PHP - 如果某事等于某事，那么回应一些东西

Question

I'm quite new to PHP now that I've started working with wordpress.

I'm trying to get something to work by using 'if'

Essentialy what I'm wanting to do is

If the status is equal to Open then return <a href="#" id="flashing">Link</a> If the status is not equal to Open then return <a href="#">Link</a>

Here's what I think would work:

<?php
   if ($status) == (open) {
     echo "id=flashing"
   }
?>

Obviosuly, I'm assuming this doesn't work but what I'm wanting to do is create a link

Any help?

Answer 1

This is a really basic PHP syntax question; please read some documentation, and look at some examples before asking for help with every piece of code you write.

There is a comprehensive online manual for PHP , with many examples. It is available in multiple translations, in case English is not your first language.

Things you have wrong in your example, with links to relevant pages of the manual:

• no semi-colon to end the statement
• no quotes around the string open
• brackets in the wrong place in the if statement

The text of your question also confuses return and echo , which have very different meanings.

Answer 2

Does this help?

if ($status == 'open') {
  echo '<a href="#" id="flashing">Link</a>';
} else {
  echo '<a href="#" >Link</a>';
}

Answer 3

Just use the following code:

<?php
    if ($status == 'Open') {
      echo '<a href="#" id="flashing">Link</a>';
    } 
    else {
      echo '<a href="#" >Link</a>';
    }
?>

Answer 4

<?php
if ($status == 'open') 
{
   echo '<a href="link" id="flashing">Link</a>';
}
else
{
   echo '<a href="link" >Link</a>';
}
?>

Answer 5

假设Open是一个字符串，它可以像这样写（或者替代其他答案）：

echo '<a href="#" '.(($status == 'Open') ? 'id="flashing"':'').'>Link</a>';