Php 如何拆分和更改json格式
我使用的是SlimPHPv2,我有以下函数Php 如何拆分和更改json格式,php,json,slim,Php,Json,Slim,我使用的是SlimPHPv2,我有以下函数 function gt($user) { $sql = "select id, id as categoryId, mobile, task from actions where status=0"; try { $db = newDB($user); $stmt = $db - > prepare($sql); $stmt - > execute(); $gs
function gt($user) {
$sql = "select id, id as categoryId, mobile, task from actions where status=0";
try {
$db = newDB($user);
$stmt = $db - > prepare($sql);
$stmt - > execute();
$gs = $stmt - > fetchAll(PDO::FETCH_OBJ);
if ($gs) {
$db = null;
echo json_encode($gs, JSON_UNESCAPED_UNICODE);
} else {
echo "Not Found";
}
} catch (PDOException $e) {
echo '{"error":{"text":'.$e - > getMessage().
'}}';
}
}
默认json输出如下所示:
[{
"id": "1",
"categoryId": "1",
"mobile": "111",
"task": "test"
},
{
"id": "2",
"categoryId": "2",
"mobile": "222",
"task": "test2"
}]
以及我正在尝试的输出。我需要生成一个ID:1(ID),然后像这样格式化它
[{
id: "1",
task: "test",
}, {
ID: "1_1", //generate this, add 1_id
categoryId: "1",
mobile: "00000",
}, {
id: "2",
task: "test2",
}, {
ID: "1_2", //generate this, add 1_id
categoryId: "2",
mobile: "11111"
}];
可能吗?
谢谢我不确定这是否正是您想要的,但您可以通过将原始JSON转换为关联数组,然后使用
Foreach()
循环将每次迭代中的数据重新构造为新的assoc数组,从而获得所需的JSON输出。之后,您可以使用JSON\u encode()
将其转换回JSON
代码:
$json = '[{
"id": "1",
"categoryId": "1",
"mobile": "111",
"task": "test"
},
{
"id": "2",
"categoryId": "2",
"mobile": "222",
"task": "test2"
}]';
$jsonArr = json_decode($json, TRUE);
$newArr = [];
foreach ($jsonArr as $v) {
$newArr[] = ['id:'=>$v['id'], 'task:' => $v['task']];
$newArr[] = ['ID:'=>'1_' . $v['id'], 'categoryId' => $v['categoryId'], 'mobile'=>$v['mobile']];
}
$newJson = json_encode($newArr);
var_dump($newJson);
[{
"id:": "1",
"task:": "test"
}, {
"ID:": "1_1",
"categoryId": "1",
"mobile": "111"
}, {
"id:": "2",
"task:": "test2"
}, {
"ID:": "1_2",
"categoryId": "2",
"mobile": "222"
}]
输出:
$json = '[{
"id": "1",
"categoryId": "1",
"mobile": "111",
"task": "test"
},
{
"id": "2",
"categoryId": "2",
"mobile": "222",
"task": "test2"
}]';
$jsonArr = json_decode($json, TRUE);
$newArr = [];
foreach ($jsonArr as $v) {
$newArr[] = ['id:'=>$v['id'], 'task:' => $v['task']];
$newArr[] = ['ID:'=>'1_' . $v['id'], 'categoryId' => $v['categoryId'], 'mobile'=>$v['mobile']];
}
$newJson = json_encode($newArr);
var_dump($newJson);
[{
"id:": "1",
"task:": "test"
}, {
"ID:": "1_1",
"categoryId": "1",
"mobile": "111"
}, {
"id:": "2",
"task:": "test2"
}, {
"ID:": "1_2",
"categoryId": "2",
"mobile": "222"
}]
编辑--更新答案
正如在注释中所讨论的,您可以将SQL数组作为对象输出。我已经使用PDO::Fetch_ASSOC
将Fetch设置为作为关联数组输出,并更改了foreach()
循环以引用ASSOC数组$gs
。这应该可以工作,但如果不行,则使用var\u dump($gs)
再次输出$gs
的结果。如果需要,您仍然需要编码为JSON,但这一行已经注释掉了
function gt($user) {
$sql = "select id, id as categoryId, mobile, task from actions where status=0";
try {
$db = newDB($user);
$stmt = $db->prepare($sql);
$stmt->execute();
$gs = $stmt->fetchAll(PDO::FETCH_ASSOC); //Fetch as Associative Array
if ($gs) {
$db = null;
$newArr = [];
foreach ($gs as $v) { //$gs Array should still work with foreach loop
$newArr[] = ['id:'=>$v['id'], 'task:' => $v['task']];
$newArr[] = ['ID:'=>'1_' . $v['id'], 'categoryId' => $v['categoryId'], 'mobile'=>$v['mobile']];
}
//$newJson = json_encode($newArr); //JSON encode here if you want it converted to JSON.
} else {
echo "Not Found";
}
} catch(PDOException $e) {
//error_log($e->getMessage(), 3, '/var/tmp/php.log');
echo '{"error":{"text":'. $e->getMessage() .'}}';
}
}
如果你注释掉我的代码和var_dump($gs);您得到JSON输出了吗?你能把JSON放在你的问题中吗(编辑任何机密的东西)我能看到问题是什么,你试图
JSON\u decode()
和不是JSON的对象。要进行测试,请添加$gs=json\u encode($gs)
作为if()
语句的第一行,对象随后将进行json_解码并应完成。显然,这不是一个解决方案,所以让我知道这是否有效。好的,让我试着更新我的答案以匹配您的代码。请参阅解释以获得充分的理解,并让我知道你们进展如何。是的,这样做很有魅力!非常感谢你!