【发布时间】:2026-02-11 21:25:01
【问题描述】:
我正在使用以下代码将来自数据库的数据制作为 json 格式
public function employeeSearch()
{
$arrayOfEmployee = array();
$arrayToPush = array();
$arrayToJSON = array();
$new_item = $this->apicaller->sendRequest(array(
"controller" => "Employee",
"action" => "employeeSearch",
"searchCriteria" => "12345"
));
$arrayOfEmployee = json_decode($new_item,true);
foreach($arrayOfEmployee as $key => $employee)
{
$arrayToPush = array('data' => $employee['FullName'], 'value' => $employee['_id']['$oid']);
array_push($arrayToJSON, $arrayToPush);
}
echo json_encode($arrayToJSON);
}
输出是
[{"data":"Aasiya Rashid Khan","value":"5aa662b0d2ccda095400022f"},
{"data":"Sana Jeelani Khan","value":"5aa75d8fd2ccda0fa0006187"},
{"data":"Asad Hussain Khan","value":"5aaa51ead2ccda0860002692"},
{"data":"Ayesha Khan Khann","value":"5aab61b4d2ccda0bc400190f"},
{"data":"adhar card name","value":"5aaba0e1d2ccda0bc4001910"}
]
现在我希望 json 元素看起来像
{
"suggestions": [
{
"value": "Guilherand-Granges",
"data": "750"
},
{
"value": "Paris 01",
"data": "750"
}
]
}
我必须在 jQuery 自动完成插件中实现这个... 请帮忙!!!
【问题讨论】:
-
您现有的输出和所需的数据结构似乎没有共享任何值。例如,
"Ayesha Khan Khann"和"Paris 01"是如何相关的?"data": "750"来自哪里(你真的想要"750",而不是750)? -
另外,你试过什么?