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我在这里有 PHP,用于根据名为“ecpt_shrating”的自定义字段中的数据将图像添加到 Wordpress 帖子中。

当放置在 PHP 后,它只是显示原始代码,尽管图像是可见的。

有点新手,对不起,非常感谢任何帮助。

  $value = get_post_meta( get_the_ID(), 'ecpt_shrating', true );
if( $value == hell ) { <img src='/wp-content/uploads/2013/02/image1.gif'>

    } elseif( $value == poor ) { <img src='/wp-content/uploads/2013/02/image2.gif'>

    } elseif( $value == nice ) { <img src='/wp-content/uploads/2013/02/image3.gif'>

    } elseif( $value == angelic ) { <img src='/wp-content/uploads/2013/02/image4.gif'>

    } elseif( $value == heaven ) { <img src='/wp-content/uploads/2013/02/image5.gif'>

    }
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1 回答 1

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您必须确保您的代码放置在<?php ?>(PHP 标记)中。您的代码将无法按原样工作:

<?php
$value = get_post_meta( get_the_ID(), 'ecpt_shrating', true );
$image = '';
if( $value == 'hell' ){
    echo '<img src="/wp-content/uploads/2013/02/image1.gif">';
} elseif( $value == 'poor' ) {
    echo '<img src="/wp-content/uploads/2013/02/image2.gif">';
} elseif( $value == 'nice' ) {
    echo '<img src="/wp-content/uploads/2013/02/image3.gif">';
} elseif( $value == 'angelic' ) {
    echo '<img src="/wp-content/uploads/2013/02/image4.gif">';
} elseif( $value == 'heaven' ) {
    echo '<img src="/wp-content/uploads/2013/02/image5.gif">';
}?>

或者:

<?php
$value = get_post_meta( get_the_ID(), 'ecpt_shrating', true );
$image = '';
switch($value){
    case "hell":
        $image = 'image1.gif';
        break;
    case "poor":
        $image = 'image2.gif';
        break;
    case "nice":
        $image = 'image3.gif';
        break;
    case "angelic":
        $image = 'image4.gif';
        break;
    case "heaven":
        $image = 'image5.gif';
        break;

}
if(!empty($image)){
    echo '<img src="/wp-content/uploads/2013/02/'.$image.'">';
}?>
于 2013-02-11T22:21:54.233 回答