问题描述
检查表是否存在于DynamoDb中的最佳方法是什么？
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.
是否处于活动状态。
* 稍后作为示例添加到错误代码为400的各种情况下
* Added later as an example to various cases for error code 400
检查表是否存在非常容易，它可以具有以下
TableStatus之一=>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 followingTableStatus => CREATING, ACTIVE, DELETING or UPDATING
但如果出现错误400，则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的命令中的语法错误，例如用tabel_name代替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）我猜但没有检查，如果我同时超过了规定
3) I would guess but didn't check, if I exceed at that same time the provisioned capacity on the table.
推荐答案
您可以查看官方PHP的 describe_table SDK。 400 表示 不存在。官方文档中有一个非常详尽的示例。在底部的删除示例中查看它的用法。
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.
这里是（摘录）来自doc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.";
}
}
?>
这篇关于检查表在DynamoDB中是否存在的最佳方法是什么？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！