BigQuery PHP API发送不在Google云端存储中的json [英] BigQuery PHP API send json that isn't in Google Cloud Storage

问题描述

我发现了一个例子，说明如何让这个json在Google云端存储中的文件：

  $ confLoad-> setSourceUris（array（
gs：// json_file_bucket /some_name.json
））; 
 $ conf = new \Google_JobConfiguration; 
 $ conf-> setLoad（$ confLoad）; 
 $ job-> setConfiguration（$ conf）; 
 $ service = new \Google_BigQueryService（$ this-> client）; 
 $ running = $ service-> jobs-> insert（$ this-> projectId，$ job）; 
  

但我不明白如何在没有外部json文件的情况下实现这一功能。

解决方案

如果您只添加一行，最好的选择是使用 TableData。 insertAll（）方法，它允许您一次导入一行或几行。更多信息请此处。

这里是python中的代码（我知道你在使用PHP，但它应该有点类似）：

  body = {rows：[
 {json：{column_name：7.7，}} 
]} 
 response = bigquery.tabledata（）。insertAll（
 projectId = PROJECT_ID，
 datasetId = DATASET_ID，
 tableId = TABLE_ID，
 body = body）.execute（）
  

另一种方法是使用媒体上传。这与将media_body参数添加到作业配置一样简单，其中media_body包含要加载的数据（并且不需要指定 sourceUris 。例如：

  job = {
'配置'：{$ b $'load'：load_config＃在此指定目标表
 
 $ b $ result = jobs.insert（
 projectId = project_id，
 body = job，
 media_body = media_body）.execute（）
  

有关PHP 这里。

I want to update an existing table I have in BigQuery with a single new row.

I found an example how this could be done if I have that json in a file in Google Cloud Storage:

$confLoad->setSourceUris(array(
    "gs://json_file_bucket/some_name.json"
));
$conf = new \Google_JobConfiguration;
$conf->setLoad($confLoad);
$job->setConfiguration($conf);
$service = new \Google_BigQueryService($this->client);
$running = $service->jobs->insert($this->projectId, $job);

But I don't understand how this can be achieved without an external json file.

解决方案

If you're adding just a single row, your best option is to use the TableData.insertAll() method, which lets you import a single row or a few rows at a time. More information is here.

Here is the code in python (I realize you're using PHP, but it should be somewhat similar):

body = {"rows":[
    {"json": {"column_name":7.7,}}
    ]}
response = bigquery.tabledata().insertAll(
    projectId=PROJECT_ID,
    datasetId=DATASET_ID,
    tableId=TABLE_ID,
    body=body).execute()

The alternative is to use 'media upload'. This is as simple as adding a media_body parameter to the job configuration, where media_body contains the data you are loading (and you don't need to specify sourceUris. For example:

job = {
    'configuration': {
      'load': load_config # specify destination table here
    }
  }
result = jobs.insert(
  projectId=project_id,
  body=job,
  media_body=media_body).execute()

There is more information about the media upload option in PHP here.

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