Question

以下鏈接給我錯誤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>'"); 

任何想法如何解決它？

Answer 1

你有額外的)和不必要的級聯，這裏是正確的版本：

$link1 = "<a href='http://www.domain.com/path/comments/index.php?submission=".urlencode($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 2

像這樣：

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

通過這樣做：

".Comment Link." 

PHP認爲Comment Link爲code，它顯然不是。

Answer 3

1. 如果「評論連接」是錨標記的文本，它應是雙引號中。但是你已經把它放在雙引號之外，以這種方式解釋爲PHP代碼。

   你正在做的是這樣的：

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

   但是，你應該做的是這樣的：

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

2. 您也有一個額外的）到底。

3. 因此您的實際代碼應該

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