php array result to string question

htakeuchi · February 25, 2013

Hello:

I would like to request an opinion so maybe I can find the reason which I am not being able to see for the following.

I am using an array that will organize the content of a directory in the oldest to newest order, once it is ordered, display it's content.

So far, so good, where I find my self stumped is when I want to send the array result to a csv file for record keeping. I am getting the word "Array" instead of the file name,... What am I doing wrong?

I hope someone can assist me in understanding what am I doing wrong.

Thanks...

Hiroshi T

<?php

$files = glob( './images/*.*' );

array_multisort(

array_map( 'filectime', $files ),

SORT_NUMERIC,

SORT_ASC,

$files

);

echo '<html xmlns="http://www.w3.org/1999/xhtml">

<head>

</head>

<body>

Picture Count: <input type="text" name="pcount"><br>

</div>

</div>

</div>

</div>

</body>

</html>';

//$count = _POST["pcount"];

$imageDB = "imageDB.csv";

$DateStamp = date("m-d-Y");

$TimeStamp = date("H:i.s");

$openDB = fopen($imageDB, 'a') or die("Cannot update Database!");

$Data2appnd = "$files,$count,$DateStamp,$TimeStamp.\n";

fwrite($openDB,$Data2appnd);

fclose($opendDB);

?>

AyKay47 · February 25, 2013

"Array" is what is output when attempting to output an array directly, which is what $files is.

To maintain the CSV structure, perhaps implodeing the $files array using a particular character would be the best solution so that you can explode it upon reading the file:

$filesStr = implode("-", $files);

You could also insert a new line per $files element or store the data inside a database, depends on your logic

Edited February 25, 2013 by AyKay47

Barand · February 25, 2013

how do you want the appended data to look?

what does your array look like?

echo '<pre>', print_r($files , 1), '</pre>';

htakeuchi · February 25, 2013

Thank you for both answers...

The goal is to have a directory that is being populated by an FTP service, where images are being deposited, with this small php code, the goal is to open each image starting from the oldest inside the directory, once this is shown, using a form, the individual verifying the image, will set a comment via the form _GET or _POST. Once this is submitted, the image is moved to a different directory, (This part has not been written yet)

The part where I was asking for help, needs to record the opened file's name and the comment form the form in a csv file, which also records date and time stamps.

I believe "Barand's" option would be the best... I will get right to it and try both options and will report back... I sincerely appreciate your expertise.

Kind regards

Hiroshi T

AyKay47 · February 25, 2013

barands post was for information gathering purposes not a solution.

If I understand your logic correctly, you first need to solve the problem of iterating through all of the photos in the directory.

Perhaps something like this in between the <form> tags:

//loop through files
foreach( $files as $image )
{
 //generate html input elements for each image
 echo "<input type='hidden' name='image[]' value='$image' />";
 echo "<input type='text' name='count[]' />";
}

Then when validating form input:

//perform any scrubbing necessary on user input
$image = $_POST['image'];
$count = $_POST['count'];
//initialize other necessary variables
$dateStamp = date("m-d-y");
$timeStamp = date("H:i.s");
$DBHandle = fopen("imageDB.csv", "a");
if($DBHandle)
{
 //iterate through $_POST values from form and insert into csv file
 foreach( $image as $key => $img )
 {
   $data = "$img,$count[$key],$dateStamp,$timeStamp\n";
   fwrite($DBHandle, $data);
 }
}

This will display all of the images and corresponding text fields in the same form. When the user submits the form the script will store the necessary data into the csv file.

Keep in mind that this is only a skeleton to get you on the right path.

Edited February 25, 2013 by AyKay47

htakeuchi · February 26, 2013

Thank you very much!

AyKay47, thank you very much for your reply, you are correct. Although, for some strange reason and very "Stupid" on my part, What I needed was to record on the CSV is the name of the oldest file being displayed (in this case $files[0]), prior to my posting here, I truly believe I requested that on the fwrite() command, but something must have not been written correctly.

Last night I managed to get what I needed by simply adding the $files[0] on the data2append section, which in exchange, writes the file name on the CSV.

I am very grateful for your assistance and for taking time to assist me.

Here is how the Code looks right now and does what I wanted.

Thanks again,

Hiroshi T

---

<?php

$files = glob( './images/*.*' );

array_multisort(

array_map( 'filectime', $files ),

SORT_NUMERIC,

SORT_ASC,

$files

);

echo '<html xmlns="http://www.w3.org/1999/xhtml">

<head>

</head>

<body>

People Count: <input type="text" name="pcount"><br>

</div>

</div>

</div>

</div>

</body>

</html>';

$imageDB = "imageDB.csv";

$DateStamp = date("m-d-Y");

$TimeStamp = date("H:i.s");

$openDB = fopen($imageDB, 'a') or die("Cannot update Database!");

$Data2appnd = "$files[0],$count,$DateStamp,$TimeStamp.\n";

fwrite($openDB,$Data2appnd);

fclose($opendDB);

?>

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php array result to string question

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