我对IPTC元数据有疑问。是否可以通过IPTC元数据(关键字)搜索不在数据库中的图像并显示它们,我将如何进行此操作?我只需要一个基本的想法。
我知道PHP有iptcparse()函数。
我已经编写了一个函数,可以通过.jpg扩展名获取图库文件夹和所有子目录中所有图像的图像名称,位置和扩展名。
我需要弄清楚如何提取元数据而不将其存储在数据库中以及如何搜索它,获取与搜索标记匹配的相关图像(它们的IPTC关键字应该匹配)以及如何显示它们。我知道我有最终结果(搜索后)我可以使用src =“$ filelocation”>回显图像标签。如果我在数组中得到最终结果。
基本上,我不确定是否需要将所有图像存储到mysql数据库中,并在实际搜索和显示结果之前提取关键字并将其存储在数据库中。另外,如果你可以引导我去任何已经能够做到这一点的画廊,那也可以提供帮助。
感谢您对此问题的任何帮助。
答案 0 :(得分:3)
目前尚不清楚特别是给你带来什么问题,但也许这会给你一些想法:
<?php
# Images we're searching
$images = array('/path/to/image.jpg', 'another-image.jpg');
# IPTC keywords to values (from exiv2, see below)
$query = array('Byline' => 'Some Author');
# Perform the search
$result = select_jpgs_by_iptc_fields($images, $query);
# Display the results
foreach ($result as $path) {
echo '<img src="', htmlspecialchars($path), '">';
}
function select_jpgs_by_iptc_fields($jpgs, $query) {
$matches = array();
foreach ($jpgs as $path) {
$iptc = get_jpg_iptc_metadata($path);
foreach ($query as $name => $values) {
if (!is_array($values))
$values = array($values);
if (count(array_intersect($iptc[$name], $values)) != count($values))
continue 2;
}
$matches[] = $path;
}
return $matches;
}
function get_jpg_iptc_metadata($path) {
$size = getimagesize($path, $info);
if(isset($info['APP13']))
{
return human_readable_iptc(iptcparse($info['APP13']));
}
else {
return null;
}
}
function human_readable_iptc($iptc) {
# From the exiv2 sources
static $iptc_codes_to_names =
array(
// IPTC.Envelope-->
"1#000" => 'ModelVersion',
"1#005" => 'Destination',
"1#020" => 'FileFormat',
"1#022" => 'FileVersion',
"1#030" => 'ServiceId',
"1#040" => 'EnvelopeNumber',
"1#050" => 'ProductId',
"1#060" => 'EnvelopePriority',
"1#070" => 'DateSent',
"1#080" => 'TimeSent',
"1#090" => 'CharacterSet',
"1#100" => 'UNO',
"1#120" => 'ARMId',
"1#122" => 'ARMVersion',
// <-- IPTC.Envelope
// IPTC.Application2 -->
"2#000" => 'RecordVersion',
"2#003" => 'ObjectType',
"2#004" => 'ObjectAttribute',
"2#005" => 'ObjectName',
"2#007" => 'EditStatus',
"2#008" => 'EditorialUpdate',
"2#010" => 'Urgency',
"2#012" => 'Subject',
"2#015" => 'Category',
"2#020" => 'SuppCategory',
"2#022" => 'FixtureId',
"2#025" => 'Keywords',
"2#026" => 'LocationCode',
"2#027" => 'LocationName',
"2#030" => 'ReleaseDate',
"2#035" => 'ReleaseTime',
"2#037" => 'ExpirationDate',
"2#038" => 'ExpirationTime',
"2#040" => 'SpecialInstructions',
"2#042" => 'ActionAdvised',
"2#045" => 'ReferenceService',
"2#047" => 'ReferenceDate',
"2#050" => 'ReferenceNumber',
"2#055" => 'DateCreated',
"2#060" => 'TimeCreated',
"2#062" => 'DigitizationDate',
"2#063" => 'DigitizationTime',
"2#065" => 'Program',
"2#070" => 'ProgramVersion',
"2#075" => 'ObjectCycle',
"2#080" => 'Byline',
"2#085" => 'BylineTitle',
"2#090" => 'City',
"2#092" => 'SubLocation',
"2#095" => 'ProvinceState',
"2#100" => 'CountryCode',
"2#101" => 'CountryName',
"2#103" => 'TransmissionReference',
"2#105" => 'Headline',
"2#110" => 'Credit',
"2#115" => 'Source',
"2#116" => 'Copyright',
"2#118" => 'Contact',
"2#120" => 'Caption',
"2#122" => 'Writer',
"2#125" => 'RasterizedCaption',
"2#130" => 'ImageType',
"2#131" => 'ImageOrientation',
"2#135" => 'Language',
"2#150" => 'AudioType',
"2#151" => 'AudioRate',
"2#152" => 'AudioResolution',
"2#153" => 'AudioDuration',
"2#154" => 'AudioOutcue',
"2#200" => 'PreviewFormat',
"2#201" => 'PreviewVersion',
"2#202" => 'Preview',
// <--IPTC.Application2
);
$human_readable = array();
foreach ($iptc as $code => $field_value) {
$human_readable[$iptc_codes_to_names[$code]] = $field_value;
}
return $human_readable;
}
答案 1 :(得分:0)
如果您没有从图像中提取这些IPTC数据,那么每次有人搜索时,您都必须:
如果你有超过几张图像,这对于表演来说真的很糟糕,我会说。
因此,在我看来,它会好得多:
这对你来说意味着更多的工作:你有更多的代码要编写......
...但这也意味着当有多个图片和许多用户进行搜索时,您的网站将有更好的生存机会。