问题在超链接中使用php变量 [英] Issue with using php variable in hyperlink

问题描述

如何在另一个php变量的超链接中使用一个php变量作为url？下面是我的代码，显然不工作。

How do you use one php variable for the url in a hyperlink around another php variable? Below is my code, which is clearly not working. I want $url to be the hyperlinked text of $songTitle.

if ($result) {
    $numberOfRows = $result->num_rows;

    for($i=0; $i < $numberOfRows; $i++) {
        $row = $result->fetch_assoc();
        echo '<tr>';
        echo '<td>' . $row['artistName'] . '</td>';
        echo '<td>'<a href=". $row['url'] . > . $row['songTitle'] . </a> '</td>';
        echo '<td>' . $row['yOR'] . '</td>';
        echo '</tr>';
    }
}

推荐答案

对于（$ i = 0; $ i <$ numberOfRows; $ i ++）{
$ row = $ result- $ p

You are not concatenating your strings properly.

for($i=0; $i < $numberOfRows; $i++) {
        $row = $result->fetch_assoc();
        echo '<tr>';
        echo '<td>' . $row['artistName'] . '</td>';
        echo '<td><a href="' . $row['url'] . '" >' . $row['songTitle'] . '</a></td>';
        echo '<td>' . $row['yOR'] . '</td>';
        echo '</tr>';
    }

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