如何使用PHP和Mysql按照Json对象值向表中插入数据

我需要一个人帮忙。我需要使用PHP和MySQL根据一些Json对象值将多个数据插入表中。我在下面解释我的代码

$commnt=[{
   'day_id':2,
   'comment':'vodka1'
},{
  'day_id':3,
  'comment':'vodka2'
}
]

$result=[{
    'day_id':1,
    'restaurant':'193'
},{
  'day_id':2,
  'restaurant':'193'
},{
  'day_id':3,
  'restaurant':'193'
}
]

在这里，我需要根据day_id将来自两个Json对象的所有数据输入下表。我正在解释下表的列

db_详细信息：

我的要求是，当第天id相同时，相应的注释字段值将以其他方式输入表中，注释字段值将保持空白。预期输出如下所示

id    day_id   restaurant   comment

1       1        193               

2       2        193         vodka1

3       3        193         vodka3

$insertintodetails=mysqli_query($connect,'INSERT INTO db_details
(day_id,restaurant,comment) values ("'. $result.'","'.$result[$i]['restaurant'].'","'.$commnt[$i]['comment'].'")');

我的问题如下

id    day_id   restaurant   comment

1       1        193               

2       2        193         vodka1

3       3        193         vodka3

$insertintodetails=mysqli_query($connect,'INSERT INTO db_details
(day_id,restaurant,comment) values ("'. $result.'","'.$result[$i]['restaurant'].'","'.$commnt[$i]['comment'].'")');

---

这里可能有很多用例，比如两个Json对象长度可能相同或不同，但是注释应该按照day_id插入，否则它将保持空白。在我的查询中，我无法按要求插入。请帮我解决这个问题

我从您的问题细节上面创建了一个示例。您可以尝试下面的代码。它应该对你有用

$newArray = $insertintodetails = array();
foreach($result as $rs)
{
  $newArray[$rs->day_id] = [
    "`day_id`=>'{$rs->day_id}'",
    "`restaurant`=>'{$rs->restaurant}'"
];
}
foreach($commnt as $rs){
  $newArray[$rs->day_id][] = "`comment`='{$rs->comment}'";
}
foreach($newArray as $rs){
  $insertintodetails[]=mysqli_query($connect,'INSERT INTO db_details SET '.implode(',',$rs));
}

//For example taken array and convert it to JSON    
$comment = json_encode(array(array('day_id' => '2', 'comment' => 'vodka1'), array('day_id' => '3', 'comment' => 'vodka2')));
$result = json_encode(array(array('day_id' => '1', 'restaurant' => '193'), array('day_id' => '2', 'restaurant' => '193'), array('day_id' => '3', 'restaurant' => '193')));

//Convert JSON to array...
$arrComment = json_decode($comment, true);
$arrResult = json_decode($result, true);

foreach($arrResult AS $keyResult => $dataResult){
    $day_id = $dataResult['day_id'];//day_id
    $restaurant = $dataResult['restaurant'];//rasturant
    $strComment = '';//comment
    //Check and extract comment value from multi dimensional comment($arrComment) array...
    if($getComment = find_comment_with_dayid($arrComment, $day_id)){
      $strComment = $getComment;
    }

    //Insert records...
    $insertintodetails=mysqli_query($connect,'INSERT INTO db_details
(day_id, restaurant, comment) values ("'. $day_id .'","'. $restaurant .'","'. $strComment .'")');
}


//Function will return comment for matched day_id
function find_comment_with_dayid($arrComment, $fieldKey) {
    foreach($arrComment as $indCommenr => $dataComment) {
        //Check for day_id, matched then return comment...
        if($dataComment['day_id'] == $fieldKey) return $dataComment['comment'];
    }
    return FALSE;
}

---

在这里，我以您的问题中报告的避免JSON问题的示例数组为例。希望这个工作顺利

$commnt不是数组，它是json对象。您的数组在哪里？对不起，请检查我更新的帖子。它仍然是json。您可以使用json_解码到array的某些转换。我将测试并让您知道。@hackout:在您的代码中，某些foreach缺少{}。它无法正常工作。请检查您的代码。会出现许多错误。不，它工作正常。{day_id:2，restaurant:193，comment:}的注释值为空。这可以通过将行合并到一个语句（而不是每行一个insert语句）来大大改进。虽然它可能对小数据集没有多大影响，但随着应用程序的扩展，它可能会非常有益。所以现在，-1，但如果你编辑你的帖子，我会重新考虑。有关更多信息，请参阅