我有一个返回某些字段的 SQL 查询,我正在使用 json_encode()
获取 JSON 格式的数据,但是我无法获取我想要的格式。
PHP代码
<?php
function data() {
$runDistanceBasedOnCityQuery = "SELECT rc.id, rc.cityId, c.cityName, rc.runId, r.distance, rc.status FROM run_city rc INNER JOIN cities c ON c.id = rc.cityId INNER JOIN run_distance r ON r.id = rc.runId ORDER BY c.cityName";
$runDistanceBasedOnCityResult = $db->prepare($runDistanceBasedOnCityQuery);
$runDistanceBasedOnCityResult->bindParam(":cityId", $cityId, PDO::PARAM_INT);
$runDistanceBasedOnCityResult->execute();
$runDistanceBasedOnCityOutput = $runDistanceBasedOnCityResult->rowCount();
if ($runDistanceBasedOnCityOutput > 0) {
while ($runDistanceBasedOnCityRow = $runDistanceBasedOnCityResult->fetch(PDO::FETCH_ASSOC)) {
$array1 = array($runDistanceBasedOnCityRow['runId'], $runDistanceBasedOnCityRow['distance'], $runDistanceBasedOnCityRow['status']);
for ($i = 0; $i < sizeof($array1); $i++) {
$array2 = array("id" => $runDistanceBasedOnCityRow['id'], "runId" => $runDistanceBasedOnCityRow['cityId'], $runDistanceBasedOnCityRow['cityName'] => $array1);
}
$finalResultRunDistanceBasedOnCity[] = $array2;
}
$responseRunDistanceBasedOnCity = $finalResultRunDistanceBasedOnCity;
} else {
$responseRunDistanceBasedOnCity = 'Runs not found';
}
$result = array("status" => true,
"runsBasedOnCity" => $responseRunDistanceBasedOnCity
);
json($result);
}
function json($data) {
header('Content-Type:application/json');
if (is_array($data)) {
echo json_encode($data);
}
}
?>
我得到的JSON格式
"runsBasedOnCity": [
{
"id": "1",
"runId": "1",
"Bengaluru": [
"2",
"10k",
"1"
]
},
{
"id": "2",
"runId": "1",
"Bengaluru": [
"1",
"5k",
"1"
]
},
{
"id": "3",
"runId": "1",
"Bengaluru": [
"5",
"3k",
"0"
]
},
{
"id": "4",
"runId": "2",
"Chennai": [
"1",
"5k",
"1"
]
},
{
"id": "5",
"runId": "2",
"Chennai": [
"2",
"10k",
"1"
]
},
{
"id": "6",
"runId": "2",
"Chennai": [
"4",
"15k",
"1"
]
}
]
我需要的格式
"runsBasedOnCity": [
{
"id": "1",
"cityId": "1",
"Bengaluru":
[
{
runId : "2",
distance : "10k",
status : "1"
},
{
runId : "1",
distance: "5k",
status : "1"
},
{
runId : "5",
distance : "3k",
status : "0"
}
]
},
{
"id": "2",
"cityId": "2",
"Chennai":
[
{
runId : "1",
distance : "5k",
status : "1"
},
{
runId : "2",
distance: "10k",
status : "1"
},
{
runId : "4",
distance : "15k",
status : "1"
}
]
}
我想不出更好的方法,我对此很陌生,请帮助我。谢谢!
最佳答案
要有效地对子数组数据进行分组,您应该实现临时键。 cityId
是一个适合分组的值——因为 cityNames
将来可能有意重复,但 cityId
绝不能在您的数据库中无意/有意重复表。
当遇到每个新的 cityId
时,有条件的 isset()
调用将确定是否应该存储一组新的/完整的数据,或者是否应该只存储数据附加到子数组。
我正在调用 array_slice()
,因为它减少了不必要的语法/代码膨胀。
遍历所有行后,您可以重新索引$result
数组,将其嵌套在runBasedOnCity
中,并添加status
元素.
我将使用 PRETTY_PRINT
展示我的演示,这样它更容易阅读,但在您的实际代码中,您应该删除该参数。另外,还有一点建议——尽量让您的变量名称简短,以提高可读性。
代码:(Demo)
$resultset = [
["id" => "1", "cityId" => "1", "cityName" => "Bengaluru", "runId" => "2", "distance" => "10k", "status" => "1"],
["id" => "2", "cityId" => "1", "cityName" => "Bengaluru", "runId" => "1", "distance" => "5k", "status" => "1"],
["id" => "3", "cityId" => "1", "cityName" => "Bengaluru", "runId" => "5", "distance" => "3k", "status" => "0"],
["id" => "4", "cityId" => "2", "cityName" => "Chennai", "runId" => "1", "distance" => "5k", "status" => "1"],
["id" => "5", "cityId" => "2", "cityName" => "Chennai", "runId" => "2", "distance" => "10k", "status" => "1"],
["id" => "6", "cityId" => "2", "cityName" => "Chennai", "runId" => "4", "distance" => "15k", "status" => "1"]
];
foreach ($resultset as $row) {
if (!isset($result[$row["cityId"]])) {
$result[$row["cityId"]] = array("id" => $row["id"], "cityId" => $row["cityId"], $row["cityName"] => array(array_slice($row,-3)));
} else {
$result[$row['cityId']][$row["cityName"]][] = array_slice($row,-3);
}
}
if (!isset($result)) { // don't need to check rowCount() at all
$result = 'Runs not found';
} else {
$result = array_values($result);
}
$result = array("status" => true, "runsBasedOnCity" => $result);
var_export(json_encode($result, JSON_PRETTY_PRINT));
输出:
'{
"status": true,
"runsBasedOnCity": [
{
"id": "1",
"cityId": "1",
"Bengaluru": [
{
"runId": "2",
"distance": "10k",
"status": "1"
},
{
"runId": "1",
"distance": "5k",
"status": "1"
},
{
"runId": "5",
"distance": "3k",
"status": "0"
}
]
},
{
"id": "4",
"cityId": "2",
"Chennai": [
{
"runId": "1",
"distance": "5k",
"status": "1"
},
{
"runId": "2",
"distance": "10k",
"status": "1"
},
{
"runId": "4",
"distance": "15k",
"status": "1"
}
]
}
]
}'
在解释了您希望如何在子数组中保留 id
值之后,解决方案如下:
代码:(Demo)
foreach ($resultset as $row) {
if (!isset($result[$row["cityId"]])) {
$result[$row["cityId"]] = array("cityId" => $row["cityId"], $row["cityName"] => array(array("id" => $row["id"])+array_slice($row,-3)));
} else {
$result[$row['cityId']][$row["cityName"]][] = array("id" => $row["id"])+array_slice($row,-3);
}
}
if (!isset($result)) { // don't need to check rowCount() at all
$result = 'Runs not found';
} else {
$result = array_values($result);
}
$result = array("status" => true, "runsBasedOnCity" => $result);
var_export(json_encode($result, JSON_PRETTY_PRINT));
关于php - 使用 PHP MySQL 创建嵌套的 JSON,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/49505372/