dongxuan1314
dongxuan1314
2015-12-01 13:32

在显示XML数据的.php文件中显示PHP变量

Basically I have a live search suggestions which uses AJAX. I'm not familiar with AJAX or XML however i've managed to get it working without any errors.

If you take a look at the code file you can see how I would like the XML data to be generated which is the title and url tags to be populated by a database of products. Instead of displaying the database string the live search suggestions is displaying everything between the XML tags for example between the title tags: ".$aa." .

Can anyone please help me work around this issue. It's most likely a silly mistake or an easy solution.

<link>
<?php
header("Content-type: text/xml");
echo "<?xml version='1.0' encoding='UTF-8'?>";
$productdetails = mysql_query("SELECT * FROM products");

while($row = mysql_fetch_array($productdetails)){
    $aa = $row['product_name'];
    $bb = $row['link'];
    echo "<title>".$aa."</title>";
    echo "<url>".$bb."</url>";
    
}
?>
</link>

</div>
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1条回答

  • dongrui6787 dongrui6787 6年前

    There were link tags before and after the code but oin the php you tried sending a header - I think this might be more like what you are trying to do.

    <?php
        header("Content-type: text/xml");
        $productdetails = mysql_query("SELECT * FROM products");
    
        echo "<?xml version='1.0' encoding='UTF-8'?>";
        while( $row = mysql_fetch_array( $productdetails ) ) {
            echo "
            <link>
                <title>".$row['product_name']."</title>
                <url>".$row['link']."</url>
            </link>";
        }
    ?>
    
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