Question

这是我的JSON输出：

array:2 [
    0 => "[{"subject":"Science","date":"2018-08-09 13:03:14","average":4},{"subject":"Science","date":"2018-09-13 13:18:58","average":4}]"
    1 => "[{"subject":"Social","date":"2018-08-14 05:59:56","average":4}]"
]

我想要什么：

[{"subject":"Science","date":"2018-08-09 13:03:14","average":4},{"subject":"Science","date":"2018-09-13 13:18:58","average":4},{"subject":"Social","date":"2018-08-14 05:59:56","average":4}]

我不确定如何实现这一目标。我尝试了array_merge，array_combine，array_map，json_encode，json_decode以及它们的各种组合！我显然缺少了一些东西。

这是我的代码：

// $classroom_subjects references a table, listing subjects
// for each subject, get the assessment data
foreach ($classroom_subjects as $subject) {
    $loop_graph_data[] = AssessmentData::select('subjects.short_name as subject', 'assessment_data.created_at as date', 'assessment_data.assessed_level as average')
        ->join('subjects', 'subjects.id', 'assessment_data.subject_id')
        ->where('assessment_data.subject_id', $subject->subject_id)
        ->where('assessment_data.student_id', $student_id)
        ->whereBetween('assessment_data.created_at', [$current_term->term_start." 00:00:01", $current_term->term_end." 23:59:59"])
        ->get()
        ->toJson();
}

// create an empty variable
$kept_data = [];

// loop through each of the assessment_data elements and only
// keep the one's that have data (no empty arrays)
foreach($loop_graph_data as $graph_data) {
    if ($graph_data != "[]") {
        $kept_data[] = $graph_data;
    }
}

我想将两个（或更多-这是动态的）数组“连接”在一起，成为一个无缝的JSON格式的字符串。

有人知道该怎么做吗？

Answer 1

像json_encode (array_merge(json_decode($a, true),json_decode($b, true)))之类的东西应该可以工作。

深入foreach

请参考以下类似问题： Merging two json in PHP

Answer 2

我最终不得不对array_flatten数据加倍。这是我使用的解决方案：

// get all the data for making a graph
// it needs to be compiled by subject
foreach ($classroom_subjects as $subject) {
    $loop_graph_data[] = AssessmentData::select('subjects.short_name as subject', 'assessment_data.created_at as date', 'assessment_data.assessed_level as average')
        ->join('subjects', 'subjects.id', 'assessment_data.subject_id')
        ->where('assessment_data.subject_id', $subject->subject_id)
        ->where('assessment_data.student_id', $student_id)
        ->whereBetween('assessment_data.created_at', [$current_term->term_start." 00:00:01", $current_term->term_end." 23:59:59"])
        ->get();
}

$kept_data = [];
foreach($loop_graph_data as $graph_data) {
    if ($graph_data != "[]") {
        $kept_data[] = $graph_data;
    }
}

$flattened = array_flatten(array_flatten($kept_data));
$flattened = json_encode($flattened);

dd($flattened);

输出：

[{“ subject”：“ Science”，“ date”：“ 2018-08-09 13:03:14“，”平均值“：4}，{”主题“：”科学“，”日期“：” 2018-09-13 13:18:58“，”平均值“：4}，{”主题“：”社交“，”日期“：” 2018-08-14 05:59:56“，”平均值“：4}]

使用PHP / Laravel联接/合并/合并JSON数据

2 个答案: