WordPress 条件自定义字段

发布于 2024-10-06 19:27:21 字数 872 浏览 0 评论 0原文

我想在 WordPress 安装的单页中输出自定义字段。基本上，我想在页面中输出相关的创意通用许可类型。因此，基本上，带有 cc-license 键的自定义字段可能只有一个值，但该值必须是六个创意通用许可证之一。

假设我已经插入了密钥 cc-license 的值。这是我的代码：

<?php 
$nilai = get_post_meta($post->ID, 'cc-license', true);
echo $nilai; //just want to check the output,its ok!
if ($nilai = 'Attribution Non-commercial Share Alike') { ?>
<a rel="license" href="http://creativecommons.org/licenses/by-sa/3.0/"><img alt="Creative Commons License" style="border-width:0" src="http://i.creativecommons.org/l/by-sa/3.0/88x31.png" /></a><br />This work is licensed under a <a rel="license" href="http://creativecommons.org/licenses/by-sa/3.0/">Creative Commons Attribution-ShareAlike 3.0 Unported License</a>                    
<?php } ?>

问题是，如果我将值更改为任何文本，它仍然会实现条件语句中的 HTML 代码。哪一部分是错误的？

原文

I want to output a custom field in my single page of my WordPress install.
Basically, I want to output the related creative common license type in the page. So basically the custom field with key cc-license may have one value only, but the value must be one of the six creative common licenses.

Assuming I already insert the value for the key cc-license. So here is my code:

<?php 
$nilai = get_post_meta($post->ID, 'cc-license', true);
echo $nilai; //just want to check the output,its ok!
if ($nilai = 'Attribution Non-commercial Share Alike') { ?>
<a rel="license" href="http://creativecommons.org/licenses/by-sa/3.0/"><img alt="Creative Commons License" style="border-width:0" src="http://i.creativecommons.org/l/by-sa/3.0/88x31.png" /></a><br />This work is licensed under a <a rel="license" href="http://creativecommons.org/licenses/by-sa/3.0/">Creative Commons Attribution-ShareAlike 3.0 Unported License</a>                    
<?php } ?>

The problem is, if I change the value to any text, it still implement the HTML code in the conditional statement. Which part is wrong?

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坠似风落 2024-10-13 19:27:21

试试这个，我想你可能刚刚错过了一个=

<?php 
    $nilai = get_post_meta($post->ID, 'cc-license', true);

    echo $nilai; //just want to check the output,its ok!

    if ($nilai == 'Attribution Non-commercial Share Alike') : ?>

        <a rel="license" href="http://creativecommons.org/licenses/by-sa/3.0/"><img alt="Creative Commons License" style="border-width:0" src="http://i.creativecommons.org/l/by-sa/3.0/88x31.png" /></a><br />
        This work is licensed under a <a rel="license" href="http://creativecommons.org/licenses/by-sa/3.0/">Creative Commons Attribution-ShareAlike 3.0 Unported License</a>  

    <?php endif;
?>

Try this, I think you might have just missed a =

<?php 
    $nilai = get_post_meta($post->ID, 'cc-license', true);

    echo $nilai; //just want to check the output,its ok!

    if ($nilai == 'Attribution Non-commercial Share Alike') : ?>

        <a rel="license" href="http://creativecommons.org/licenses/by-sa/3.0/"><img alt="Creative Commons License" style="border-width:0" src="http://i.creativecommons.org/l/by-sa/3.0/88x31.png" /></a><br />
        This work is licensed under a <a rel="license" href="http://creativecommons.org/licenses/by-sa/3.0/">Creative Commons Attribution-ShareAlike 3.0 Unported License</a>  

    <?php endif;
?>

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