hellonoko

November 12, 2007

My below code works perfectly fine for uploading images. But when I change the file type to audio/mpeg for mp3s does it does not work.

Does not error either.

Any ideas?

Thanks for your time.

<?php

//include "admin_menu.php";

ini_set('display_errors', 1);
error_reporting(E_ALL);

?>


<title>Upload Image</title><form name="form1" method="post" action="upload_image.php" enctype="multipart/form-data">
<input name="imagefile" type="file" size="100">

<input type="submit" name="Submit" value="Upload"> 


<?php

//$submit = $_POST['Submit'];
if (empty($_FILES['imagefile']))
	{
		echo "<br><br>";
		echo "Please choose a image to upload...";
		die;
	}

if(!empty( $_POST['Submit'] ))
{

	//If the Submitbutton was pressed do:

	if ($_FILES['imagefile']['type'] == "audio/mpeg")
	{

		$_FILES['imagefile']['name'] = str_replace(" ", "_", $_FILES['imagefile']['name']);

		copy ($_FILES['imagefile']['tmp_name'], "music/".$_FILES['imagefile']['name']) or die ("Could not copy"); 

		echo "<br><br>"; 
        	//echo "<b>Name: </b>".$_FILES['imagefile']['name'].""; 
		//echo "<br>";
        	//echo "<b>Size: </b>".$_FILES['imagefile']['size'].""; 
        	//echo "<br>";
		//echo "<b>Type: </b>".$_FILES['imagefile']['type'].""; 
        	//echo "<br><br>";
		echo "<b>File Uploaded...</b>"; 
        }

 	else 
 	{
            echo "<br><br>";
            //echo "Could Not Copy... (".$_FILES['imagefile']['name'].")<br>";
		echo "Could Not Copy... <br>";
		die;
     	}
}   

echo "<br>";
//$image_to_thumbnail = $_FILES['imagefile']['name'];
//echo $image_to_thumbnail;

//include "resize_image.php";

//echo "<br>";
//echo "<img src=http://www.artbytoni.com/images/thumbs/t_".$image_to_thumbnail." />";
//echo "<br>";
//echo "<img src=http://www.artbytoni.com/images/full/".$image_to_thumbnail." />";
?> 

</form>

September 30, 2007

Can anyone advise as to why my below nested query code is not working?

Error is : Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\www\accordion\accordion\phpaccordion.php on line 340

Thanks!

			<?php

			$query = "SELECT * FROM items";

			$result = mysql_query($query);
			$rows = mysql_num_rows($result);

			for ($i=0; $i <$rows; $i++)
			{
				$row = mysql_fetch_array($result);

				//echos each item from DB

				echo		"<h1 class='accordion_toggle'>"  .$row[title]."</h1>";
				echo		"<div style='height: auto; display: none;' class='accordion_content'>";
				echo		"<h2>".$row[title]."</h2>";

				echo		"<form id=add_note' name='add_note' method='get' action='phpaccordion.php'>";
				echo			"<input name='action' type='hidden' value='add_note'>";
				echo			"<input name='note_content' type='text' value='new note'>";
				echo			"<input name='child_of' type='hidden' value=".$row[id].">";
				echo			"<input type='image' src='images/plus.gif' alt='New Note'>";
				echo		"</form>";

				echo		"<p>".$row[content]."</p>";


							$query2 = "SELECT * FROM main_item WHERE childof = '$row[id]'";

							$result2 = mysql_query($query2);
							$rows2 = mysql_num_rows($result2);

							for ($i2=0; $i2 <$rows2; $i2++)
							{
								$row2 = mysql_fetch_array($result2);

								echo "+".$row2[content]."";

							}	
				echo		"<form id=".$row[id]." name='delete' method='get' action='phpaccordion.php'>";
				echo		"<input name='item_to_delete' type='hidden' value=".$row[id]." />";
				echo		"<input type='hidden' name='action' value='delete'>";
				echo		"<p><input type='image' SRC='images/x.gif' ALT='Delete'></p>";
				//echo		"<input type='submit' value='Delete'>";
				echo		"<p></P>";
				echo		"</form>";
				echo		"</div>";

			}
		?>

September 27, 2007

I have a feeling im formatting this query string wrong and thats why it is not working.

Any ideas?

