PHP、MYSQL、带表排序器的HTML表
试图添加到我正在创建的页面。我对jquery知之甚少,所以我猜这就是我的错所在。我已经在页面的PHP、MYSQL、带表排序器的HTML表,php,jquery,mysql,tablesorter,Php,Jquery,Mysql,Tablesorter,试图添加到我正在创建的页面。我对jquery知之甚少,所以我猜这就是我的错所在。我已经在页面的区域添加了所需的代码,并对表进行了必要的更改。我的表仍然像使用HTML一样呈现。想法 <html> <head> <title>Inventory</title> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/2.0.3/jq
区域添加了所需的代码,并对表进行了必要的更改。我的表仍然像使用HTML一样呈现。想法
<html>
<head>
<title>Inventory</title>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/2.0.3/jquery.min.js"></script>
<script type="text/javascript" src="http://tablesorter.com/__jquery.tablesorter.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){ $("table").tablesorter(); });
</script>
</head>
<body>
<?php
$con=mysqli_connect("localhost","user","pass","db_name");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$query = "SELECT
products.name,
products.sku,
inventory.quantityfry,
inventory.quantityjuv,
inventory.quantityadult,
inventory.notes,
inventory.location,
inventory.owner
FROM
products
INNER JOIN
inventory
ON
products.sku=inventory.sku";
$result = mysqli_query($con,$query) or die(mysqli_error($con));
echo "<table border='1' id='table' class='tablesorter'>
<thead>
<tr>
<th>Species</th>
<th>SKU</th>
<th>Fry Count</th>
<th>Juvie Count</th>
<th>Adult Count</th>
<th>Notes</th>
<th>Location</th>
<th>Owner</th>
</tr>
</thead>";
while ($row = mysqli_fetch_assoc($result)) {
echo "<tbody>";
echo "<tr>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['sku'] . "</td>";
echo "<td>" . $row['quantityfry'] . "</td>";
echo "<td>" . $row['quantityjuv'] . "</td>";
echo "<td>" . $row['quantityadult'] . "</td>";
echo "<td>" . $row['notes'] . "</td>";
echo "<td>" . $row['location'] . "</td>";
echo "<td>" . $row['owner'] . "</td>";
echo "</tr>";
echo "</tbody>";
}
mysqli_free_result($result);
echo "</table>";
mysqli_close($con);
?>
</body>
</html>
库存
$(document).ready(function(){$(“table”).tablesorter();});
谢谢 三件事:
,否则IE会变成怪癖模式,让你的网站看起来很糟糕tbody
中,请按如下方式更正代码(这只是一个片段):
echo”
种
SKU
鱼苗数
朱维伯爵
成虫计数
笔记
地方
物主
";
while($row=mysqli\u fetch\u assoc($result)){
回声“;
回显“$row['name']”;
回显“$row['sku']”;
回显“$row['quantityfry']”;
回显“$row['quantityjuv']”;
回显“$row['Quantity成人”“”;
回显“$row['notes']”;
回显“$row['location']”;
回显“$row['owner']”;
回声“;
}
mysqli_免费_结果($result);
回声“;
您正在为每一行添加一个tbody标记。很容易集成您可能还需要
$(“#table”).datasorter()
。不知道datasorter是否需要一个特定的元素,或者将自己附加到由$('table')
返回的所有表。为什么不直接在源代码处对其进行排序呢?将echo放置在“;回声“代码>和echo”“;回声“代码>在之外,而循环。
echo "<table border='1' id='table' class='tablesorter'>
<thead>
<tr>
<th>Species</th>
<th>SKU</th>
<th>Fry Count</th>
<th>Juvie Count</th>
<th>Adult Count</th>
<th>Notes</th>
<th>Location</th>
<th>Owner</th>
</tr>
</thead><tbody>";
while ($row = mysqli_fetch_assoc($result)) {
echo "<tr>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['sku'] . "</td>";
echo "<td>" . $row['quantityfry'] . "</td>";
echo "<td>" . $row['quantityjuv'] . "</td>";
echo "<td>" . $row['quantityadult'] . "</td>";
echo "<td>" . $row['notes'] . "</td>";
echo "<td>" . $row['location'] . "</td>";
echo "<td>" . $row['owner'] . "</td>";
echo "</tr>";
}
mysqli_free_result($result);
echo "</tbody></table>";