我有2个数据库表。表2与表1有关。
在php中我正在构建一个多维数组来保存数据键/值。
因为PHP需要有唯一的“键”MY JSON看起来像这样：
[
{
"packs": {
"9": {
"characters": {
"40": {
"id": "40",
"title": "Jack Bauer",
"pic": "68bcbe014c.jpg",
"gender": "Male"
},
"41": {
"id": "41",
"title": "Chloe O'Brian",
"pic": "ffb6acc8e3.jpg",
"gender": "Male"
},
"42": {
"id": "42",
"title": "Tony Almeida",
"pic": "23f199e223.jpg",
"gender": "Male"
}
}
},
"7": {
"characters": {
"7": {
"id": "7",
"title": "Elvis Presley",
"pic": "78300767ad.jpg",
"gender": "Male"
},
"16": {
"id": "16",
"title": "Madonna",
"pic": "70663a42f7.jpg",
"gender": "Male"
},
"17": {
"id": "17",
"title": "Lady Gaga",
"pic": "c5099c619b.jpg",
"gender": "Male"
},
"21": {
"id": "21",
"title": "Pink Floyd",
"pic": "52ddce314a.jpg",
"gender": "Male"
},
"22": {
"id": "22",
"title": "Led Zeppelin",
"pic": "84cd58ada3.jpg",
"gender": "Male"
},
"31": {
"id": "31",
"title": "The Beatles",
"pic": "bd22a4d648.jpg",
"gender": "Male"
},
"32": {
"id": "32",
"title": "Foo Fighters",
"pic": "250fb6ecec.jpg",
"gender": "Male"
},
"33": {
"id": "33",
"title": "Bananarama",
"pic": "da7c2b56cf.jpg",
"gender": "Male"
},
"35": {
"id": "35",
"title": "Boney-M",
"pic": "3cbdada38b.jpg",
"gender": "Male"
},
"38": {
"id": "38",
"title": "The Spice Girls",
"pic": "4751f0fbb7.jpeg",
"gender": "Male"
},
"39": {
"id": "39",
"title": "Girls Aloud",
"pic": "644dcf71ca.jpg",
"gender": "Male"
}
}
},
"8": {
"characters": {
"9": {
"id": "9",
"title": "Keith Lemon",
"pic": "ff6ef10853.jpg.jpg",
"gender": "Male"
},
"23": {
"id": "23",
"title": "Fearne Cotton",
"pic": "0d038b6516.jpg",
"gender": "Male"
},
"24": {
"id": "24",
"title": "Holly Willoughby",
"pic": "836fc4184c.jpg",
"gender": "Male"
},
"30": {
"id": "30",
"title": "Rufus Hound",
"pic": "062bee9602.jpg",
"gender": "Male"
}
}
},
"3": {
"characters": {
"5": {
"id": "5",
"title": "Tom Cruise",
"pic": "ff296fafb9.jpg",
"gender": "Male"
},
"10": {
"id": "10",
"title": "Linda Lovelace",
"pic": "ac1bea43d3.jpg",
"gender": "Male"
},
"15": {
"id": "15",
"title": "Gwyneth Paltrow",
"pic": "43a22d7240.jpg",
"gender": "Male"
},
"44": {
"id": "44",
"title": "Errol Flynn",
"pic": "cea17c1275.jpg",
"gender": "Male"
},
"45": {
"id": "45",
"title": "Halle Berry",
"pic": "752b5c92c5.jpg",
"gender": "Male"
}
}
},
"2": {
"characters": {
"4": {
"id": "4",
"title": "Donald Duck",
"pic": "8d367f41b1.jpg",
"gender": "Male"
},
"6": {
"id": "6",
"title": "Mickey Mouse",
"pic": "8d9629c115.jpg",
"gender": "Male"
},
"28": {
"id": "28",
"title": "Pluto",
"pic": "fb2c0e2dd0.jpg",
"gender": "Male"
},
"29": {
"id": "29",
"title": "Minnie Mouse",
