php - 使用 PHP 迭代此 json 代码的最佳方法

Question

我有这个json代码：

{
   "data": [
      {
         "name": "John",
         "id": "11"
      },
      {
         "name": "Max",
         "id": "22"
      },
      {
         "name": "Martin",
         "id": "33"
      },
      {
         "name": "Adrian",
         "id": "44"
      }
   ]
}

现在，我需要以这种方式打印所有名称和 ID：

John 11
Max 22
Martin 33
Adrian 44

最简单的方法是什么？

Answer 1

像这样：

$arr = json_decode($string, true); // true to retrieve an associative array
foreach($arr['data'] as $sub) {
  echo $sub['name'],' ',$sub['id'],PHP_EOL;
}

输出：

John 11
Max 22
Martin 33
Adrian 44

Answer 2

<?php
$a='{ "data": [ { "name": "John", "id": "11" }, { "name": "Max", "id": "22" }, { "name": "Martin", "id": "33" }, { "name": "Adrian", "id": "44" } ] }';
$b=json_decode($a);
$data=$b->{'data'};
foreach($data as $item) {
  print $item->{'name'}.' '.$item->{'id'}."\n";
}

?>

另外，Javascript 版本：

objectValueList=function(x) {
  var tags=[];
  for (i in x) {
    if(x.hasOwnProperty(i)) {
      tags.push(x[i]);
    }
  }
  return tags;
}


var a='{ "data": [ { "name": "John", "id": "11" }, { "name": "Max", "id": "22" }, { "name": "Martin", "id": "33" }, { "name": "Adrian", "id": "44" } ] }';
var b=JSON.parse(a);
var data=objectValueList(b.data);
for (x in data) {
  console.log(data[x].name+" "+data[x].id);
}

Answer 3

使用json_decode().

$data=json_decode($string);

foreach($data as $person){
    foreach($person as $key=>$val){
        foreach($val as $tuple)
            echo $tuple." ";
        echo "\r\n"; 
        }
    }   
}