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PHP中Json的正确格式_Php_Json - Fatal编程技术网

PHP中Json的正确格式

PHP中Json的正确格式,php,json,Php,Json,我正在尝试这样获得JSON输出 {"allterms":[{"group":{"Name":"Test 1"},{"group":{"Name":"Test2","Id":"298"}}] 我现在的代码是 while($r = mysql_fetch_assoc($rs)) { $rows['allterms']['group'][] = $r; } 这给了我这个 {"allterms":{"group":[{"Name":"Test1", "Id":"1740"},{"Name":"

我正在尝试这样获得JSON输出

{"allterms":[{"group":{"Name":"Test 1"},{"group":{"Name":"Test2","Id":"298"}}]
我现在的代码是

while($r = mysql_fetch_assoc($rs)) {
$rows['allterms']['group'][] = $r;   
}
这给了我这个

{"allterms":{"group":[{"Name":"Test1", "Id":"1740"},{"Name":"Test2","Id":"631"}}]

如何调整代码,使每个项目都有一个父项

这样更改循环:

while($r = mysql_fetch_assoc($rs)) {
    $rows['allterms'][]['group'] = $r;   
}
这将产生:

array(
    'allterms' => array(
        0 => array(
            'group' => array(...),
        ),
        1 => array(
            'group' => array(...),
        )
        ...
    )
作为json,它将是:

{
  "allterms": [
    {
      "group": {
        {
          "Name": "Test1",
          "Id": "1740"
        },
    {
      "group": {
        {
          "Name": "Test2",
          "Id": "631"
        }
     }
   ]
}
您可以使用PHP中的函数

while($r = mysql_fetch_assoc($rs)) {
    array_push($rows['allterms'], $r);   
}

试试
$rows['allterms']['group']
,应该会给你一个对象数组。非常感谢Kingkero,因为它不是作为答案形成的,所以我不能接受它是正确的,所以我接受了下面的AD7six。但是谢谢!