如何在php中将JSON对象转换为纯逗号分隔字符串？

如何从php中的json对象创建此类字符串，如下所示：

var $json = [{name:'abaneel',age:23},
    {name: 'john', age: 32},
    {name: 'Dev' , age:22}
];

---

以下是您需要做的：

$simple string ="abaneel,23,john,32,Dev,22";

---

希望这将有助于

以下是您需要做的事情：

$simple string ="abaneel,23,john,32,Dev,22";

$json = '[
  {"name": "abaneel", "age" : 23},
  {"name": "john", "age" : 32},
  {"name": "Dev", "age" : 22}
]';
$implode = array();
$multiple = json_decode($json, true);
foreach($multiple as $single)
    $implode[] = implode(', ', $single);

echo implode(', ', $implode);   //this will output abaneel, 23, john, 32, Dev, 22
/*
    You can see that it will be hard to distinguish that where line(one array is ending);
*/
echo implode(' | ', $implode);  //will output abaneel, 23 | john, 32 | Dev, 22

希望这会有所帮助

$json = '[
  {"name": "abaneel", "age" : 23},
  {"name": "john", "age" : 32},
  {"name": "Dev", "age" : 22}
]';
$implode = array();
$multiple = json_decode($json, true);
foreach($multiple as $single)
    $implode[] = implode(', ', $single);

echo implode(', ', $implode);   //this will output abaneel, 23, john, 32, Dev, 22
/*
    You can see that it will be hard to distinguish that where line(one array is ending);
*/
echo implode(' | ', $implode);  //will output abaneel, 23 | john, 32 | Dev, 22

我是堆栈溢出的新手，我不明白你所说的“标记作为答案”是什么意思。。。。但它的工作很好。谢谢你，我的意思是你应该接受这个作为答案，如果你把答案悬停在投票按钮下，会有一个勾号，请勾选。我是堆栈溢出新手，我不明白你所说的标记作为答案是什么意思。。。。但它的工作很好。谢谢你，我的意思是你应该接受这个答案，如果你把答案悬停在投票按钮下，会有一个勾号，请勾选。