我只知道这部分代码对性能Web应用程序不利,因为它在执行过程中花费了很多时间使用两个SQL查询而不是一个SQL查询。我想阅读您对这部分代码的建议。另外,我想阅读您的代码解决方案。该代码工作正常。
我感谢您的帮助。
先感谢您
<html>
<body>
<select name="category">
<option value="category">category name</option>
<?php
$sql = "SELECT category.name as cat, article.name as art from category
JOIN article ON category.id = article.id";
$query = mysqli_query($conn, $sql);
while($row = mysqli_fetch_array($query)){
echo "
<option value='".$row["cat"]."'>".$row["cat"]."</option>";
}
mysqli_close($conn);
?>
</select>
<select name="article">
<option value="articlename">article name</option>
<?php
$sql = "SELECT category.name as cat, article.name as art from category
JOIN article ON category.id = article.id";
$query = mysqli_query($conn, $sql);
while($row = mysqli_fetch_array($query)){
echo "
<option value='".$row["art"]."'>".$row["art"]."</option>";
}
mysqli_close($conn);
?>
</select>
</body>
</html>