PHP代码解析错误：语法错误，意外

Question

我想为此查询建立一个链接

这是我的代码

$getfood = "<a href="./student.php?id=">" . $food['clickid'] . "</a>" $food['clickid'];

我收到错误：Parse error: syntax error, unexpected '/'

如何以正确的方式编写代码..

应该像<a href="./student.php?id=33">33</a> 33

Answer 1

  

“它应该像<a href="./student.php?id=33">33</a> 33”

以下是代码：

<?php
$food = array("clickid" => 33);
$getfood = "<a href=\"./student.php?id=" .
    $food['clickid'] . "\">". $food['clickid'] . "</a> " . $food['clickid'];
print $getfood;

打印：

<a href="./student.php?id=33">33</a> 33

记住，我逃脱了引号并用一个点连接了每一个字符串。

Answer 2

你需要逃避双引号字符。

$getfood = "<a href=\"./student.php?id=\">" . $food['clickid'] . "</a>" . $food['clickid'];