"pic": "378760ff77.jpg",
"gender": "Male"
},
"36": {
"id": "36",
"title": "Cinderella",
"pic": "a7e4888213.jpg",
"gender": "Male"
},
"37": {
"id": "37",
"title": "Snow White",
"pic": "a9cf05a857.jpg",
"gender": "Male"
}
}
},
"4": {
"characters": {
"3": {
"id": "3",
"title": "Bill Clinton",
"pic": "03c6567ddb.jpg",
"gender": "Male"
},
"11": {
"id": "11",
"title": "Margaret Thatcher",
"pic": "91c9fa9fd0.jpg",
"gender": "Male"
},
"13": {
"id": "13",
"title": "David Cameron",
"pic": "a689984360.jpg",
"gender": "Male"
},
"14": {
"id": "14",
"title": "Nick Clegg",
"pic": "3243e298e5.jpg",
"gender": "Male"
},
"26": {
"id": "26",
"title": "George Bush JR",
"pic": "46296f6b0e.jpg",
"gender": "Male"
},
"27": {
"id": "27",
"title": "Ed Milliband",
"pic": "66f1449994.jpg",
"gender": "Male"
}
}
},
"5": {
"characters": {
"8": {
"id": "8",
"title": "Stephen Hawking",
"pic": "b8c4f17530.jpg",
"gender": "Male"
},
"18": {
"id": "18",
"title": "Alan Turing",
"pic": "82b4d84e35.jpg",
"gender": "Male"
},
"19": {
"id": "19",
"title": "Albert Einstein",
"pic": "a6cd74dbaa.jpg",
"gender": "Male"
},
"34": {
"id": "34",
"title": "Brian Cox (prof)",
"pic": "92b6005de9.jpg",
"gender": "Male"
},
"43": {
"id": "43",
"title": "Richard Feynman",
"pic": "5de10d1128.jpg",
"gender": "Male"
}
}
},
"6": {
"characters": {
"12": {
"id": "12",
"title": "Jeff Koons",
"pic": "8e3ca5f497.jpg",
"gender": "Male"
},
"20": {
"id": "20",
"title": "Salvador Dali",
"pic": "b5bafb7934.jpg",
"gender": "Male"
},
"25": {
"id": "25",
"title": "Rembrandt",
"pic": "73e2710029.jpg",
"gender": "Male"
},
"49": {
"id": "49",
"title": "Vincent Van Gough",
"pic": "6ee455ab28.jpg",
"gender": "Male"
}
}
}
}
}
我正在尝试解决我的iOS开发人员想要排序的问题......显然“{”包“：{”9“：”9错了，不遵循键：值JSON结构。我如何使用DB表1中每个结果集的唯一标识符为此创建正确的PHP数组或对象？
结果的结构应如下：
data_db_1：1
data_db_2：ID data_db_2：标题 data_db_2：PIC data_db_2：性别
data_db_1：2
data_db_2：ID data_db_2：标题 data_db_2：PIC data_db_2：性别
data_db_1：3
data_db_2：ID data_db_2：标题 data_db_2：PIC data_db_2：性别
data_db_1：4 data_db_2：ID data_db_2：标题 data_db_2：PIC data_db_2：性别
看起来像这样：
{
"id": "9",
"title": "24",
"credits": "100",
"character": [
{
"id": "50",
"title": "Jack Bauer",
"pic": "68bcbe014c.jpg",
"gender": "Male"
},
{
"id": "50",
"title": "Jack Bauer",
"pic": "68bcbe014c.jpg",
"gender": "Male"
},
{
"id": "50",
"title": "Jack Bauer",
"pic": "68bcbe014c.jpg",
"gender": "Male"
},
{
"id": "50",
"title": "Jack Bauer",
"pic": "68bcbe014c.jpg",
"gender": "Male"
}
]
}
我无法在网上找到与此相关的内容。我能想到解决它的唯一方法是使用json_encode（）和echo以及一堆concat从PHP数组中单独创建每个部分。当然json_encode（）应该能够从一个简单的php数组，一个复杂的php多维数组或一个PHP对象创建正确的JSON而不会出汗......
我的php函数按要求执行此操作：
$data_ar = output_pack_data($pack_id);
echo "[".json_encode($data_ar)."]"; // first bit of dirty here
function output_pack_data($pack_id = false)
{
global $db;
$output_keys = true; //false;
if($pack_id != false)
{
$q = "WHERE id='{$pack_id}' AND active = '1'"; // select the relevant pack ID
}
else
{
$q = "WHERE active='1' ORDER BY order_num ASC"; // select all pack ids
}
$rs = $db->rs("whoami_packs",$q);
$pack_obj = new stdClass(); // declare new std class object here......
$data = array(); // prep new array to hold the data
while($rs && $r = $db->fetch($rs)) // loop thorugh each pack id
{
$pack_obj->packs->id = $r->id; // add the pack title
$pack_obj->packs->title = $r->title; // add the pack title
$pack_obj->packs->credits = $r->credits; // add the required credits to access
$packs_rs = $db->rs("whoami_characters","WHERE pack_id='{$r->id}' AND active = '1'"); // get the character data relevant for this pack
$i=0;
while($packs_rs && $pack_r = $db->fetch($packs_rs)) // loop through the character data
{
$id = $r->id;
if($output_keys == false)
{
$data['packs'][$r->id]['characters'][$pack_r->id][$pack_r->title][$pack_r->pic][$pack_r->gender] = true; // build the array
}
else
{
$data['packs'][$r->id]['characters'][$pack_r->id]['id'] = $pack_r->id; // build the array
$data['packs'][$r->id]['characters'][$pack_r->id]['title'] = str_out($pack_r->title); // build the array
$data['packs'][$r->id]['characters'][$pack_r->id]['pic'] = $pack_r->pic;
$data['packs'][$r->id]['characters'][$pack_r->id]['gender'] = ($pack_r->gender = 'm') ? "Male" : "Female";
}
}
}
return $data; // return the array
}
答案 0 :(得分：1)
函数中有一些多余的代码。当你有非连续（数字）索引时，json_encode不会使用[x，y，z]编码而是使用对象表示法。
显然，您的同事希望ID作为每个对象的属性而不是对象的关键。因此，只需通过其id删除引用数组元素，但首先创建完整数组，然后通过$ parent []将其附加到其父元素。这样你就得到了一个带有连续数字id的数组 - ＆gt; json_encode（）创建一个数组符号
（未经测试，我懒得从原始问题中提供的json输出构建sql测试数据）
function output_pack_data($pack_id=false)
{
global $db;
$data = array();
if($pack_id != false) {
$q = "WHERE id='{$pack_id}' AND active = '1'";
}
else {
$q = "WHERE active='1' ORDER BY order_num ASC";
}
$rs_packs = $db->rs("whoami_packs", $q);
while($rs_packs && ($rec_pack=$db->fetch($rs_packs)) ) {
$pack = array(
'id'=>$rec_pack->id,
'title'=>$rec_pack->title,
'credits'=>$rec_pack->credits,
'characters'=>array()
);
$rs_chars = $db->rs("whoami_characters","WHERE pack_id='{$pack->id}' AND active = '1'");
while($rs_chars && ($rec_char=$db->fetch($rs_chars)) ) {
$pack['characters'][] = array(
'id' => $rec_char->id,
'title' => str_out($rec_char->title),
'pic' => $rec_char->pic,
'gender' => 'm'==$rec_char->gender ? "Male" : "Female"
);
}
$data[] = $pack;
}
return $data;
}
请阅读https://en.wikipedia.org/wiki/Join_%28SQL%29和http://www.w3schools.com/sql/sql_join.asp
答案 1 :(得分：0)
对不起，这是评论，而不是答案（再次）。
我认为iOS开发人员（比如我自己）所说的是你使用每个对象的id来定义每个对象，而不是键。
