Rust

S3 Get Bucket Objects with CommonPrefixes

See more Amazon S3 Examples

Demonstrates how to get a list of bucket objects using the prefix and delimiter query params to get an XML result with CommonPrefixes. Rust

// This example assumes the Chilkat HTTP API to have been previously unlocked.
// See Global Unlock Sample for sample code.

let http = chilkat::Http::new();

// Insert your access key here:
http.set_aws_access_key("AWS_ACCESS_KEY");

// Insert your secret key here:
http.set_aws_secret_key("AWS_SECRET_KEY");

// In this example, my bucket is "chilkat100".
// It contains a number of folders, one of which is named "images".
// I want to get a list of all sub-folders under the "images" folder
let Ok(str_xml) = http.s3_list_bucket_objects("chilkat100?prefix=images/&delimiter=/") else {
    println!("{}", http.last_error_text());
    return;
};

println!("Response status code = {}", http.last_status());

let xml = chilkat::Xml::new();
if xml.load_xml(&str_xml).is_err() {
    println!("{}", xml.last_error_text());
    return;
}

// If the response status code was not 200, then the XML response is not a 
// listing of objects, but instead contains error information.
if http.last_status() != 200 {
    println!("{}", xml.get_xml().unwrap_or_default());
    println!("Failed.");
    return;
}

// A sample response is shown below.
println!("{}", xml.get_xml().unwrap_or_default());
println!("----");

// Here is the list of sub-folders (i.e. CommonPrefixes)

// <?xml version="1.0" encoding="UTF-8"?>
// <ListBucketResult xmlns="http://s3.amazonaws.com/doc/2006-03-01/">
//     <Name>chilkat100</Name>
//     <Prefix>images/</Prefix>
//     <Marker/>
//     <MaxKeys>1000</MaxKeys>
//     <Delimiter>/</Delimiter>
//     <IsTruncated>false</IsTruncated>
//     <CommonPrefixes>
//         <Prefix>images/africa/</Prefix>
//     </CommonPrefixes>
//     <CommonPrefixes>
//         <Prefix>images/sea_creatures/</Prefix>
//     </CommonPrefixes>
// </ListBucketResult>

let mut prefix = String::new();

// The XML can be parsed like this:
let mut i = 0;
let count_i = xml.num_children_having_tag("CommonPrefixes");
while i < count_i {
    xml.set_i(i);
    prefix = xml.get_child_content("CommonPrefixes[i]|Prefix").unwrap_or_default();
    println!("Prefix = {}", prefix);
    i = i + 1;
}