file upload

[searls03]

I have this code:

<?php
if($_POST['submit']){
// Where the file is going to be placed
$target_path = "images/".$_SESSION['username']."/";

foreach ($_FILES["uploadedfile"]["name"] as $key => $value) {
$uploadfile = $target_path . basename($_FILES[uploadedfile][name][$key]);
//echo $uploadfile;
if (move_uploaded_file($_FILES['uploadedfile']['tmp_name'][$key], $uploadfile)) { 
echo $value . ' uploaded<br>'; }
}

}

?>

and I want to insert:

 $sql = mysql_query("UPDATE pics SET link='$target_path' WHERE id='$userid1'")or die(mysql_error());

so that when it uploads the file, it will also put info, such as a link to the file in the dbase.  how can I make $target_path = "images/".$_SESSION['username']."/ (the name of the file)"; I am stumped as to where to put the code or how to do this.  please help.

[phpSensei]

Hi Sean,

 

If you want to insert use INSERT not UPDATE. I also don't understand the problem here, you already seem to have figured it out as you explained your situation.

[searls03]

I want to insert that piece of code into the first piece.  I do not know where to put it or how to make it so that it is the file name on the end of $target_path.  I need to know where to place the second code inside of the first.  haha

[phpSensei]

Hi Searls,

 

That piece of code will go inside the Foreach loop, inbetween this line

 

 

if (move_uploaded_file($_FILES['uploadedfile']['tmp_name'][$key], $uploadfile)) { 
/// insert piece of code
echo $value . ' uploaded<br>'; 
}
}

[searls03]

And what do I usr for the file name piece?

[phpSensei]

Hi Searls,

 

The file name would be in full path $uploadfile = $target_path . basename($_FILES[uploadedfile][name][$key]);

[searls03]

So how exactly is the SQL statement written then????

[phpSensei]

Do some research man, Its not that hard.

 

 

Google "Mysql INSERT Tutorial"

[voip03]

$sql = "INSERT INTO tableName  (imageName, userid)  VALUES('" .$uploadfile ."','" .$_SESSION['username']."')";

http://www.w3schools.com/PHP/php_mysql_insert.asp

[phpSensei]

$sql = "INSERT INTO tableName  (imageName, userid)  VALUES('" .$uploadfile ."','" .$_SESSION['username']."')";

http://www.w3schools.com/PHP/php_mysql_insert.asp

 

You're simply doing all the work for him, let him learn and try it before asking us to type for him.

 

P.S, Stay away from w3schools.

[searls03]

I have this and it still does not update database, it uploads though........

<?php
if($_POST['submit']){
// Where the file is going to be placed
$target_path = "images/".$_SESSION['username']."/";

foreach ($_FILES["uploadedfile"]["name"] as $key => $value) {
$uploadfile = $target_path . basename($_FILES[uploadedfile][name][$key]);
//echo $uploadfile;
if (move_uploaded_file($_FILES['uploadedfile']['tmp_name'][$key], $uploadfile)) {  $sql = mysql_query("UPDATE pics SET link='$target_path', name='$uploadedfile'  WHERE id='$userid1'")or die(mysql_error());
echo $value . ' uploaded<br>'; }
}

}

?>

[voip03]

thanks  phpSensei

pl let me know why w3school not good?

[conan318]

 

P.S, Stay away from w3schools.

 

whats wrong with 3 schools?

 

php manual is pretty good

[phpSensei]

Even before this link this should have signal some warnings...

 

http://w3fools.com/

[chintansshah]

I have this and it still does not update database, it uploads though........

<?php
if($_POST['submit']){
// Where the file is going to be placed
$target_path = "images/".$_SESSION['username']."/";

foreach ($_FILES["uploadedfile"]["name"] as $key => $value) {
$uploadfile = $target_path . basename($_FILES[uploadedfile][name][$key]);
//echo $uploadfile;
if (move_uploaded_file($_FILES['uploadedfile']['tmp_name'][$key], $uploadfile)) {  $sql = mysql_query("UPDATE pics SET link='$target_path', name='$uploadedfile'  WHERE id='$userid1'")or die(mysql_error());
echo $value . ' uploaded<br>'; }
}

}

?>

 

Can you print your query on screen, if you get correct value then you should identify some other problem.

I think, your query should be

$sql = mysql_query("UPDATE pics SET link='".$target_path."', name='".$uploadedfile."'  WHERE id='".$userid1."'")or die(mysql_error());