您的示例和开发人员的预期输出都是正确的，但这是问题的实际信息结构。
如果您前往http://json.org/example.html查看JSON数据结构的一些示例，可能会更清楚一点。
让我们用XML来看一下。如果将以下JSON转换为XML：
[
{
"1": {
"id": "1",
"title": "Jack Bauer",
"pic": "68bcbe014c.jpg",
"gender": "Male"
}
},
{
"2": {
"id": "2",
"title": "Jack Bauer",
"pic": "68bcbe014c.jpg",
"gender": "Male"
}
}
]
您将离开：
<?xml version="1.0" encoding="UTF-8" ?>
<1>
<id>1</id>
<title>Jack Bauer</title>
<pic>68bcbe014c.jpg</pic>
<gender>Male</gender>
</1>
<2>
<id>2</id>
<title>Jack Bauer</title>
<pic>68bcbe014c.jpg</pic>
<gender>Male</gender>
</2>
</xml>
简单地说，虽然在两种情况下都是正确的结构，但数据放置不正确。 ＆lt; 1＆gt;和＆lt; 2＆gt;节点应该是相同的密钥，这个原则遵循大多数标准（如RSS）。
根据开发人员的要求，他希望（在JSON和XML示例中）：
{
"character": [
{
"id": "1",
"title": "Jack Bauer",
"pic": "68bcbe014c.jpg",
"gender": "Male"
},
{
"id": "2",
"title": "Jack Bauer",
"pic": "68bcbe014c.jpg",
"gender": "Male"
}
]
}
和XML：
<?xml version="1.0" encoding="UTF-8" ?>
<character>
<id>1</id>
<title>Jack Bauer</title>
<pic>68bcbe014c.jpg</pic>
<gender>Male</gender>
</character>
<character>
<id>2</id>
<title>Jack Bauer</title>
<pic>68bcbe014c.jpg</pic>
<gender>Male</gender>
</character>
</xml>
第二个例子的结构更加清晰，因为键（字符）定义了每个值，而不是每个值本身的id。
答案 2 :(得分：0)
所以看起来json_encode（）会忽略键值，如果它们是一个简单的递增计数器....
这个功能：
function output_pack_data($pack_id = false)
{
global $db;
$output_keys = true; //false;
if($pack_id != false)
{
$q = "WHERE id='{$pack_id}' AND active = '1'"; // select the relevant pack ID
}
else
{
$q = "WHERE active='1' ORDER BY order_num ASC"; // select all pack ids
}
$rs = $db->rs("whoami_packs",$q);
$data = array(); // prep new array to hold the data
$i = 0; // AHA!
while($rs && $r = $db->fetch($rs)) // loop through each pack id
{
$data[$i]['pack_id'] = $r->id;
$data[$i]['title'] = $r->title; // add the pack title
$data[$i]['credits'] = $r->credits; // add the required credits to access
$chars_rs = $db->rs("whoami_characters","WHERE pack_id='{$r->id}' AND active = '1'"); // get the character data relevant for this pack
$ii = 0; // AHA!
while($chars_rs && $char_r = $db->fetch($chars_rs)) // loop through the character data
{
if($output_keys == false)
{
$data[$i]['characters'][$ii][$char_r->title][$char_r->pic][$char_r->gender] = true; // build the array
}
else
{
$data[(int)$i]['characters'][(int)$ii]['id'] = str_out($char_r->id); // build the array
$data[(int)$i]['characters'][(int)$ii]['title'] = str_out($char_r->title); // build the array
$data[(int)$i]['characters'][(int)$ii]['pic'] = $char_r->pic;
$data[(int)$i]['characters'][(int)$ii]['gender'] = ($char_r->gender = 'm') ? "Male" : "Female";
}
$ii++;
}
$i++;
}
return $data; // return the array
}
将返回所需的结果。奇怪的是，在文档中没有提到它！哦，好的，现在排序！希望这可以帮助其他人解决这个问题....注意使用（int）来确保$ i或$ ii计数器是正确的整数。在这种情况下使用DB ID将不起作用，它们将被添加到输出中。
答案 3 :(得分：0)
使用选项JSON_FORCE_OBJECT（自PHP 5.3.0起可用，请参阅here）将强制json_encode创建有效的JSON对象