我知道这很简单,但我被困住了,我真的很想得到一些帮助
这是我正在生成的一个 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"}]
如何提取与字段 1 关联的值?
如何访问每个不同数组的元素?
试试这段代码
$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);
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"}];
如果你得到文本形式的响应,那么使用 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
}
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);
以 php 的方式
$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'];
}
在 php 中,你可以:
$json = json_decode($_GET['variable'];
http://php.net/manual/en/function.json-decode.php
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']);
});
这是一个简单的 JSON 对象数组。 所以有了jQuery。
var jsonArray = JSON.parse("your string");
for(var int i=0 ; i < jsonArray.length() ; i++){
var jsonObject = jsonArray[i];
Console.log(jsonObject.field1)
}