我想在用户上传后提取docx，我想显示内容。但我似乎不知道该怎么称呼它……因为它一直显示"文件未找到"。如果我定义:
$document = try.docx
所以我知道它不能调用上传的文件。这是源代码:
<?php
include 'configure.php';
if(isset($_FILES['uploaded_file']))
{
$document = $_FILES ['uploaded_file']['tmp_name'];// here was the issue.. tried many way but still failed
function extracttext($filename)
{
$ext = explode('.', $filename);
$ext=end ($ext);
if($ext == 'docx')
$dataFile = "word/document.xml";
else
$dataFile = "content.xml";
$zip = new ZipArchive;
if (true === $zip->open($filename))
{
if (($index = $zip->locateName($dataFile)) !== false)
{
$text = $zip->getFromIndex($index);
$xml = new DOMDocument;
$xml->loadXML($text, LIBXML_NOENT | LIBXML_XINCLUDE | LIBXML_NOERROR | LIBXML_NOWARNING);
return strip_tags($xml->saveXML());
}
$zip->close();
}
return "File not found";
}
echo extracttext($document);
}
?>
$_FILES['uploaded_file']['tmp_name']不包含文档的名称，它包含类似'/tmp/asdjashdkjashda'的内容，您必须使用:$_FILES["uploaded_file"]["name"]来提取扩展名。
你的代码不工作，因为你是从'/tmp/asdjashdkjashda'提取扩展名，而不是'docx'，所以你总是在寻找$dataFile = "content.xml";(只适用于odt文件)。
所以，要获得扩展使用['name']，打开zip使用['tmp_name']:
$document_path = $_FILES ['uploaded_file']['tmp_name'];
$document_name=$_FILES ['uploaded_file']['name'];
function extracttext($filename,$filepath){
$ext = explode('.', $filename);
[...]
if (true === $zip->open($filepath))
[...]
}
echo extracttext($document_name,$document_path);