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POHow to access zip files in PHP code
 

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  1. POHow to access zip files in PHP code
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    <p>Hey Guys sorry about the bad formatting and thnks for your support. I will try to explain better this time. I have a zip file named Content_4_2.zip at location 'C:\xampp\htdocs\drupalTheme'. This zip contains 2 xml files. Now the web service method to which i want to pass this zip has the definition as follows.</p> <pre><code>&lt;soap:Body&gt; &lt;AddFileToProject xmlns="http://tempuri.org/"&gt; &lt;Ticket&gt;string&lt;/Ticket&gt; &lt;ProjectID&gt;string&lt;/ProjectID&gt; &lt;Filename&gt;string&lt;/Filename&gt; &lt;FileData&gt;base64Binary&lt;/FileData&gt; &lt;SourceLanguageID&gt;string&lt;/SourceLanguageID&gt; &lt;TargetLanguageIDs&gt; &lt;string&gt;string&lt;/string&gt; &lt;string&gt;string&lt;/string&gt; &lt;/TargetLanguageIDs&gt; &lt;Metadata&gt; &lt;Meta Name="string" Value="string" /&gt; &lt;Meta Name="string" Value="string" /&gt; &lt;/Metadata&gt; &lt;IsReferenceMaterial&gt;boolean&lt;/IsReferenceMaterial&gt; &lt;/AddFileToProject&gt; &lt;/soap:Body&gt; </code></pre> <p>The FileData 'base64Binary' is the content i want to send i.e. the zip file. In Java i was able to convert the zip file into a bytearray and the web service was called sucessfully.Now i know that there is no such conversion in PHP. So my approach is 1.Get the name of the zip file from the directory 2.Pass the contents of the zip to web service.</p> <p>I have tried this:</p> <pre><code> $zipFile = 'C:/xampp/htdocs/drupalTheme/Content_4_2.zip'; $zip = zip_open($zipFile); $addFile = $ServicesLink-&gt;AddFileToProject(array('Ticket'=&gt;$ticket,'ProjectID'=&gt;$projectID,'Filename'=&gt;$fileName,'FileData'=&gt; $zip, 'SourceLanguageID'=&gt; $srcLang, 'TargetLanguageIDs'=&gt; $finalTarArray,'Metadata'=&gt; null,'IsReferenceMaterial'=&gt;false)); </code></pre> <p>and this appraoch </p> <pre><code> $zipFile = 'C:/xampp/htdocs/drupalTheme/Content_4_2.zip'; $file = file_get_contents('./Content_4_2.zip',true); $byteArr = str_split($file); foreach ($byteArr as $key=&gt;$val) { $byteArr[$key] = ord($val); } $addFile = $ServicesLink-&gt;AddFileToProject(array('Ticket'=&gt;$ticket,'ProjectID'=&gt;$projectID,'Filename'=&gt;$fileName,'FileData'=&gt; $byteArr, 'SourceLanguageID'=&gt; $srcLang, 'TargetLanguageIDs'=&gt; $finalTarArray,'Metadata'=&gt; null,'IsReferenceMaterial'=&gt;false)); </code></pre> <p>Both result in corrupt zip files sent over.</p> <p>Would like to have your views on the same.</p>
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    1. USAngela Sim
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    1. COWelcome to SO! There are libraries available for this: Googling `drupal zip` or `php zip` should give you everything you need.
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      1. USPekka 웃
    2. CONow that it's reformatted, I'm not sure what your question is. Do you need to unzip the zip file, or is this about somehting else?
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      1. USPekka 웃
    3. CO@Angela Sim: Your question is not entirely clear - You say _"In Java i was able to convert the zip file into a bytearray"_, which sounds like you just converted (and sent) the whole zip file. But in your test example, you try to unzip the archive first and sent the contained files. Which file(s) do you want to sent - the zip archive as a whole, or the contained files separately?
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      1. USHenrik Opel
 

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