检查 DynamoDb 中是否存在表的最佳方法是什么?
What is the best way to check if table exists in DynamoDb?
如果代码在 PHP 中，我将不胜感激.
I would appreciate it if the code would be in PHP.
无论是否活跃.
* 稍后作为示例添加到错误代码 400 的各种情况
* Added later as an example to various cases for error code 400
检查表是否存在很容易，它可以有以下之一TableStatus => 正在创建、活动、删除或更新
It's very easy to check if the table exist, it can have one of the following TableStatus => CREATING, ACTIVE, DELETING or UPDATING
但如果我收到错误 400，它可能意味着不止一件事.
but in case i get error 400 it can mean more than one thing.
1) 错误地将空字符串作为表名发送.
1) sent null string as a table name by mistake.
[x-aws-body] => {"TableName":""})
[x-aws-body] => {"TableName":""} )
[body] => CFSimpleXML Object
(
[__type] => com.amazon.coral.validate#ValidationException
[message] => The paramater 'tableName' must be at least 3 characters long and at most 255 characters long
)
[status] => 400
2) 发送到 DynamoDB 的命令中存在语法错误，例如写入 table_name 而不是 table_name.
2) syntax error in the command sent to DynamoDB, for example writting tabel_name instead of table_name.
[x-aws-body] => {"TabelName":"test7"})
[x-aws-body] => {"TabelName":"test7"} )
[body] => CFSimpleXML Object
(
[__type] => com.amazon.coral.validate#ValidationException
[message] => The paramater 'tableName' is required but was not present in the request
)
[status] => 400
3) 我会猜测但没有检查，如果我同时超过了桌子上的预置容量.
3) I would guess but didn't check, if I exceed at that same time the provisioned capacity on the table.
你可以看看官方PHP SDK的describe_table".400 表示不存在" 官方文档中有一个相当广泛的示例.看看它在删除"示例中是如何使用的，就在底部.
You can have a look at "describe_table" of the official PHP SDK. 400 means "does not exist" There is a pretty extensive example in the official documentation. Look at how it is used in the "delete" example, right at the bottom.
http://docs.amazonwebservices.com/amazondynamodb/latest/developerguide/LowLevelPHPTableOperationsExample.html
这是文档中的(剥离的)示例
Here is the (stripped) example from the doc
<?php
require_once dirname(__FILE__) . '/sdk/sdk.class.php';
$dynamodb = new AmazonDynamoDB();
$table_name = 'ExampleTable';
$response = $dynamodb->describe_table(array('TableName' => $table_name));
if((integer) $response->status !== 400)
{
$error_type = $response->body->__type;
$error_code = explode('#', $error_type)[1];
if($error_code == 'ResourceNotFoundException')
{
echo "Table ".$table_name." exists.";
}
}
?>
这篇关于检查 DynamoDB 中是否存在表的最佳方法是什么?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！