所以我正在学习如何使用jQuery,PHP,AJAX和MySQL。我想从MySQL服务器获取一个表并显示它。为此,我有2个PHP文件,一个是登录到MySQL数据库并回显一个html表的服务器端文件,另一个是用户看到的index.php文件。我使用jQuery的ajax加载方法在单击按钮时更新页面,并返回服务器端PHP的echo表。当我这样做时,表的边框永远不会被加载。有人可以帮我这个吗? Screen Shot
indexTest.php(连接到MySQL服务器/服务器端的php文件):
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
else {
$sql = "select name, species from pet";
$result = $conn->query($sql);
$counter = mysqli_num_rows($result);
if ($counter > 0) {
echo "<table id = 'pets'>
<tr>
<td><strong>names</strong></td>
<td><strong>species</strong></td>
</tr>";
while ($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['species'] . "</td>";
echo "</tr>";
}
echo "</table>";
}
}
$conn->close();
的script.js:
$(document).ready(function(){
$("button").click(function(){
$("#fromAjax").load("indexTest.php", true);
});
});
index.php(我正在加载表中的那个!):
<!DOCTYPE html>
<html>
<head>
<script src='jQuery-2.1.3.js'></script>
<script type='text/javascript' src='script.js'></script>
<link rel = "stylesheet" href = "stylesheet.css" type = "stylesheet/css"></link>
</head>
<body>
<div id="myDiv"><p id = "fromAjax">This will be replaced with the Table</p></div>
<button>Change Content</button>
</body>
</html>
我的CSS:
*{
font-family; "Impact", "Times New Roman";
}
#fromAjax{
color: red;
}
#pets{
color: blue;
border: 5px solid black;
}
答案 0 :(得分:1)
如同在提到的评论中一样,从表格的ID中删除#:
echo "<table id = 'pets'>
然后添加以下CSS
#pets td{ border: 5px solid pink; }
你的桌子上应该有一个边框:-)