Thanks

if ( isset($current_category) )
{
	$query = "SELECT count(*) FROM gallery WHERE category == ".$current_category."";
}
else
{
	$query = "SELECT count(*) FROM gallery";
}

$result = mysql_query($query);
$query_data = mysql_fetch_row($result);
$numrows = $query_data[0];

echo $numrows;
exit();

April 23, 2007

The .JPG is just filler for a where a imagine link will go later.

Got it working.

Thanks for the help!

April 23, 2007

Here is my code again. Not sure what im missing. But it is not inserting. No errors either...

Thanks.

<?php

include 'dbconnect.php';

$title = $HTTP_GET_VARS[title];
$description = $HTTP_GET_VARS[description];
$media = $HTTP_GET_VARS[media];
$price = $HTTP_GET_VARS[price];
$imagelink = $HTTP_GET_VARS[image_link];

$action = $HTTP_GET_VARS[action];

echo $title;
echo "<br>";
echo $description;
echo "<br>";
echo $media;
echo "<br>";
echo $price;
echo "<br>";
echo $image_link;
echo "<br>";

if ($action == "Add")
{
	$query = "INSERT INTO gallery (title, description, media, price, image) VALUES ($title, $description, $media, $price, '.jpg')" or die (mysql_error('Error, insert query failed'));
	$result = mysql_query($query);
}


?>

<form name="add_gallery_item" method="get" action="add_gallery_item.php">
  <label>
  title:
<input type="text" name="title" id="title">
  </label>
  <p>description:  
    <label>
    <input type="text" name="description" id="description">
    </label>
  </p>
  <p>media:
    <label>
    <input type="text" name="media" id="media">
    </label>
</p>
  <p>price:
    <label>
    <input type="text" name="price" id="price">
    </label>
  </p>
  <p>imagelink:
    <label>
    <select name="image_link" id="image_link">
    </select>
    </label>
  </p>
  <p>
    <label>
    <input type="submit" name="action" id="add_item" value="Add">
    </label>
  </p>
</form>

April 22, 2007

Andy:

Thanks. Will give that a try.

April 22, 2007

Soft quotes hmm? Ill give it a try. Thanks.

April 22, 2007

Yea I didn't actually have the second echo in there.

April 22, 2007

Sometimes.

April 22, 2007

My bellow code is causing carriage returns/<br>'s between the end of the form and the last line:

echo $row[category];

This also happens if I join the last two lines:

echo "</form>" . echo $row[category];

echo "<form id=delete_category name=delete_category method=post action=categories.php>";
	echo "<input name=action type=submit value=Delete >"; 
	echo "<input name=id type=hidden value=".$row[id].">";
	echo "</form>";
	echo $row[category];

Any suggestions?

Thanks.

April 16, 2007

Using the below code:

<?php

$dictionary = fopen('US.dic','r');

while (!feof($dictionary))
{
$lines[] = fgets($dictionary, 24004096);
}

fclose($dictionary);

?>

I receive this error:

Fatal error: Allowed memory size of 8388608 bytes exhausted (tried to allocate 24004097 bytes) in C:\www\dic\d.php on line 8

The file is 1,185K

I am using WAMP for server. Do I need to change some settings or is there a better way to load large files.

Thanks

April 16, 2007

I need to read a .DIC file into an array so that I can manipulate it in PHP.

If I open this file in OpenOffice its a word on each line list so I could format it into a different type of file if that would be necessary.

Any suggestions?

Thanks.

February 24, 2007

I am not sure what your pointing out here Archadian?

echo "<br><br>";
		echo ""; 
        	echo "<b>Name: </b>".$_FILES['imagefile']['name'].""; 
		echo "<br>";
        	echo "<b>Size: </b>".$_FILES['imagefile']['size'].""; 
        	echo "<br>";
		echo "<b>Type: </b>".$_FILES['imagefile']['type'].""; 
        	echo "<br><br>";
		echo "<b>Copy Done....</b>";

The problem is that my code was working. And then for no reason it stopped working. When i added the code at the bottom of the page that is commented out.

Jessi:

I may just go for a restart and reinstall of WAMP

February 24, 2007

Returns:

Parse error: parse error, unexpected T_IF in C:\www\artbytoni.com\admin\upload_image.php on line 21

After adding:

<?php

$submit = $_POST['submit']

if(isset( $submit ))
{
	//If the Submitbutton was pressed do:

And the error reporting.

