用php创建json数组
我试图创建json数组,用php代码填充数据表 字段应为:用php创建json数组,php,arrays,json,Php,Arrays,Json,我试图创建json数组,用php代码填充数据表 字段应为: { "data": [ { "RecordID": 1, "OrderID": "61715-075", "Country": "China", "ShipCountry": "CN", }, { "RecordID": 2, "OrderID": "63629-4697", "Country": "
{
"data": [
{
"RecordID": 1,
"OrderID": "61715-075",
"Country": "China",
"ShipCountry": "CN",
},
{
"RecordID": 2,
"OrderID": "63629-4697",
"Country": "Indonesia",
"ShipCountry": "ID",
},
{
"RecordID": 3,
"OrderID": "68084-123",
"Country": "Argentina",
"ShipCountry": "AR",
}
]
}
我试着这样做只是为了测试田地
$array=array('1','2','3','4')
但它又回来了
{"data":{"RecordID":1,"Country":"Indonesia","CompanyName":"Indonesia"}}{"data":{"RecordID":1,"Country":"Indonesia","CompanyName":"Indonesia"}}{"data":{"RecordID":1,"Country":"Indonesia","CompanyName":"Indonesia"}}{"data":{"RecordID":1,"Country":"Indonesia","CompanyName":"Indonesia"}}
您正在构建多个JSON字符串。要使一个数组包含所有数据,您所要做的就是对最外层的数组进行编码(在您的例子中,这似乎是
$array
)
因此,这样做就足够了:
echo json_encode($array)代码>
在准备好数据之后,您必须调用json\u encode
函数,因此在本例中,在循环之后。您正在构建多个json字符串。要使一个数组包含所有数据,您所要做的就是对最外层的数组进行编码(在您的例子中,这似乎是$array
)
因此,这样做就足够了:
echo json_encode($array)代码>
准备好数据后,必须调用json_encode
函数,因此在本例中,在循环之后。尝试-$temp['data']=array…
&json_encode
在循环之外。尝试-$temp['data']=array…
&json_encode
在循环之外。
{"data":{"RecordID":1,"Country":"Indonesia","CompanyName":"Indonesia"}}{"data":{"RecordID":1,"Country":"Indonesia","CompanyName":"Indonesia"}}{"data":{"RecordID":1,"Country":"Indonesia","CompanyName":"Indonesia"}}{"data":{"RecordID":1,"Country":"Indonesia","CompanyName":"Indonesia"}}