PHP代码块中的php动态链接

Question

我最近学会了通过php创建动态链接，但我今天的问题是我不知道如何在php代码块中做到这一点，例如我知道如何做到这一点：

     <div><a href="blah.php?id=<?php echo $id; ?>">content</a></div>

但我是在php代码块的head标签上做这个的，所以基本上我是在为动态内容创建一个动态链接，基本上我要在主页上动态呈现图像，这些图像是指向产品的链接。这就是我所拥有的：

    <?php 
     include_once "scripts/connect_to_mysql.php";
     //Select query for latest items
     $dynamic_newest = "";
     $sql = mysql_query("SELECT * FROM products ORDER BY date_added DESC LIMIT 24");
     $productCount = mysql_num_rows($sql); // count the output amount
     if ($productCount > 0) {
          while($row = mysql_fetch_array($sql)){
              $id = $row["id"];
              $pid = $id;
              $category = $row["category"];
              $subcategory = $row["subcategory"];
              $dynamic_newest .= "<a href="THIS IS WHAT I DONT KNOW HOW TO DO"><img src='inventory_images/$pid.jpg' width='100px' height='100px' /></a>";
        }
          } else {
          $dynamic_newest = "<h1>There are no products to display yet.</h1>";
          }
   ?>

这可能真的很简单，但我找不到它，也许我没有问到magic google正确的问题。提前感谢！

Answer 1

我不确定这是否是你要找的，但不管怎样..

$dynamic_newest .= "<a href='blah.php?id=$pid'><img src='inventory_images/$pid.jpg' width='100px' height='100px' /></a>";

Answer 2

逃避问题，改变：

 $dynamic_newest .= "<a href="THIS IS WHAT I DON'T KNOW HOW TO DO"><img src='inventory_images/$pid.jpg' width='100px' height='100px' /></a>";

至：

 $dynamic_newest .= "<a href=\"blah.php?id=$id\"><img src='inventory_images/$pid.jpg' width='100px' height='100px' /></a>";

Answer 3

虽然您现在不使用它，但我相信您将来会发现它的用处。稍后，当您需要编写大量预格式化的非php文本块并想要向其中添加php变量时，您可以这样做：

echo<<<YOUR_IDENTIFIER

<h1>Hi!</h1>
<form action='post'>
Welcome to my webpage =).
You can place code here as if it were not inside a PHP block,
but you can also use PHP variables. That means you can even
insert quotes like this --> "", though since this is still HTML,
it would be more accurate to use &gt; and &quot;. Your cleanest
bet and best-practice coding method is to {$encapsulate} php
variables in squiggley brackets. You can even
{$encapsulateArrays['likethis']}.
</form>

YOUR_IDENTIFIER;

echo "Back to regular PHP code.";

确保YOUR_IDENTIFIER;前没有空格或制表符。

要回答你最初的问题(与我上面的长篇大论无关)，请确保正确地使用反斜杠\任何引号进行转义，以免意外终止字符串文字。别忘了用{}封装你的变量，即使你在做常规的echo "My {$variable} here";，它也不会影响解析，但当你在做下一个项目时，2个月后的调试会变得容易得多。