourites VALUES ('$uid','$value') ON DUPLICATE KEY UPDATE value = VALUES(value + 1");
$query->execute();
$conn = null;
}
?>
任何帮助将不胜感激,谢谢。
编辑
@PHPglue
我确定您要的是getjson返回结果的页面(如果是),这是html页面
<!DOCTYPE HTML>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<meta name="viewport" content="width=device-width, intial-scale=1">
<title>Homepage</title>
<link href="styleSheet.css" rel="stylesheet" type="text/css">
<script type="text/javascript" src="jquery-3.1.1.js"></script>
<script type="text/javascript" src="soc.js"></script>
</head>
<body>
<div id="page">
<nav>
<ul>
<li><a href="index.html">Home Page</a></li>
<li><a href="careers.html">Careers</a>
<ul>
<li><a href="#">Careers and Apprenticeships</a></li>
</ul>
</li>
</ul>
</nav>
<header>
<h1><u>Careers and Apprenticeships</u></h1>
</header>
<div class="container">
<div class="row">
<div class="col-sm-12">
<input class="form control" type="text" id="searchterm" placeholder="Put your search term here...">
<button class="btn btn-primary" id="dosearch">Search!</button>
</div>
</div>
</div>
<div class="container">
<div class="row">
<div class="col-sm-12">
<table class="table table-striped" id="soctable">
<thead>
<tr>
<th>SOC Code</th>
<th>Job Title</th>
<th>Description</th>
<th>Qualifications</th>
<th>Tasks</th>
</tr>
</thead>
<tbody>
</tbody>
</table>
</div>
</div>
</div>
</div>
<footer>
</footer>
</body>
</html>
编辑2
大家好,我发现ajax在工作,问题是由ON DUPLICATE KEY UPDATE引起的,任何人都可以检查语法并告诉我是否遇到任何问题。 就像我上面说过的,我希望当搜索相同的术语时将该值更新为2,依此类推。
尝试将sql行修改为:
$query=$conn->prepare("INSERT INTO favourites
VALUES ('$uid','$value')
ON DUPLICATE KEY UPDATE value = value + 1");
希望这可以帮助 :)
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