尝试显示超链接时解析错误

Question

以下链接给出了错误Parse error: syntax error, unexpected T_STRING。

$link1 = "<a href='http://www.domain.com/path/comments/index.php?submission=".urlencode($submission)."&submissionid=".urlencode($submissionid)."&url=".urlencode($url)."&countcomments=".urlencode($countcomments)."&submittor=".urlencode($submittor)."&submissiondate=".urlencode($submissiondate)."&dispurl=".urlencode($dispurl)."'>'".Comment Link."'</a>'");

知道怎么解决吗？

Answer 1

像这样：

$link1 = "<a href='http://www.domain.com/path/comments/index.php?submission=".urlencode($submission)."&submissionid=".urlencode($submissionid)."&url=".urlencode($url)."&countcomments=".urlencode($countcomments)."&submittor=".urlencode($submittor)."&submissiondate=".urlencode($submissiondate)."&dispurl=".urlencode($dispurl)."'>Comment Link</a>";

通过这样做：

".Comment Link."

PHP认为Comment Link是code，显然不是。{/ p>

Answer 2

你有额外的)和不必要的连接，这是正确的版本：

$link1 = "<a href='http://www.domain.com/path/comments/index.php?submission=".urlencode($submission)."&submissionid=".urlencode($submissionid)."&url=".urlencode($url)."&countcomments=".urlencode($countcomments)."&submittor=".urlencode($submittor)."&submissiondate=".urlencode($submissiondate)."&dispurl=".urlencode($dispurl).">Comment Link</a>";

echo $link1;

Answer 3

1. 如果“Comment Link”是锚标记的文本，它应该在双引号内。但是你把它放在双引号之外，以便它被解释为PHP代码。

   你做的事情如下：

   $link = "<a href='" . $pageurl . "'>" . Comment Link . "</a>"

   但你应该做的事情如下：

   $link = "<a href='" . $pageurl . "'>Comment Link</a>"

2. 你最后还有一个额外的。

3. 所以你的实际代码应该是

   $link1 = "<a href='http://www.domain.com/path/comments/index.php?submission=".urlencode($submission)."&submissionid=".urlencode($submissionid)."&url=".urlencode($url)."&countcomments=".urlencode($countcomments)."&submittor=".urlencode($submittor)."&submissiondate=".urlencode($submissiondate)."&dispurl=".urlencode($dispurl)."'>Comment Link</a>";