将 JSON 转换为 PHP 数组 - Converting JSON to PHP Array

Converting JSON to PHP Array

本文关键字：PHP 数组转换 JSON | 更新日期: 2023-09-27

我正在向PHP发送一个包含多个项目的JSON帖子。我的 JSON 看起来像这样：

 [
    {
        "request": "submitTicket",
        "id": "3",
        "delivLoc": "1 COLORADO CITY",
        "estimatedBarrels": "123.0",
        "facilityID": "T666778",
        "highDegreeF": "0.0",
        "highOilFeet": "0"
    },
    {
        "request": "submitTicket",
        "id": "4",
        "delivLoc": "1 COLORADO CITY",
        "estimatedBarrels": "143.0",
        "facilityID": "T666778",
        "highDegreeF": "0.0",
        "highOilFeet": "0"
    },
    {
        "request": "submitTicket",
        "id": "5",
        "delivLoc": "1 COLORADO CITY",
        "estimatedBarrels": "122.0",
        "facilityID": "T666778",
        "highDegreeF": "0.0",
        "highOilFeet": "0"
    }
]

我一直在尝试在 PHP 中使用 json_decode（），但它返回 null，导致 foreach 循环失败。为什么解码不起作用？

实际上，这个JSON在到达PHP时会返回斜杠。我在这里发布之前将它们取出，并尝试使用 stripslashes（）运行它。

试试这个。

$data ='[
{"request":"submitTicket","id":"3","delivLoc":"1 COLORADO CITY","estimatedBarrels":"123.0","facilityID":"T666778","highDegreeF":"0.0","highOilFeet":"0"},
{"request":"submitTicket","id":"4","delivLoc":"1 COLORADO CITY","estimatedBarrels":"143.0","facilityID":"T666778","highDegreeF":"0.0","highOilFeet":"0"},
{"request":"submitTicket","id":"5","delivLoc":"1 COLORADO CITY","estimatedBarrels":"122.0","facilityID":"T666778","highDegreeF":"0.0","highOilFeet":"0"}
]';
$jsonarray=json_decode($data,true);
print_r($jsonarray);

我在听取了大家的建议后想通了。在尝试解码 JSON 之前，我不得不从帖子中删除斜杠。我像这样解码为 PHP 数组：

$data = json_decode(stripslashes($_POST['json']));

我想昨晚太晚了，我有点忽略了这一点。感谢大家的帮助和快速回复。

试试这个。

$json = '{"a"：1，"b"：2，"c"：3，"d"：4，"e"：5}';var_dump（json_decode（$json））;var_dump（json_decode（$json， true））;

输出：

 object(stdClass)#1 (5) {
    ["a"] => int(1)
    ["b"] => int(2)
    ["c"] => int(3)
    ["d"] => int(4)
    ["e"] => int(5)
}
array(5) {
    ["a"] => int(1)
    ["b"] => int(2)
    ["c"] => int(3)
    ["d"] => int(4)
    ["e"] => int(5)
}

我可以看到的是正在传递的实际数据可能不是有效的 json。

只需在变量中分配数据并像这样使用

var data = [{
        "request": "submitTicket",
        "id": "3",
        "delivLoc": "1 COLORADO CITY",
        "estimatedBarrels": "123.0",
        "facilityID": "T666778",
        "highDegreeF": "0.0",
        "highOilFeet": "0"
    },
    {
        "request": "submitTicket",
        "id": "4",
        "delivLoc": "1 COLORADO CITY",
        "estimatedBarrels": "143.0",
        "facilityID": "T666778",
        "highDegreeF": "0.0",
        "highOilFeet": "0"
    },
    {
        "request": "submitTicket",
        "id": "5",
        "delivLoc": "1 COLORADO CITY",
        "estimatedBarrels": "122.0",
        "facilityID": "T666778",
        "highDegreeF": "0.0",
        "highOilFeet": "0"
    }
]

然后在数据部分使用ajax（我假设您正在使用ajax）传递它，如下所示

$.ajax({
    url:[your url],
    data:data,
    ....
});

希望这会起作用