Question

我正在使用网站上的API将数据推送到服务器。每当我的arduino检测到脉冲时，它都会将其发送给COSM。通过使用触发器，数据将传输到我的服务器。我最初将数据写入.txt文件和Json对象，但现在我想开始实现mongo以拥有不同的集合。

由于某种原因，当我尝试将其作为Mongo集合添加时添加Mongo连接后，数据未传输到我的服务器。我希望能够直接在Mongo中写下信息，避免创建文件。

任何建议都非常受欢迎，以下是代码：

<?php

// Creating a data base in Mongodb to store my information from my $jsonFile
$connection = new MongoClient(); //Connect to mongo
$db = $connection -> qandm; // select my DB which is called qandM
$collection = $db -> pulseData;

//Gets all the information from Cosm
$pulse = $_POST['body'];

//access and open a file
//echo $pulse;

//Converts the data into PHP arrays
$pulseObj = json_decode($pulse, true);

//Parse through the specific information from the array and gets each piece of information in an array
$userName = $pulseObj["triggering_datastream"]["id"];
$dataTime= $pulseObj["triggering_datastream"]["at"];
$dataValue= $pulseObj["triggering_datastream"]["value"]["value"];


//Writes all the data coming from COSM
$file = fopen("data.txt", "a+");//a+ give the option to have the cursor at the end to access the file read and write it
    /*  $pulse.="\r\n"; */
fwrite($file, $pulse);//takes incoming data and writes it in the file
fclose($file);

//Opens a new .txt file and writes the values that we selected before into our file
$string = $userName." ".$dataTime." ".$dataValue." \r\n";
//error_log allows me to see in my Apache log server the information that I'm printing
error_log($string);

//Write all the information I parsed in my three variables in a new file
$file2 = fopen("rawData.txt", "a+");
fwrite($file2,$string);
fclose($file2);

//json sample

//Inputs the data from the time and the pulse value into a json object
$json = array("User" => $userName, "timestamp"=> $dataTime, "value"=> $dataValue);

//Opens a new json object
$jsonFile = fopen("data.json", "a+");

//Writes the data of our new arrayed information into the open json object
fwrite($jsonFile, json_encode($json));
fclose($jsonFile);




//A loop to populate
foreach($json as $data){
    $collection->insert($data);

}

//find the data I just stored
$cursor = $collection->find();

//Output it in a UL
echo "<p> My Pulse </p>";

echo '<ul>';
foreach($cursor as $doc){

    echo' <li> My pulse is: '.$doc['value'];

}
 echo '</ul>';


/*$data = "data.txt";
$fh = fopen($data, "w") or die ("can't open file");
$data = json_encode($_POST);
fwrite($fh, $data);
fclose($fh);*/

//print_r($file);



?>

Answer 1

这可能是您问题的根源：

//A loop to populate
foreach($json as $data){
    $collection->insert($data);

}

您正在迭代$ json数组并将值（但不是键）传递给MongoCollection的insert方法。根据{{3}}，此方法需要一个对象或一个数组。根据我的理解，您要尝试执行此操作的代码，foreach循环应替换为以下内容：

$collection->insert($json);

这将创建一个类似于您的阵列的mongo对象。您的代码当前正在尝试将每个数组元素的值作为单独的条目插入。

在Mongodb中存储实时数据

1 个答案: