PHP 錯誤 - 警告：count()：參數必須是實現 Countable 的數組或對象

Question

完整的錯誤是：“警告：count()：參數必須是在第 54 行的 /app/public/wp-content/themes/site/inc/lists_posts.php 中實現 Countable 的數組或對象”

它說的是第 54 行，這是第一個 if 語句開始的地方..知道如何解決這個問題嗎？

<?php foreach($projects as $key => $project){
$display = true;
$terms = get_the_terms($project->ID,"diadem_project_category");

if(count($terms) == 1){
    // function located in function.php
    $display = check_project_display($terms[0]->term_id);
}

if($display){
    
    $permalink = get_permalink($project->ID);
    $project_caption = get_field('project_caption',$project->ID);
    $type = get_field('type',$project->ID);
    $subscript_number = get_field('subscript_number',$project->ID);
    $download_pdf = get_field('download_pdf',$project->ID);
    $thumbnail = get_the_post_thumbnail_url($project->ID);
    $image_url = wp_get_attachment_image_src( get_post_thumbnail_id($project->ID), 'medium');
    $categories = wp_get_post_terms($project->ID, $category_name, array("fields" => "all"));
    
    $target ="";
    
    $slug = "";
    if($categories){ 
        foreach($categories as $cat){ 
            $slug .= $cat->slug." "; 
            
            if($category_name == "diadem_insight_category"){
                $project_caption = $cat->name;
            }
            
        } 
    }
    
    $project->post_title = str_replace("'","’",$project->post_title);

    
?>

Answer 1

有時在獲取數據時，會發生一些錯誤，返回false而不是結果集。 所以在檢查數據長度之前，我們也應該比較輸出類型。

if($terms !== false && count($terms) == 1){
    // function located in function.php
    $display = check_project_display($terms[0]->term_id);
}

而且我認為您可能會得到n術語，而不僅僅是一個。 在這種情況下

if($terms !== false && count($terms) >= 1){
    // function located in function.php
    $display = check_project_display($terms[0]->term_id);
}

Answer 2

首先使用var_dump($terms)或echo gettype($terms)檢查變量$terms的數據類型。它必須是數組或對象，然后才不會發生警告。您使用的count()方法期望參數數據類型應該是array或object 。