foreach names

slimjim

Does anyone know How I can get each image name so that I can insert each name into its own database field ?

Also I am trying to figure out how to echo an image name that does not get uploaded based on wrong ext and size...

I been at this all day so if you can help me out, that would be great.

<?

if($submit == "Send"){

$uploaddir = "upload"; 
//files to upload to

$allowed = array("image/jpeg", "image/pjpeg", "image/jpg", "image/gif"); 
//images allowed

$max_size = "50000"; // about 50kb 
//max size of file

$unique_id = md5(uniqid(time()));
//give image name a id

foreach ($_FILES["files"]["error"] as $key => $error) { 
   if ($error == UPLOAD_ERR_OK) { 
       $tmp_name = $_FILES["files"]["tmp_name"][$key]; 
       $name = $_FILES["files"]["name"][$key];
       $filename = $unique_id.'_'.$_FILES["files"]["name"][$key];
       $fame = move_uploaded_file($tmp_name, "$uploaddir/$filename"); 

if(in_array($_FILES['files']['type'][$key], $allowed))
{ 
$okk = "1";
} else {
	$hmm="Wrong image type</b>";
}
if($_FILES['files']['size'][$key] < $max_size)
{
$ok = "2";	
} else {
	$hmm="Wrong image size<BR><b>Image must not be larger then $max_size k";
}

if (($ok == "2") && ($okk == "1")){
       move_uploaded_file($tmp_name, "$uploaddir/$filename");
       echo "Success Image Upload";

} else {
	echo "$hmm";
   } 
}
}
}
?>
<html>

<head>
<title>test page</title>
</head>
<body>
<form action="upload_test2.php" method="post" enctype="multipart/form-data"> 
<p>Files: <BR>
<input type="file" name="files[]" /><BR>
<input type="file" name="files[]" /><BR>
<input type="file" name="files[]" /><BR>
<input type="file" name="files[]" /><BR>
<input type="file" name="files[]" /><BR>
<input type="file" name="files[]" /><BR>
<!-- You can add more fields by using the same syntax above. --> 
<input type="submit" name="submit" value="Send" /> 
</p> 
</form>
</body>
</html>

thorpe

so that I can insert each name into its own database field

why would you want to do that? you are heading for a poorly designed db. may i suggest you rethink it.

edcellgavin

your code is working fine, just comment out this line

$fame = move_uploaded_file($tmp_name, "$uploaddir/$filename");

regarding the image name and using your example, try echoing

$_FILES['files']['name'][n]

where n is the index of your file form element (0-4)

slimjim

it seems that if I use wrong extention in first input, i get the correct error
If i put in correct ext in first and wrong in second, they both get uploaded, which should not be able to.

Same goes with any of the inputs, I only catch the wrong ext file if it is in first one, if i use correct files in others at same time.

So I am not able to prevent wrong types if I use all inputs on form.

Can anyone help me please.

edcellgavin

try this coz you haven't re-initialized the $okk and $ok variable inside the loop

foreach ($_FILES["files"]["error"] as $key => $error) { 
   if ($error == UPLOAD_ERR_OK) { 
       $tmp_name = $_FILES["files"]["tmp_name"][$key]; 
       $name = $_FILES["files"]["name"][$key];
       $filename = $unique_id.'_'.$_FILES["files"]["name"][$key];
    print $_FILES['files']['type'][$key] . "<br>";
if(in_array($_FILES['files']['type'][$key], $allowed))
{ 
$okk = "1";
} else {
	$okk = 0;
    $hmm="Wrong image type</b>";
}
if($_FILES['files']['size'][$key] < $max_size)
{
$ok = "2";    

} else {
	$ok = 0;
    $hmm="Wrong image size<BR><b>Image must not be larger then $max_size k";
}

if (($ok == "2") && ($okk == "1")){
       move_uploaded_file($tmp_name, "$uploaddir/$filename");
       echo "Success Image Upload";

} else {
    echo "$hmm";
   } 
}
}

slimjim

If I am uploading 6 successful files and all file names were changed using " filename "

How do I get each file name that was a successfull upload so that each name can be inserted into thier own field in database ?

edcellgavin

do you want to save only the filename in the dabase or the file itself?
could you please give a more concrete example... tnx

slimjim

Only the file name of each image..

slimjim

If i use this :
$_FILES['files']['name'][$key]

that really doenst give me a way to get each image name so that I can do an insert , although i can echo each name, but that doesnt help me..

slimjim

forgot to say, each image gets a new name like this below...

d84f4c19223e7708537dd5cfc2d7aa53_food.gif

edcellgavin

why not store the value of $filename in the database. that value is the value of youf filename in your upload directory (privided that you have had a successful upload)? $filename is the value of the file itself, right?

slimjim

Yes that is what I want to do, is store that $filename value

But how do i get each file name is my problem...

If I just use $filename
I end up with all filenames in one database field

Need to get each $filename seperatly....

