如何用变量回显超链接？ [英] How to echo a hyperlink with a variable?

问题描述

正确地获取这些超链接并将HTML与PHP混合似乎是我不断的悲伤源泉。这个超链接怎么了？

Getting these hyperlinks right and mixing HTML with PHP seems to be a constant source of sorrow for me. Whats wrong with this hyperlink?

echo '<a href = "googler.php?query='.$suggestion[$ss_count].'">$suggestion[$ss_count]</a><br>;

推荐答案

它应该是

echo '<a href = "googler.php?query='.$suggestion[$ss_count].'">'.$suggestion[$ss_count].'</a><br>';

在单引号内，变量不会被插值，因此您必须从字符串文字中提取链接文本。

Within single quotes variables are not interpolated so you have to pull the link text out of the string literal.

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