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PHP echo中无法识别Javascript location.href_Javascript_Php - Fatal编程技术网

PHP echo中无法识别Javascript location.href

PHP echo中无法识别Javascript location.href,javascript,php,Javascript,Php,我正试图让onclick=“javascript:location.href='where.ever'在php echo中工作,以便在显示下拉列表时能够单击整行 我有以下代码: echo "<div class='dropdown-item-row row' onclick='javascript:location.href='" . $row['username'] . "'> <div class='dd-pic'>

我正试图让
onclick=“javascript:location.href='where.ever'
在php echo中工作,以便在显示下拉列表时能够单击整行

我有以下代码:

echo "<div class='dropdown-item-row row' onclick='javascript:location.href='" . $row['username'] . "'>
            <div class='dd-pic'>
                <a href='" . $row['username'] . "'>
                    <img src='" . $row['profile_pic'] ."'>
                </a>
            </div>
            <div class='dd-right'>
                <div class='dd-title'>
                    <a href='" . $row['username'] . "'>" . $row['first_name'] . " " . $row['last_name'] . "</a>
                </div>
                <div class='dd-user-url'>
                    <a href='" . $row['username'] . "'>" . $row['username'] ."</a>
                </div>
                <p class='dd-common-friends'>" . $mutual_friends ."</p>
            </div>
        </div>";
我在回声中表达的方式有问题吗


任何帮助都将不胜感激。

始终查看结果输出。您的输出如下所示:

<div class='dropdown-item-row row' onclick='javascript:location.href='where.ever''>

另外,如果退出PHP模式输出HTML,您可能会发现事情更整洁。使用
返回PHP模式以输出值。这种技术通常在IDE中对代码着色效果更好!

我尝试在整个div中使用tage,但也不起作用。回音将被放入另一个div中,但是,这可能是问题?我不知道,但这里根本不需要JavaScript。
<div class='dropdown-item-row row' onclick='javascript:location.href='where.ever''>
?>
<a href="<?=$row['username']?>" class="dropdown-item-row row">
    <div.....
    .....>
</a>
<?php