February 24, 2007

The problem is that the code was working perfectly.

Then suddenly started displaying code on page.

And then when I added in the full <?php tag stopped displaying but stopped working.

This is on both my machine and my host.

I will add the proper POST in.

February 24, 2007

$submit ?

A submit button?

I posted full code of the page in my first post.

<input type="submit" name="Submit" value="Upload"> 


<?php

if(isset( $Submit ))
{
	//If the Submitbutton was pressed do:

February 24, 2007

I turned them on in the WAMP menu.

It is my computer but at the same time I am trying code once it works on my host. 1and1.

February 23, 2007

Turned on 'short open tag'

Assume that it?

Still no luck.

February 23, 2007

Well that fixed it. Spitting out code on the page. But now the page is broken and does not function.

-Ian

February 23, 2007

my php page is suddenly displaying half code half what it should display.

I made a few minor changes to the code but nothing really.

Running this off of WAMP. Not getting any errors.

What causes this?

"; echo ""; echo "Name: ".$_FILES['imagefile']['name'].""; echo "

"; echo "Size: ".$_FILES['imagefile']['size'].""; echo "

"; echo "Type: ".$_FILES['imagefile']['type'].""; echo "

"; echo "Copy Done...."; } else { echo "

"; echo "Could Not Copy, Wrong Filetype (".$_FILES['imagefile']['name'].")

"; } } // echo "

"; // $image_to_thumbnail = $_FILES['imagefile']['name']; // echo $image_to_thumbnail; //include "resize_image.php"; ?>

Acctual code:

<?php

include "admin_menu.php";

?>


<title>Upload Image</title><form name="form1" method="post" action="" enctype="multipart/form-data">
<input name="imagefile" type="file" size="100">

<input type="submit" name="Submit" value="Upload"> 


<?

if(isset( $Submit ))
{
	//If the Submitbutton was pressed do:

	if ($_FILES['imagefile']['type'] == "image/jpeg")
	{
		copy ($_FILES['imagefile']['tmp_name'], "../images/full/".$_FILES['imagefile']['name']) or die ("Could not copy"); 

		echo "<br><br>";
		echo ""; 
        	echo "<b>Name: </b>".$_FILES['imagefile']['name'].""; 
		echo "<br>";
        	echo "<b>Size: </b>".$_FILES['imagefile']['size'].""; 
        	echo "<br>";
		echo "<b>Type: </b>".$_FILES['imagefile']['type'].""; 
        	echo "<br><br>";
		echo "<b>Copy Done....</b>"; 
        }

 else 
 {
            echo "<br><br>";
            echo "Could Not Copy, Wrong Filetype (".$_FILES['imagefile']['name'].")<br>";
     }
}   

//	echo "<br><br>";
//	$image_to_thumbnail = $_FILES['imagefile']['name'];
//	echo $image_to_thumbnail;

//include "resize_image.php";
?> 

</form>

February 23, 2007

I am trying to make the below image resizing code save to a different directory than the one the .php file is in.

I have tried a few things but without success. Any ideas?


<?php

//Name you want to save your file as
$file = 'original.jpg';

$save = 't_'.$file;

echo "Creating file: $save";

$size = 0.45;

// header('Content-type: image/jpeg') ;
list($width, $height) = getimagesize($file) ;

$modwidth = $width * $size;
$modheight = $height * $size;
$tn = imagecreatetruecolor($modwidth, $modheight) ;
$image = imagecreatefromjpeg($file) ;

imagecopyresampled($tn, $image, 0, 0, 0, 0, $modwidth, $modheight, $width, $height) ;

// Here we are saving the .jpg, you can make this gif or png if you want
//the file name is set above, and the quality is set to 100%

imagejpeg($tn, $save, 100) ;
?>

Thanks,

Ian

February 18, 2007

I have two tables.

One is a list of URLs and information about them uniqueID/url/categoryCode/description

The other is the different categories uniqueID/name

I am trying to go through my list of URLS and dispay the correstponding category name

For example if i have the URL www.phpfreaks.net in my urls table and its categoryID is 4

I want to the reference to my category table and if one of the fields id # is 4 which represents.. Web Design.

I want to be able to display Web Design rather than 4.

Below is my code wich I am sure is not the best way to do it and for some reason it only works for the first item.

