Php 如何合并这两个JSON数组？

My php生成两个JSON数组，如：

[{"category":183,"private_review_ids":[63,59,62]},
{"category":363,"private_review_ids":[331]}, 
{"category":371,"private_review_ids":[341]},
{"category":379,"private_review_ids":[350]}]
[{"category":363,"public_review_ids":[331]},
{"category":373,"public_review_ids":[343]},
{"category":384,"public_review_ids":[356]},
{"category":183,"public_review_ids":[347]}]

我如何合并这些数组，使它们只是下面表单的一个数组。它不是简单地合并数组，而是可能将JSON对象中的一个键（

public\u review\u id

）的值转换为另一个键（

private\u review\u id

）。这是我想要的JSON数组的格式：

[{"category":183,"private_review_ids":[63,59,62],"public_review_ids":[347] },
{"category":363,"private_review_ids":[331],"public_review_ids":[]}, 
{"category":371,"private_review_ids":[341],"public_review_ids":[]},
{"category":379,"private_review_ids":[350]},"public_review_ids":[]},
{"category":373,"private_review_ids":[],"public_review_ids":[343]},
{"category":384,"private_review_ids":[],"public_review_ids":[356]}]

如您所见，如果该值同时出现在

private\u review\u id

和

public\u review\u id

中，则它应该只出现在

private\u review\u id

键中

我试着使用

array\u unique

和

array\u merge

但实际上没有成功

这是我的密码：

<?php
require('myfile.php');

    //here is the user_id, which is the corresponding user_id for username +5555555

$user_id = "21";
//Select all related info in the review_shared table 
//where the contact_id column is equal to $user_id.

//a value in the contact_id column means a review is shared with a person, $user_name,
//who owns that number, $user_id
$sql = "SELECT * FROM review_shared WHERE contact_id = ?";
$stmt2 = $con->prepare($sql) or die(mysqli_error($con));
$stmt2->bind_param('i', $user_id) or die ("MySQLi-stmt binding failed ".$stmt2->error);
$stmt2->execute() or die ("MySQLi-stmt execute failed ".$stmt2->error);
$result2 = $stmt2->get_result();

//fetch all rows associated with the respective contact_id value
//in review_shared table
while ($row = $result2->fetch_assoc()) {

    //get the corresponding cat_id in the row
    $cat_id = $row["cat_id"];

    //get the corresponding review_id in the row
    $review_id = $row["review_id"];
    //make an array called $results
    $results[$row['cat_id']][] = $review_id; 

}

$jsonData = array_map(function($catId) use ($results) {
    return [
        'category' => $catId,
        'private_review_ids' => $results[$catId],
        ];
}, array_keys($results));
echo json_encode($jsonData);


    //**********************

//select all rows where public_or_private column = 2
//in review table
$sql2 = "SELECT * FROM review WHERE public_or_private = 2";
$result2 = mysqli_query($con,$sql2);

    //fetch all associated rows where public_or_private column = 2
    while ($row = $result2->fetch_assoc()) {

    //get the corresponding review_id in the row
    $review2_id = $row["review_id"];

    //get the corresponding cat_id in the row
    $cat2_id = $row["cat_id"];

    //make an array called $results
    $results2[$row['cat_id']][] = $review2_id;              

    }

    $jsonData2 = array_map(function($cat2Id) use ($results2) {
    return [
        'category' => $cat2Id,
        'public_review_ids' => $results2[$cat2Id],
        ];
}, array_keys($results2));
echo json_encode($jsonData2);


?>

---

如果重构代码，可以立即将查询结果附加到所需的数据结构中，从而消除对数据进行4次（每次两次）迭代的需要

如果执行公共和私有审查查询，使其结果为变量

$publicReviews

和

$privateReviews

，则：

<?php

// Public and private review query results
$publicReviews = $stmt1->get_result();
$privateReviews = $stmt2->get_result();

// Prepare combined reviews array
$reviews = [];

// Iterate through private review results and append to combined reviews
while (($row = $privateReviews->fetch_assoc())) {
    $category_id = $row['cat_id'];
    $review_id = $row['review_id'];

    $reviews[$category_id]['category'] = $category_id;
    $reviews[$category_id]['private_review_ids'][] = $review_id;
    $reviews[$category_id]['public_review_ids'] = [];
}

// Iterate through public review results and append to combined reviews
while (($row = $publicReviews->fetch_assoc())) {
    $category_id = $row['cat_id'];
    $review_id = $row['review_id'];

    $reviews[$category_id]['category'] = $category_id;

    // Create empty private reviews array, where it doesn't exist
    if (! isset($reviews[$category_id]['private_review_ids'])) {
        $reviews[$category_id]['private_review_ids'] = [];
    }

    // Add review id to public reviews where it doesn't exist in private reviews
    if (! in_array($review_id, $reviews[$category_id]['private_review_ids'])) {
        $reviews[$category_id]['public_review_ids'][] = $review_id;
    }
}

echo json_encode(array_values($reviews));

---

我在第43行的/var/www/html/CategorySearch.php中得到
解析错误：语法错误，意外的“？”
这是
$reviews[$cat\u id]['public\u review\u id']=$reviews[$cat\u id]['public\u review\u id']？？[];当我删除额外的
？
时，我得到
解析错误：语法错误，意外“；”在第43行的/var/www/html/CategorySearch.php中
您没有使用PHP7+？看起来我有php版本5.5.9-1ubuntu4.21。我应该更新吗？但不是那么老，2017年2月。在任何情况下，我有一个问题得到的结果在变量$Puffic评论，让我看看更多……PHP 5.5是EOL在2016年7月，所以是的，你可能要考虑升级。至于变量，
$publicReviews
，为了清晰起见，我只是在回答中重命名了它们。您只需按正确的顺序获取查询结果，即可使
while
循环正常工作。我将更新我的答案，使其在没有空coalese运算符的情况下工作。Tx，JSON应该是一个数组，是吗？现在的输出是这样的：
{“383”：{“private_-review_-id”：[353]，“public_-review_-id”：[]}，“203”：{“private_-review_-id”：[149]，“public_-review_-id”：[201]，“public_-review_-id”：[]}
我希望它像：
[{“category”：“383”，“private_-review-id”：[353]，“public_-review_-id”：[353]，“public_-review_-id”：，“私人审查ID”：[149]，“公共审查ID”：[]}，“类别”：“239”，“私人审查ID”：[201]，“公共审查ID”：[]}]