php中的数组格式(使用laravel框架)
我正在尝试用php格式化数组。它不会以预期的json格式返回。以下是详细信息php中的数组格式(使用laravel框架),php,laravel,laravel-5,Php,Laravel,Laravel 5,我正在尝试用php格式化数组。它不会以预期的json格式返回。以下是详细信息 $categories = category::where('cat_flag','Y')->pluck('cat_name as category_name')->toArray(); $items = item::leftjoin('categories','items.cat_id' ,'=', 'categories.id') ->where('it
$categories = category::where('cat_flag','Y')->pluck('cat_name as category_name')->toArray();
$items = item::leftjoin('categories','items.cat_id' ,'=', 'categories.id')
->where('item_flag','Y')
->get([
'item_name',
'items.cat_id as category_id',
'cat_name as category_name'
])->toArray();
$formatedArray = [];
foreach ($categories as $category) {
foreach ($items as $item) {
if ($item['category_name'] == $category) {
$formatedArray['cat_name'] = $category;
$formatedArray['datas'][] = $item;
}
}
}
我的预期输出如下。但数组格式部分存在逻辑错误。我不知道如何正确格式化。提前谢谢
[
{
"cat_name" : "food",
"datas" : [
{
"item_name": "item2",
"category_id": "1"
},
{
"item_name": "item4",
"category_id": "1"
}
]
},
{
"cat_name" : "drinks",
"datas" : [
{
"item_name": "coca cola",
"category_id": "4"
}
]
}
]
但我的输出显示为。所有项目都属于同一类别
{
"cat_name": "cat4",
"datas": [
{
"item_name": "item2",
"category_id": "1"
},
{
"item_name": "item22",
"category_id": "2"
}
]
}
嘿,哥们,试试这个,你们正在比较字符串和对象
foreach ($categories as $category) {
foreach ($items as $item) {
if ($item['category_name'] == $category) {
if(!empty($formatedArray)){
foreach(formatedArray as $array){
if(!empty($array) && $array['cat_name'] == $category){
array_push($formatedArray['datas'], $item)
}else{
$newArray = [];
$newArray['cat_name'] = $category;
$newArray['datas'][] = $item;
array_push($formatedArray, $newArray);
}
}
}else{
$formatedArray['cat_name'] = $category;
$formatedArray['datas'][] = $item
}
}
}
}
嘿,哥们,试试这个,你们正在比较字符串和对象
foreach ($categories as $category) {
foreach ($items as $item) {
if ($item['category_name'] == $category) {
if(!empty($formatedArray)){
foreach(formatedArray as $array){
if(!empty($array) && $array['cat_name'] == $category){
array_push($formatedArray['datas'], $item)
}else{
$newArray = [];
$newArray['cat_name'] = $category;
$newArray['datas'][] = $item;
array_push($formatedArray, $newArray);
}
}
}else{
$formatedArray['cat_name'] = $category;
$formatedArray['datas'][] = $item
}
}
}
}
我认为以下代码就是问题所在: 在每次迭代中,$formattedArray内容将被替换为新值
$formatedArray = [];
foreach ($categories as $category) {
foreach ($items as $item) {
if ($item['category_name'] == $category) {
$formatedArray['cat_name'] = $category;
$formatedArray['datas'][] = $item;
}
}
}
将上述内容替换为:
$formatedArray = $thisArray = [];
foreach ($categories as $category) {
foreach ($items as $item) {
if ($item['category_name'] == $category) {
$thisArray['cat_name'] = $category;
$thisArray['datas'][] = $item;
$formatedArray[] = $thisArray;
$thisArray = [];
}
}
}
我认为以下代码就是问题所在: 在每次迭代中,$formattedArray内容将被替换为新值
$formatedArray = [];
foreach ($categories as $category) {
foreach ($items as $item) {
if ($item['category_name'] == $category) {
$formatedArray['cat_name'] = $category;
$formatedArray['datas'][] = $item;
}
}
}
将上述内容替换为:
$formatedArray = $thisArray = [];
foreach ($categories as $category) {
foreach ($items as $item) {
if ($item['category_name'] == $category) {
$thisArray['cat_name'] = $category;
$thisArray['datas'][] = $item;
$formatedArray[] = $thisArray;
$thisArray = [];
}
}
}
格式化错误是什么?显示时没有错误。我无法将“cat_name”:“food”设置为该对象。它只设定了一次。但我想说的是格式。谢谢回复,如果你应该$category->category\u name而不仅仅是$category,那可能是posint格式错误是什么?显示时没有错误。我无法将“cat_name”:“food”设置为该对象。它只设定了一次。但我想说的是格式。感谢回复如果你应该$category->category\u name而不仅仅是$category,那可能是posin谢谢回复。这就是为什么我将数组转换成$category['category_name']。但是它显示了非法的字符串偏移量“category_name”@Mithun更新了看这个你的意思是这样吗谢谢你给我时间。但同样非法的字符串偏移量“category_name”您能给我展示一下数据库中两个表$Categories=array:2[0=>“cat2”1=>“cat4”]的几个示例吗?谢谢您的回复。这就是为什么我将数组转换成$category['category_name']。但是它显示了非法的字符串偏移量“category_name”@Mithun更新了看这个你的意思是这样吗谢谢你给我时间。但同样非法的字符串偏移量“category_name”您能给我展示一下数据库中两个表$Categories=array:2[0=>“cat2”1=>“cat4”]的几个示例吗?谢谢您的回复。但它将同一类别划分为多个类别。我需要相同类别下的相同项目{“类别名称”:“cat2”,“数据”:[{“项目名称”:“项目2”,“类别id”:“1}]},{“类别名称”:“cat2”,“数据”:[{“项目名称”:“项目22”,“类别id”:“1”}]},但我需要像下面这样的评论{“catu name”:“cat2”,“datas”:[{“item_name”:“item2”,“category_id”:“1”},{“item_name”:“item22”,“category_id”:“1”}]}谢谢回复。但它将同一类别划分为多个类别。我需要相同类别下的相同项目{“类别名称”:“cat2”,“数据”:[{“项目名称”:“项目2”,“类别id”:“1}]},{“类别名称”:“cat2”,“数据”:[{“项目名称”:“项目22”,“类别id”:“1”}]},但我需要下面的注释{“cat_name”:“cat2”,“datas”:[{“item_name”:“item2”,“category_id”:“1”},{“item_name”:“item22”,“category_id”:“1”}]}