php - 使用 PHP 解析 JSON 数组

Question

我知道这很简单，但我坚持下去，我真的需要一些帮助

这是我正在生成的 JSON 字符串。

[{"field1":3,"field2":"5","field3":"value3","field4":"value4"},{"field1":3,"field2":"8","field3":"value3","field4":"value4"},{"field1":3,"field2":"6","field3":"value3"}]

如何提取与 field1 关联的值？

以及如何访问每个不同数组的元素？

Answer 1

试试这个代码

$jsonString = '[{"field1":3,"field2":"5","field3":"value3","field4":"value4"},{"field1":3,"field2":"8","field3":"value3","field4":"value4"},{"field1":3,"field2":"6","field3":"value3"}]';

$json = json_decode($jsonString, true);

print_r($json);

Answer 2

试试这个（在 Javascript 中）

var jsonData=[{"field1":3,"field2":"5","field3":"value3","field4":"value4"},{"field1":3,"field2":"8","field3":"value3","field4":"value4"},{"field1":3,"field2":"6","field3":"value3"}];

如果您以文本形式获得响应，则使用 jsonData=JSON.parse(yourResponseText);

for(var i=0;i<jsonData.length;i++){
alert('your required val:'+jsonData[i].field1);
}

在php中

$data = json_decode($json);//$json is your json data

foreach ($data as $item) {
  echo $item->field1
}

Answer 3

https://stackoverflow.com/a/3627901/485790

console.log(jQuery.parseJSON(' [{"field1":3,"field2":"5","field3":"value3","field4":"value4"},{"field1":3,"field2":"8","field3":"value3","field4":"value4"},{"field1":3,"field2":"6","field3":"value3"}]').field1);

Answer 4

以php方式

$jsonString = '[{"field1":3,"field2":"5","field3":"value3","field4":"value4"},{"field1":3,"field2":"8","field3":"value3","field4":"value4"},{"field1":3,"field2":"6","field3":"value3"}]';

$json = json_decode($jsonString, true);

foreach($json as $item){
   echo $item['field1'];
}

Answer 5

在 php 中，您可以：

$json = json_decode($_GET['variable'];

http://php.net/manual/en/function.json-decode.php

Answer 6

试试这个，

var json = '[{"field1":3,"field2":"5","field3":"value3","field4":"value4"},{"field1":3,"field2":"8","field3":"value3","field4":"value4"},{"field1":3,"field2":"6","field3":"value3"}]';
    $.each(jQuery.parseJSON(json), function () {
        alert(this['field1']);
        alert(this['field2']);
        alert(this['field3']);
        alert(this['field4']);

});

Answer 7

这是一个简单的 json 对象数组。所以用jquery

var jsonArray = JSON.parse("your string");
   for(var int i=0 ; i < jsonArray.length() ; i++){
     var jsonObject = jsonArray[i];
     Console.log(jsonObject.field1)
   }