PHP合并含results的JSON数组去重失败，求有效解决方法

Hey there! Let's break down exactly what's going wrong with your current code and walk through a reliable solution to merge your three JSON arrays and remove duplicates properly.

What's Wrong with Your Current Approach #

Let's unpack the issues in your code step by step:

• Extra brackets & missing "results" field: When you use array_merge_recursive($array1, $array2, $array3), since all three input arrays have a results key, the function merges the values of those keys into a single nested array. Then when you run array_values(array_unique(...)) on the entire merged array (which only has one results entry), you end up wrapping the entire list in an extra set of brackets—and lose the original results key in the process.
• Duplicates not being removed: array_unique($array, SORT_REGULAR) compares entire array elements as-is. In your data, some cat_id values are integers (e.g., 4 in Array1) while others are strings (e.g., "5" in Array2). These are treated as distinct values by PHP, so duplicates aren't caught. Even if types matched, array_unique isn't the most reliable way to deduplicate objects/arrays based on a unique identifier like cat_id.

Step-by-Step Solution #

Here's a straightforward, robust way to merge your arrays and deduplicate them based on cat_id (while fixing type inconsistencies):

1. First, decode your JSON strings (if you haven't already) #

// Your original JSON data
$json1 = '{"results":[{"cat_id":2,"cat_name":"veterinarian"},{"cat_id":3,"cat_name":"electrician"}, {"cat_id":4,"cat_name":"dentist"}]}';
$json2 = '{"results":[{"cat_id":"8","cat_name":"dental hygienist"},{"cat_id":"5","cat_name":"stocktaker"}, {"cat_id":"9","cat_name":"builder"}]}';
$json3 = '{"results":[{"cat_id":4,"cat_name":"dentist"},{"cat_id":5,"cat_name":"stocktaker"}, {"cat_id":3,"cat_name":"electrician"}]}';

// Decode JSON into associative arrays
$array1 = json_decode($json1, true);
$array2 = json_decode($json2, true);
$array3 = json_decode($json3, true);

2. Merge all the results items into a single list #

Instead of merging the top-level arrays, extract and combine just the results contents:

$allCategories = array_merge($array1['results'], $array2['results'], $array3['results']);

3. Deduplicate using cat_id (with type normalization) #

We'll use cat_id as a unique key in a temporary array to eliminate duplicates. First, normalize cat_id to a consistent type (string or integer) to avoid mismatches between numeric strings and integers:

$uniqueCategories = [];

foreach ($allCategories as $category) {
    // Convert cat_id to string to handle both integer and string values consistently
    $normalizedId = (string)$category['cat_id'];
    // Only keep the first occurrence of each category (remove the if to keep the last occurrence)
    if (!isset($uniqueCategories[$normalizedId])) {
        $uniqueCategories[$normalizedId] = $category;
    }
}

// Convert the associative array back to an indexed array
$uniqueCategories = array_values($uniqueCategories);

4. Wrap back into the original results structure and output #

// Reconstruct the JSON structure with the unique categories
$finalResult = ['results' => $uniqueCategories];

// Output the final JSON
echo json_encode($finalResult);

Alternative: Using array_reduce for Conciseness #

If you prefer a more functional approach, you can use array_reduce instead of the foreach loop:

$uniqueCategories = array_reduce($allCategories, function($carry, $category) {
    $normalizedId = (string)$category['cat_id'];
    $carry[$normalizedId] = $category;
    return $carry;
}, []);

$uniqueCategories = array_values($uniqueCategories);
$finalResult = ['results' => $uniqueCategories];
echo json_encode($finalResult);

Expected Output #

Running this code will give you the clean, deduplicated JSON you want:

{"results":[{"cat_id":2,"cat_name":"veterinarian"},{"cat_id":3,"cat_name":"electrician"},{"cat_id":4,"cat_name":"dentist"},{"cat_id":"8","cat_name":"dental hygienist"},{"cat_id":"5","cat_name":"stocktaker"},{"cat_id":"9","cat_name":"builder"}]}

内容的提问来源于stack exchange，提问作者CHarris