How To Pass In Variable from PHP to Java

Question

I have a simple PHP form which allows user to selet a file for image processing as well as a Java program to generate some imagesas output. The PHP will make a copy of the image in a specified folder but I need to pass in the image name from PHP to Java in order for the java program to run for the uploaded file. Is there any way to do it? These are part of my current codes for PHP...

<form method="post" action="imageview.php" enctype="multipart/form-data">
Select Image File to Upload: <input type="file" name="image" />
<br><input type="submit" name="upload" value="Upload File!">
</form>

<?php
if (file_exists("photosmp/".$imagename))
{
    echo $imagename." already exists.";
    echo "It is stored in ".$imagetmpname;
}
else
{
    $newfilename = 'duplicate_'.$imagename;
    move_uploaded_file($imagetmpname,"photosmp/". $newfilename);
    $image = "photosmp/$newfilename";
}
?>

How are you calling Java from PHP?

anubhava
– anubhava

2014-08-11 05:24:56 +00:00
Commented Aug 11, 2014 at 5:24 — anubhava
– anubhava, Commented Aug 11, 2014 at 5:24
echo exec ("java PROG");

eyve
– eyve

2014-08-11 05:28:22 +00:00
Commented Aug 11, 2014 at 5:28 — eyve
– eyve, Commented Aug 11, 2014 at 5:28

anubhava · Accepted Answer · 2014-08-11 05:29:58Z

2

Call your Java program by passing a command line parameter:

echo exec ('java PROG "' . $image . '"');

answered Aug 11, 2014 at 5:29

anubhava

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5 Comments

eyve Over a year ago

is there any way to change this code in order to let the image file from PHP to be taken into this Java prog? FileDialog fd = new FileDialog(new Frame(), "Open",FileDialog.LOAD); fd.setFile("*.jpg"); fd.setDirectory(".\\"); fd.setLocation(50, 50); fd.setVisible(true); String FilePath = fd.getDirectory(); String FileName = fd.getFile(); String imageUpload = FilePath + FileName; String imageCopy = FilePath + "copy_" + FileName ;

anubhava Over a year ago

You must be having a main(String[] args) method in Java. Just access argv[1] as your image name which is being passed from command line using exec function of PHP.

eyve Over a year ago

may i know the way to go about it? cause im new in Java... this is my main(String[ args) method

public static void main(String[] args) throws IOException {     FileDialog fd = new FileDialog(new Frame(), "Open...", FileDialog.LOAD);     fd.setFile("*.jpg");     fd.setDirectory(".\\");     fd.setLocation(50, 50);     fd.setVisible(true);		     String FilePath = fd.getDirectory();     String FileName = fd.getFile();     String imageUpload = FilePath + FileName;     String imageCopy = FilePath + "copy_" + FileName ;

anubhava Over a year ago

See I cannot understand code from comments section but as I suggested you can use String imageName = ''; if (args.length>1) imageName = args[1]; in your Java code and use imageName variable for your processing.

anubhava Over a year ago

Glad to know it worked out, can you mark the answer as accepted by clicking on tick mark on top-left of my answer.

Hrishikesh Mishra · Accepted Answer · 2014-08-11 05:38:26Z

0

There are various system command are avalaible to fork a java program, like

exec(), system(), pctl etc

You can use these.

answered Aug 11, 2014 at 5:38

Hrishikesh Mishra

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