连接HTML和PHP以建立图像链接 [英] Concatenating HTML and PHP to make an image link

问题描述

我正在尝试使用PHP和HTML将图像制作成链接.主要思想是从Twitter上获取用户的图像和屏幕名称，然后通过构建URL并在其末尾添加其屏幕名称，使图像成为指向其个人资料的可点击链接.但是我收到一条错误消息:

I'm trying to make an image into a link using PHP and HTML. The main idea is to grab user's images and screen names from Twitter, then make the image into a clickable link to their profile by building the URL and adding their screen name on the end. But I get a error message:

解析错误:语法错误，意外的T_CONSTANT_ENCAPSED_STRING，期望为'，'或';'在第71行的C:\ wamp \ www \ fyp \ tweeter3.php中.

Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING, expecting ',' or ';' in C:\wamp\www\fyp\tweeter3.php on line 71.

这是第71行(它是foreach循环的一部分):

This is line 71 (it's part of a foreach loop):

<?php echo "<a href = ".$url"><img src = ".$userImage." class = ".$class."></a>"; ?>

我无法查明其中存在语法错误. 这些是我的变量:

There's a syntax error in there I just can't pinpoint. These are my variables:

$userScreenName = $user -> screen_name;
$userImage = $user -> profile_image_url;
$class = "myImgClass";
$url = "https://twitter.com/".$userScreenName;

您能发现错误吗?

推荐答案

您在$ url和HTML引号后缺少一个点以生成有效代码:

You are missing a dot after $url and the HTML quotes to generate valid code:

<?php echo "<a href = '".$url."'><img src = '".$userImage."' class = '".$class."'></a>"; ?>

没有引号，您将得到:

<a href = the url><img src = user image class = the class></a>

带引号:

 <a href = 'the url'><img src = 'user image' class = 'the class'></a>

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