将api转换为PHP获取格式

Question

I'm a beginner in web design and I was trying this API
http://docs.imagga.com/

and it mention that auto-tagging need to use get request
and it give this:

-> it's a screen shot picture

but I have no idea how to send request

(after I saw the below answer given by Barmar) I have tried to type this code:

 $(document).ready(function(){
jQuery.ajax({
    url: 'http://api.imagga.com/v1/tagging',         
    type: "GET",                          
    data: { url: 'http://upload.wikimedia.org/wikipedia/commons/9/95/Mountain-bike-racing.jpg'} ,
    beforeSend: function(xhr) {
        xhr.setRequestHeader ("Authorization", " Basic my key");
    },
    success: function(response) {
            alert(response);
        }

});  });

but it comes to show that

• XMLHttpRequest cannot load http://api.imagga.com/v1/tagging . No 'Access-Contro l-Allow-Origin' header is present on the requested resource

thanks for your help

Answer 1

It should be something like this:

$(document).ready(function(){
    $.ajax({
        url: 'http://api.imagga.com/v1/tagging',         
        type: "GET",                          
        data: { url: 'http://upload.wikimedia.org/wikipedia/commons/9/95/Mountain-bike-racing.jpg',
        beforeSend: function(xhr) {
            xhr.setRequestHeader ("Authorization", "Basic YourAPIKey");
        },
        success: function(response) {
                alert(response);
        }
    });
});

You use the data: parameter to set URL parameters in GET requests. To implement authentication, you have to add an Authorization header, this is done using the Javascript setRequestHeader method on the XHR object.

Answer 2

I figured this question

the index.html file looks like

 $("button").click(function(){
$.get('post_url.php', function(rep){

and the post_url.php looks like

 function send_post($url){
$options = array(
    'http' => array(
        'method' => 'GET',
        'header' => 'Authorization: Basic key \r\n'
    )
);
$context = stream_context_create($options);
$result = file_get_contents($url, false, $context);

return $result;
};