PHP-将fgetcsv响应转换为特定的json

我有一个带标题的.CSV文件：

Description, BusinessSurname, IsCustomer, IsSupplier, AddressType, Business 
Address, IsInternational.

第一排：

Contact1, Contact1, True, True, Business, 123 Fake St, False

剩余的行并不重要,它更像是示例.我那里有几行数据.

我需要将其转换为json格式：

{
Description:'Desc_47AE3208-87F5-4BBA-BE40-AA4130AB4768',
SurnameBusinessName:'Name_Business',
IsCustomer:true,
IsSupplier:true,
Addresses:
[
{AddressType:'Business',Line1:'addr1_bus',IsInternational:false},
{AddressType:'Postal',Line1:'addr1_pos',IsInternational:true}
]
}

我尝试了几种不同的方法,但是没有一种方法专门给我提供这样的带有嵌套地址的json.我可以省略第二个地址(邮政地址).

如果我使用此代码：

        $filename = 'contacts1.csv';
        $handle = fopen($filename, 'r');
        $count = 0;
        while (($data = fgetcsv($handle)) !== FALSE) {
            $count++;
            if ($count == 1){
                continue;
            }
        $json = json_encode($data, true);
        echo $json;
        };

我得到这个例子：

["Contact1","Contact1","TRUE","TRUE","Business","123 High Street Sydney NSW 2000","FALSE"]
["Contact2","Contact2","TRUE","TRUE","Business","124 High Street Sydney NSW 2000","FALSE"]
["Contact3","Contact3","TRUE","TRUE","Business","125 High Street Sydney NSW 2000","FALSE"]

有没有办法获取我需要的json,如果我不能自动获取所需的json,有没有办法我可以提取每一行的值,并分配给变量,然后手动为每一个创建所需的json行,使用for循环,while循环等？例如：

{
Description: $description,
SurnameBusinessName: $BusinessSurname,
etc...
}

解决方法:

我找到了答案,但是我放弃了这个API端点,转而使用一个没有嵌套的简单端点.由于json格式的严格性,此API会接受,我基本上将csv中的每个值分配给了一个变量,然后手动创建了所需的json.尽管下面的代码用于Accounts端点,但是我可以对任何端点使用完全相同的方法,因为我有点手工创建json.我认为这是它唯一可行的方法,因为某些字段是诸如AccountName之类的字符串,一些是诸如AcountType之类的整数.

$file = 'acc.csv';
    $mode = 'r';
    $handle = fopen($file, $mode);
    while(($csv = fgetcsv($handle)) !==FALSE){
        foreach($csv as $row => $value){
            $data = $row.$value;
            switch ($row){
                case '0':
                    $accounttype = $value;
                    break;
                case '1':
                    $accountname = $value;
                    break;
                case '2':
                    $hint = $value;
                    break;
                case '3':
                    $status = $value;
                    break;
                case '4':
                    $sortorder = $value;
                    break;
                case '5':
                    $accountcode = $value;
                    break;
                case '6':
                    $parentaccountcatid = $value;

                    $json = "
                    {
      AccountType:" . $accounttype . ",
      AccountName:'" . $accountname . "',
      Hint:'" . $hint . "',
      Status:" . $status . ",
      SortOrder:" . $sortorder . ",
      AccountCode:'" . $accountcode . "',
      ParentAccountingCategoryID:'" . $parentaccountcatid . "'
     }"; 
    //echo $json; 

标签：fgetcsv,php,arrays,json
来源： https://codeday.me/bug/20191010/1886338.html