So that each name gets inserted into its own field in database

Somehow

edcellgavin

if were talking of one record - one insert, you could place the $filename in array at every loop execution and save it afterwards

foreach ($_FILES["files"]["error"] as $key => $error) {  

if (($ok == "2") && ($okk == "1")){ 
$arrFilenames[]  = $filename;
}
}

otherwise, do you insert inside the loop

foreach ($_FILES["files"]["error"] as $key => $error) { 
$query = "INSERT INTO TableName(filename) VALUES('$filename')";
}

slimjim

I have a table called images

in the table I have fields called :
image1
image2
image3

I need to get each $filename for each image uploaded into its own field

So what way should I use ?

edcellgavin

try this

$arrFilenames = array();
// inside the loop
foreach ($_FILES["files"]["error"] as $key => $error) {  

// code here
if (($ok == "2") && ($okk == "1")){ 
// successful upload
$arrFilenames[]  = $filename;
}
}

//after loop check if there are successfully uploded files
if (count($arrFilenames) > 0) {
	// insert uploaded
	$fields = "";
	$values = "";
	$sep = "";
	foreach ($arrFilenames AS $key=>$filename) {
		$key++; // increment by one since array key starts at 0
		$fields .= $sep . "image" . $key;
		$values .= $sep . $filename;  

		$sep = ",";
	}
	$query = "INSERT INTO images FIELDS($fields) VALUES($values)";
}

slimjim

where do I add the

       move_uploaded_file($tmp_name, "$uploaddir/$filename");

Because Now I am really confused on where to place what using my code above

edcellgavin

<?php
$arrFilenames = array();
	foreach ($_FILES["files"]["error"] as $key => $error) { 
   if ($error == UPLOAD_ERR_OK) { 
       $tmp_name = $_FILES["files"]["tmp_name"][$key]; 
       $name = $_FILES["files"]["name"][$key];
       $filename = $unique_id.'_'.$_FILES["files"]["name"][$key];
    print $_FILES['files']['type'][$key] . "<br>";
if(in_array($_FILES['files']['type'][$key], $allowed))
{ 
$okk = "1";
} else {
    $okk = 0;
    $hmm="Wrong image type</b>";
}
if($_FILES['files']['size'][$key] < $max_size)
{
$ok = "2";    

} else {
    $ok = 0;
    $hmm="Wrong image size<BR><b>Image must not be larger then $max_size k";
}

if (($ok == "2") && ($okk == "1")){
       move_uploaded_file($tmp_name, "$uploaddir/$filename");
       $arrFilenames[]  = $filename; 
       echo "Success Image Upload";

} else {
    echo "$hmm";
   } 
}
}

//after loop check if there are successfully uploded files
if (count($arrFilenames) > 0) {
    // insert uploaded
    $fields = "";
    $values = "";
    $sep = "";
    foreach ($arrFilenames AS $key=>$filename) {
        $key++; // increment by one since array key starts at 0
        $fields .= $sep . "image" . $key;
        $values .= $sep . $filename;  

        $sep = ",";
    }
$db = mysql_connect('host', 'username', 'password');
mysql_select_db("database", $db);
    $query = "INSERT INTO images FIELDS($fields) VALUES($values)";
mysql_query($query, $db);
}
?>

slimjim

This is uploading, but not inserting into db

$uploaddir = "upload"; 
//files to upload to

$allowed = array("image/jpeg", "image/pjpeg", "image/jpg", "image/gif"); 
//images allowed

$max_size = "50000"; // about 50kb 
//max size of file

$unique_id = md5(uniqid(time()));
//give image name a id

$arrFilenames = array(); 
    foreach ($_FILES["files"]["error"] as $key => $error) {  

   if ($error == UPLOAD_ERR_OK) {  

       $tmp_name = $_FILES["files"]["tmp_name"][$key];  

       $name = $_FILES["files"]["name"][$key]; 
       $filename = $unique_id.'_'.$_FILES["files"]["name"][$key]; 
    print $_FILES['files']['type'][$key] . "<br>"; 
if(in_array($_FILES['files']['type'][$key], $allowed)) 
{  

$okk = "1"; 
} else { 
    $okk = 0; 
    $hmm="Wrong image type</b>"; 
} 
if($_FILES['files']['size'][$key] < $max_size) 
{ 
$ok = "2";     

} else { 
    $ok = 0; 
    $hmm="Wrong image size<BR><b>Image must not be larger then $max_size k"; 
} 

if (($ok == "2") && ($okk == "1")){ 
       move_uploaded_file($tmp_name, "$uploaddir/$filename"); 
       $arrFilenames[]  = $filename;  

       echo "Success Image Upload"; 

} else { 
    echo "$hmm"; 
   }  

} 
} 

//after loop check if there are successfully uploded files 
if (count($arrFilenames) > 0) { 
    // insert uploaded 
    $fields = ""; 
    $values = ""; 
    $sep = ""; 
    foreach ($arrFilenames AS $key=>$filename) { 
        $key++; // increment by one since array key starts at 0 
        $fields .= $sep . "image" . $key; 
        $values .= $sep . $filename;   

        $sep = ","; 
    } 
    $query = "INSERT INTO images ($fields) VALUES($values)"; 
} 

}

edcellgavin

i've edited my last post. check the '$query = "INSERT INTO..." ' part

slimjim

Ok I thnk I am tired, I will try this again tomorrow. Talk to you later and good night and Thank You alot.
4am here..