在php中分配变量的数组值
在php中分配变量的数组值:在php中分配变量的数组值,php,Php,在php中分配变量的数组值: Array ( [0] => stdClass Object ( [Id] => 116249 [Amount] => 51.62 [Currency] => INR [ExchangeRate] => 1 [InvoiceDate
Array
(
[0] => stdClass Object
(
[Id] => 116249
[Amount] => 51.62
[Currency] => INR
[ExchangeRate] => 1
[InvoiceDate] => 2015-12-16T00:00:00Z
}
[1] => stdClass Object
(
[Id] => 116250
[Amount] => 55.20
[Currency] => KWD
[ExchangeRate] => 1
[InvoiceDate] => 2015-12-16T00:00:00Z
}
[2] => stdClass Object
(
[Id] => 116251
[Amount] => 59.42
[Currency] => USD
[ExchangeRate] => 1
[InvoiceDate] => 2015-12-16T00:00:00Z
}
}
foreach ($invoice as $key => $value)
{
$Id=$value->Id;
$Amount=$value->Amount;
$Currency=$value->Currency;
$ExchangeRate=$value->ExchangeRate;
$invoiceDate = str_replace('Z', '', str_replace('T', ' ',$value->InvoiceDate));
}
$jsonData ='[{"Id": "'.$Id.'",
"Amount": "'.$Amount.'",
"Currency": "'.$Currency.'",
"ExchangeRate": "'.$ExchangeRate.'",
"invoiceDate": "'.$invoiceDate.'"}];
我尝试了这段代码最后数组值只打印数组[2]值,我需要像这样的输出
[{"Id": "116249",
"Amount": "51.62",
"Currency": "INR",
"ExchangeRate": "1",
"invoiceDate": "2015-12-16T00:00:00Z"},
{"Id": "116250",
"Amount": "55.20",
"Currency": "KWD",
"ExchangeRate": "1",
"invoiceDate": "2015-12-16T00:00:00Z"},
{"Id": "116251",
"Amount": "59.42",
"Currency": "USD",
"ExchangeRate": "1",
"invoiceDate": "2015-12-16T00:00:00Z"}]
您希望从数组创建json字符串。您所需要做的就是运行以下命令
jsonData = json_encode($invoice);
如果您希望使用循环创建数组以仅生成特定值,则可以执行以下操作:
$tmp = array();
foreach ($invoice as $key => $value)
{
$tmp["id"] = $value->Id;
$tmp["amount"] = $value->Amount;
$tmp["Currency"] = $value->Currency;
$tmp["ExchangeRate"] = $value->ExchangeRate;
$tmp["invoiceDate"] = str_replace('Z', '', str_replace('T','',$value->InvoiceDate));
}
jsonData = json_encode($tmp);
*编辑*
如果您使用mysql日期作为invoiceData变量。然后最好使用PHP函数。您希望从数组创建json字符串。您所需要做的就是运行以下命令
jsonData = json_encode($invoice);
如果您希望使用循环创建数组以仅生成特定值,则可以执行以下操作:
$tmp = array();
foreach ($invoice as $key => $value)
{
$tmp["id"] = $value->Id;
$tmp["amount"] = $value->Amount;
$tmp["Currency"] = $value->Currency;
$tmp["ExchangeRate"] = $value->ExchangeRate;
$tmp["invoiceDate"] = str_replace('Z', '', str_replace('T','',$value->InvoiceDate));
}
jsonData = json_encode($tmp);
*编辑*
如果您使用mysql日期作为invoiceData变量。那么您最好使用PHP函数。我同意利亚姆的观点,即
json\u encode()
可能是解决您的问题的最佳解决方案。但是,如果您正在寻找DIY解决方案,它非常简单:
$jsonData = '[';
foreach ($invoice as $key => $value)
{
$Id=$value->Id;
$Amount=$value->Amount;
$Currency=$value->Currency;
$ExchangeRate=$value->ExchangeRate;
$invoiceDate = str_replace('Z', '', str_replace('T', ' ',$value->InvoiceDate));
// Append the next invoice's JSON data to all of the JSON data
$jsonData .='{"Id": "'.$Id.'",
"Amount": "'.$Amount.'",
"Currency": "'.$Currency.'",
"ExchangeRate": "'.$ExchangeRate.'",
"invoiceDate": "'.$invoiceDate.'"},';
}
// Clean up and close off the JSON data:
$jsonData = rtrim($jsonData, ','); // remove the trailing comma
$jsonData .= ']'; // Add the closing bracket.
代码中存在的问题是,在foreach循环运行之后,只分配了一次JSON数据。相反,您需要将每个发票的JSON数据附加到其他发票的JSON数据中
尽管如此,
json\u encode()
可能是更好的解决方案,因为它只有一行,更容易阅读,也更容易理解。我同意利亚姆的观点,json\u encode()
可能是解决您的问题的最佳方案。但是,如果您正在寻找DIY解决方案,它非常简单:
$jsonData = '[';
foreach ($invoice as $key => $value)
{
$Id=$value->Id;
$Amount=$value->Amount;
$Currency=$value->Currency;
$ExchangeRate=$value->ExchangeRate;
$invoiceDate = str_replace('Z', '', str_replace('T', ' ',$value->InvoiceDate));
// Append the next invoice's JSON data to all of the JSON data
$jsonData .='{"Id": "'.$Id.'",
"Amount": "'.$Amount.'",
"Currency": "'.$Currency.'",
"ExchangeRate": "'.$ExchangeRate.'",
"invoiceDate": "'.$invoiceDate.'"},';
}
// Clean up and close off the JSON data:
$jsonData = rtrim($jsonData, ','); // remove the trailing comma
$jsonData .= ']'; // Add the closing bracket.
代码中存在的问题是,在foreach循环运行之后,只分配了一次JSON数据。相反,您需要将每个发票的JSON数据附加到其他发票的JSON数据中
尽管如此,
json\u encode()
可能是更好的解决方案,因为它只有一行,更容易阅读,也更容易理解。你的问题是什么?我需要这样的输出,具体是什么?你能编辑你的问题告诉我们你有什么,你在做什么和你的问题吗?对不起,不要让我们把你的一切都拖出来。在你的问题上要精确:你的输入是什么,期望的输出是什么。你的问题真的只是“如何从数组中创建json字符串”?我试过只打印最后一个数组值的代码你的问题是什么?我需要这样的输出,具体是什么?你能编辑你的问题告诉我们你有什么,你在做什么和你的问题吗?对不起,不要让我们把你的一切都拖出来。在你的问题上要精确:你的输入是什么,期望的输出是什么。你的问题真的只是“如何从数组中创建json字符串”吗?我试过代码最后一个数组值只打印brother,我想是它的json_encode()。不是jsone_encode():)brother,我想是它的json_encode()。不是jsone_encode():)