Question

我正在尝试从python将图像文件上传到piczasso.com。我的代码如下：

import urllib,urllib2

url='http://piczasso.com/api_handler.php'
values = { 'data':open('./imagefile.jpg'),
           'tags':'',
           'size':'None'
}

data = urllib.urlencode(values)
req = urllib2.Request(url,data)
try:
        response = urllib2.urlopen(req).read()
        print response
except urllib2.URLError, e:
        print e.code
        print e.read()

print "done"

然而，我得到“无效的网址！”作为输出。然而，我可以使用php页面上传图像，如下所示。

<?php

/* 

Simple sample script which uses Piczasso API

Documentation about PicZasso API can be found here: http://www.piczasso.com/api.php

*/



if(isset($_GET["image"])){

    /* API sends output in JSON format */

    $image = json_decode(rawurldecode($_GET["image"]), true);

    /*

    $image array includes information about the image URL and sample links to it:

        $image["direct"] = Image URL

        $image["thumb"] = Thumbnail URL

        $image["image_page"] = Link to image

        $image["bbcode"] = bbcode for forums

        $image["html"] = html code

    Sample table below with image links.

    */

    $image["html"] = str_replace("+", "&nbsp;", htmlentities($image["html"]));

    echo "

    <img src=\"{$image["thumb"]}\" alt=\"\" />

    <table>

        <tr><td>HTML for Websites:</td><td><input type=\"text\" name=\"codes1\" value=\"{$image["html"]}\" /></td></tr>

        <tr><td>IMG Code for Forums:</td><td><input type=\"text\" name=\"codes2\" value=\"{$image["bbcode"]}\" /></td></tr>

        <tr><td>URL for E-Mail:</td><td><input type=\"text\" name=\"codes3\" value=\"{$image["image_page"]}\" /></td></tr>

        <tr><td>Direct Link for Layouts:</td><td><input type=\"text\" name=\"codes4\" value=\"{$image["direct"]}\" /></td></tr>

    </table>

    ";

} else {

    if(isset($_GET["error"])){

    /* Possible errors. Complete error list: http://www.piczasso.com/api_errorcodes.txt */

        $errorcodes = array(

            1000 => "No image selected",

            1001 => "Image failed to upload",

            1002 => "Invalid image type",

            1003 => "Image is larger than 16MB",

            1004 => "Cannot read image info",

            1005 => "Upload failed during process",

            1006 => "Database error",

            1007 => "Banned IP"

        );

        echo $errorcodes[$_GET["error"]];

    }

    /* Sample upload form below. Image can be resized also by sending for example this kind of fields:

        <tr><td>Resize x:</td><td><input type=\"text\" name=\"size_x\"/></td></tr>

        <tr><td>Resize y:</td><td><input type=\"text\" name=\"size_y\"/></td></tr>

    */ 

    echo "

    <form action=\"http://piczasso.com/api_handler.php\" method=\"post\" enctype=\"multipart/form-data\">

    <input type=\"hidden\" name=\"MAX_FILE_SIZE\" value=\"99999999\" />

    <table>

        <tr><td>File:</td><td><input type=\"file\" name=\"file\" /></td></tr>

        <tr><td>Tags:</td><td><input type=\"text\" name=\"tags\" /></td></tr>

        <tr><td>Resize:</td><td>

        <select name=\"size\">

            <option value=\"1\" selected=\"selected\" >None</option>

            <option value=\"2\" >100x75 Avatar</option>

            <option value=\"3\" >150x112 Thumbnail</option>

            <option value=\"4\" >320x240 Websites</option>

            <option value=\"5\" >640x480 For forums</option>

            <option value=\"6\">800x600 15-inch monitor</option>

            <option value=\"7\" >1024x768 17-inch monitor</option>

        </select></td></tr>

        <tr><td><input type=\"submit\" value=\"send\" /></td></tr>

    </table>

    </form>

    ";

}

?>

我的python代码有什么问题？

请帮忙谢谢

Answer 1

在PHP代码中，表单包含图像大小选项。值是数字，而显示的选项是文本。 “无”的值为1。

values = { 'data':open('./imagefile.jpg'),
           'tags':'',
           'size':'1'
}

Answer 2

import urllib2_file
import urllib2

data = {
        'apikey': '**********************************—0',#your APIKey
        'imgurl' : "www.example.net/img/dog1.jpg",
         'catid','0',
        'subject':'',
        'labeland':'',
        'labelor':'',
        'labelnot':''
       }
f = urllib2.urlopen('http://api0.example.com:/api/v1.0/apisim_search', data)
content = f.read()
print(content)

使用urllib2上传图像文件时出现无效的URL消息

2 个答案: