Question

我想显示json响应中的特定值，但某些功能无法正常工作

这是json：

 {"visitorCount":282,"totalPasserbyCount":4303,"totalVisitorCount":282,"totalConnectedCount":249,"connectedPercentage":88,"conversionRate":6,"averageDwell":89.85931558935361,"averageDwellByLevels":{"FIVE_TO_THIRTY_MINUTES":{"average":14.980392156862745,"count":52},"THIRTY_TO_SIXTY_MINUTES":{"average":56.42,"count":49},"ONE_TO_FIVE_HOURS":{"average":88.17123287671232,"count":146},"FIVE_TO_EIGHT_HOURS":{"average":378.3333333333333,"count":9},"EIGHT_PLUS_HOURS":{"average":538.5714285714286,"count":7}},"topManufacturers":{"name":"Apple","count":21,"manufacturerCounts":{"Apple":21,"Hon Hai Precision":7,"Intel":19,"Liteon":7,"Samsung":20},"avgManufacturerCounts":{"Apple":4,"Hon Hai Precision":1,"Intel":4,"Liteon":1,"Samsung":4}},"peakSummary":{"peakHour":0,"peakDate":null,"peakWeek":0,"interval":null,"peakHourCount":224,"averageHourlyCount":0.0,"hourlyCounts":null,"maxHour":0,"maxDay":0,"peakDayCount":0,"peakHourDay":"2018-08-03"},"peakWeekSummary":null,"peakMonthSummary":null}

这就是我尝试显示visitorCount号码的方式

$response = curl_exec($curl);
$err = curl_error($curl);

curl_close($curl);

if ($err) {
 echo "cURL Error #:" . $err;
} else {

  json_decode($response,true);

echo $response['visitorCount'];

打开页面时，我唯一得到的是{
而且，如果我只回显$response而没有做任何事情，我会得到所有的json响应

Answer 1

Sub MoveFirstNonZeroCell() Dim R As Range With ThisWorkbook.Sheets("Sheet1") For Each R In .Range("3:3") If R.Value <> 0 Then R.Cut Destination:=R.Offset(, 1) Exit For End If Next R End With End Sub需要一个变量才能起作用。快速示例

json_decode

希望这会有所帮助。

Answer 2

以下是解决您的问题的完整代码：

$response = curl_exec($curl);
$err = curl_error($curl);

curl_close($curl);

if ($err) {
 echo "cURL Error #:" . $err;
} else {
 $response_data = json_decode($response,true);
 echo $response_data['visitorCount'];
}

Answer 3

尝试一下

$response = curl_exec($curl);
$err = curl_error($curl);

curl_close($curl);

if ($err) {
echo "cURL Error #:" . $err;
} else {

$response = json_decode($response,true);

echo $response['visitorCount'];

}

您在当前代码的底部缺少右括号。

如果上面的代码不起作用，请尝试进行var dump，以查看您的返回结果：

var_dump($response);

从JSON响应中获取特定值，什么也不显示PHP

3 个答案: