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我有存儲爲一個PHP變量「$ decoded_traces」 解碼的蹤跡JSON對象如下:如何添加到一個JSON對象
{
"Coords": [
{
"Accuracy": "65",
"Latitude": "53.27771684322928",
"Longitude": "-9.01197836634846",
"Timestamp": "Fri Jul 05 2013 11:39:15 GMT+0100 (IST)"
},
{
"Accuracy": "65",
"Latitude": "53.27771684322928",
"Longitude": "-9.01197836634846",
"Timestamp": "Fri Jul 05 2013 11:39:15 GMT+0100 (IST)"
},
{
"Accuracy": "65",
"Latitude": "53.277716230919715",
"Longitude": "-9.01207806014157",
"Timestamp": "Fri Jul 05 2013 11:41:16 GMT+0100 (IST)"
},
{
"Accuracy": "65",
"Latitude": "53.277716230919715",
"Longitude": "-9.01207806014157",
"Timestamp": "Fri Jul 05 2013 11:41:16 GMT+0100 (IST)"
},
{
"Accuracy": "65",
"Latitude": "53.2776809358951",
"Longitude": "-9.012088286120322",
"Timestamp": "Fri Jul 05 2013 11:41:22 GMT+0100 (IST)"
}
]
}
我需要在另一個名爲元/項/屬性添加'Image'就是二進制blob的文本。所以我需要的是:
{
"Coords": [
{
"Accuracy": "65",
"Latitude": "53.27771684322928",
"Longitude": "-9.01197836634846",
"Timestamp": "Fri Jul 05 2013 11:39:15 GMT+0100 (IST)"
},
{
"Accuracy": "65",
"Latitude": "53.27771684322928",
"Longitude": "-9.01197836634846",
"Timestamp": "Fri Jul 05 2013 11:39:15 GMT+0100 (IST)"
},
{
"Accuracy": "65",
"Latitude": "53.277716230919715",
"Longitude": "-9.01207806014157",
"Timestamp": "Fri Jul 05 2013 11:41:16 GMT+0100 (IST)"
},
{
"Accuracy": "65",
"Latitude": "53.277716230919715",
"Longitude": "-9.01207806014157",
"Timestamp": "Fri Jul 05 2013 11:41:16 GMT+0100 (IST)"
},
{
"Accuracy": "65",
"Latitude": "53.2776809358951",
"Longitude": "-9.012088286120322",
"Timestamp": "Fri Jul 05 2013 11:41:22 GMT+0100 (IST)"
}
],
"Images": [
{
"Image1": "binary",
"Image2": "binary2"
}
]
}
我試過幾件事,但我真的有點迷路。顯然,添加後,我需要再次編碼,以便我可以回顯它。所有幫助讚賞
你的意思是json_decode?此外,我已經有一個變量解碼:\t \t \t \t $ decoded_traces = json_decode($ traces,true); – user2363025
如果它已經被解碼,你可以跳過這一步。 – Barmar
所以你的問題只是「我如何添加到數組?」 'json_decode()'返回的數組沒有什麼特別之處。 – Barmar