<?php

include 'dbconnect.php';

$query = "SELECT * FROM urls";
$result = mysql_query($query);
$rows = mysql_num_rows($result); 	

$c_query = "SELECT * FROM categories";
$c_result = mysql_query($c_query);
$c_rows = mysql_num_rows($c_result); 	



for ($i=0; $i <$rows; $i++)
{

	$row = mysql_fetch_array($result);
	//echo $category_result;

	//$id_from_url_info = 10;

	for ($c=0; $c <$c_rows; $c++)
	{
		$c_row = mysql_fetch_array($c_result);

		if ( $row[category] == $c_row[id])
		{
			echo $c_row[category];
		}

	}

	echo	$row[category];
	echo	"<br>";
	echo	$row[url];
	echo	"<br>";
	echo	$row[description];
	echo	"<br><br>";	

}

?>

Here is one example someone gave me. But that would only return a blank page.

<?php

include 'dbconnect.php';


$sql="SELECT * FROM urls, categories WHERE categories.id = urls.category";

$result = mysql_query($sql);

while($row = mysql_fetch_array($result ))
{
	echo $row['category'];
	echo "<br>";
	echo $row['name'];
	echo "<br>";
	echo $row['url'];


}

?>

Any ideas?

Thanks,

Ian

February 18, 2007

That looks like what I want but it only returns :::::::

I also needed to turn then query into this:

	$sql="SELECT * FROM urls, categories WHERE categories.id = urls.category";

But that returned the same thing.

Basically there is a list of URLS and their information in one table for example:

ID: 9

URL: www.phpfreaks.net

category: 4

desription: a useful site

when I run through the list of entries in the URL table and i reach the CATEGORY field (4)

I want to cross reference to the CATEGORIES table where the entries look like.

ID: 4

Category: coding websites

So that instead of displaying 4 as the category for www.phpfreaks.net my code looks up the name for category 4 is and displays.. coding websites

This way later on if i want to charge the category name i can.

Does that explain it clearly?

Thanks,

Ian

February 18, 2007

well i couldnt see how the first one works and i couldn't get the second one to work.

here is what i am trying to do. but it only works on the first time through the loop for some reason.

and im sure this is really bad way to do this.

<?php

include 'dbconnect.php';

$query = "SELECT * FROM urls";
$result = mysql_query($query);
$rows = mysql_num_rows($result); 	

$c_query = "SELECT * FROM categories";
$c_result = mysql_query($c_query);
$c_rows = mysql_num_rows($c_result); 	



for ($i=0; $i <$rows; $i++)
{

	$row = mysql_fetch_array($result);
	//echo $category_result;

	//$id_from_url_info = 10;

	for ($c=0; $c <$c_rows; $c++)
	{
		$c_row = mysql_fetch_array($c_result);

		if ( $row[category] == $c_row[id])
		{
			echo $c_row[category];
		}

	}

	echo	$row[category];
	echo	"<br>";
	echo	$row[url];
	echo	"<br>";
	echo	$row[description];
	echo	"<br><br>";	

}

?>

February 18, 2007

I have two tables in my database: categories and urls

categories has the fields ID and CATEGORY where ID is a unique identifier.

urls has the fields ID CATEGORY URL and DESCRIPTION where ID is a unique identifier and CATEGORY is the the identifier from the categories table.

I am not sure how to write a script so that when I list all the entries in URLS i can display the matching label for the category ID rather than the #.

Thanks

Ian

Sign In

hellonoko

Posts

Joined

Last visited

Content Type

Profiles

Forums

Posts posted by hellonoko

upload script works for images but not mp3

nested queries not working

using variables in the query string

[SOLVED] INSERT QUERY... not inserting.

[SOLVED] INSERT QUERY... not inserting.

code inserting <br> in output

code inserting <br> in output

code inserting <br> in output

code inserting <br> in output

code inserting <br> in output

Allowed Memory Error when loading large file.

Reading a .DIC dictionary file to array.

web page displaying code?

web page displaying code?

web page displaying code?

web page displaying code?

web page displaying code?

web page displaying code?

web page displaying code?

web page displaying code?

thumbnail code / saving image to different directories

pulling coresponding data from tables

matching fields from tables

matching fields from tables

matching fields from tables

Browse

Activity

Important Information