我正在使用Bootstrap和jQuery datatables插件。现在它同时显示默认样式和Bootstrap样式。例如,分页显示默认分页按钮内的引导按钮。
pi@raspberrypi:~ $ sudo mysql -u root
Welcome to the MariaDB monitor. Commands end with ; or \g.
Your MariaDB connection id is 17
Server version: 10.1.23-MariaDB-9+deb9u1 Raspbian 9.0
Copyright (c) 2000, 2017, Oracle, MariaDB Corporation Ab and others.
Type 'help;' or '\h' for help. Type '\c' to clear the current input statement.
MariaDB [(none)]>
我认为脚本顺序是正确的。可能是什么问题呢?我是否必须在数据表上设置一些东西才能使用bootstrap?
答案 0 :(得分:2)
尝试使用以下链接替换bootstrap和datatables css:https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css https://cdn.datatables.net/1.10.16/css/dataTables.bootstrap.min.css
这些是此处示例中给出的样式表:https://datatables.net/examples/styling/bootstrap.html
它看起来像(在示例中)
的链接src="https://cdn.datatables.net/1.10.16/js/jquery.dataTables.min.js">
必须在
之前引用 src="https://cdn.datatables.net/1.10.16/js/dataTables.bootstrap.min.js">
通过查看代码,您遗漏了标记中的<tbody>
标记。数据表的This is essential。您的标记/ PHP应如下所示:
<table class="table-bordered table-hover table-fixed" id="mytable">
<thead>
<tr>
<th>Confimar</th>
<th>Árbol Completo</th>
<th>Títol</th>
<th>Descripció</th>
<th>Code</th>
<th>Parent Code</th>
</tr>
</thead>
<tbody>
<?php
if ($stmt = sqlsrv_query($conn,$query))
{
while($tmp = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_NUMERIC))
{
//remember to echo out a row for each line returned in the loop
echo '<tr>';
echo setTableFile($tmp);
echo '</tr>'
}
}
?>
</tbody>
</table>