PHP - Upload Script Not Working Online But A Ok On Server?

Code: [Select]

Hi,



I am calling the sms service provider server with the following code


$url = $urlAccount_home."?username=".$userAccount."&password=".$passAccount."&number=".$number."&sender=".$userSenderSMS."&msg=".$msg;


             if (!($fp =fopen($url,"r")))
echo "<br><center>خطا في الاتصال </center><br>";

$result =fread($fp,10);
fclose($fp);








When sending sms to 1 person no problem, but when sending sms to more than 1 person example for a group,

i will poll the serve in loop...

what am afraid is calling the server more than once continuously, wont affect the service?

Hi, I have a some online backup software call ahsay, I' am looing to create a API call with a form in PHP

Please check out this link www.123-backup.com/obs-admin-guide.pdf

this is the guide to create a api for the  online backup (page 69)

Would some be intrested in creating this api call and php form for me ?

p.s please see attached image this is something i'am looking for.

if so how much would it be please email me if you want more info!

Sorry if this is a silly question lol.

I want to know if there is an online tool I can use to check and see what PHP code will ouput server side.

For example:

<?php
$tomorrow  = mktime(0, 0, 0, date("m")  , date("d")+1, date("Y")); ?>

Is it possible to simply fetch from the server(mySql) and then dump it online through a PHP page?
Is there some link that you can share?

Hi everyone,

I have a page that i use to upload images to my website, i got a bit fed up of uploading one at a time so i decided to add multiple file fields to the form to upload multiple images at the same time.

Im having a few problems, iv read up he http://www.php.net/manual/en/features.file-upload.multiple.php and it seems all i have to do is add [] to the form names to turn them into arrays. However when i come to upload the images, i keep getting the "$error[] = "Incorrect format!...." error from the code below. I cant seem to figure out what the problem is.

Could anybody please point me in the right direction?


<?php 
session_start();

$id = $_SESSION['id'];

$connect = mysql_connect("localhost","leemp5_admin","p7031521");
mysql_select_db("leemp5_database");

$query = mysql_query("SELECT * FROM users WHERE id='$id'");
$row = mysql_fetch_assoc($query);

$username = $row['username'];

$submit = $_POST['submit'];

$type = $_FILES['image']['type'];
$size = $_FILES['image']['size'];
$max_size = "1000"; 
$width = "100"; 
$height = "100";
 
$error = array();

function make_thumb($image_name,$filename,$new_width,$new_height) 
{

$ext=getExtension($image_name); 

if(!strcmp("jpg",$ext) || !strcmp("jpeg",$ext)) 

$source_image=imagecreatefromjpeg($image_name);   

if(!strcmp("png",$ext)) 

$source_image=imagecreatefrompng($image_name);

if(!strcmp("gif",$ext)) 

$source_image=imagecreatefromgif($image_name); 

$old_x=imageSX($source_image);

$old_y=imageSY($source_image);   

$ratio1=$old_x/$new_width; 

$ratio2=$old_y/$new_height; 

if($ratio1>$ratio2)

{ 

$thumb_width=$new_width; 

$thumb_height=$old_y/$ratio1; 

} 

else 

{ 

$thumb_height=$new_height;

$thumb_width=$old_x/$ratio2; 

} 
  

$destination_image=ImageCreateTrueColor($thumb_width,$thumb_height);   

imagecopyresampled($destination_image,$source_image,0,0,0,0,$thumb_width,$thumb_height,$old_x,$old_y);   

if(!strcmp("jpg",$ext) || !strcmp("jpeg",$ext))

{

imagejpeg($destination_image,$filename); 

}

if(!strcmp("png",$ext)) 

{

imagepng($destination_image,$filename); 

}

if(!strcmp("gif",$ext)) 

{

imagegif($destination_image,$filename); 

}

imagedestroy($destination_image); 

imagedestroy($source_image); 
}   


function getExtension($str)
{ 

$i = strrpos($str,"."); 

if (!$i) { return ""; } 

$l = strlen($str) - $i; 

$ext = substr($str,$i+1,$l); 

return $ext; 
}   

if($submit) 
{ 

$image=$_FILES['image']['name']; 

if ($image) 

{ 

$filename = stripslashes($_FILES['image']['name']);   

$extension = getExtension($filename); 

$extension = strtolower($extension); 

if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif"))

{ 

$error[] = "Incorrect format! Please make sure your image is a .jpg, .jpeg, .png or .gif file.";

} 

else

{ 

$size=getimagesize($_FILES['image']['tmp_name']); 

$sizekb=filesize($_FILES['image']['tmp_name']); 
  

if ($sizekb > $max_size*1024) 

{ 

$error[] = "Your image is too big! The maximum upload size is 1MB.";

}   

else

{

$image_name=time().'.'.$extension; 
 

$newname="uploads/" . $username . "/images/".$image_name; $copied = copy($_FILES['image']['tmp_name'], $newname); 

if (!$copied) 

{ 

$error[] = "There was an error uploading your image. Please try again!";

} 

else 

{ 

$thumb_name='uploads/' . $username . '/images/thumbs/thumb_'.$image_name; 
 

$thumb=make_thumb($newname,$thumb_name,$width,$height); 

}

}

}

}

else

{

$error[] = "Please select an image to upload!";

}

if(empty($error))

{

echo "Upload Successfully!<br />"; 

echo '<img src="'.$thumb_name.'">';

mysql_query("INSERT INTO images VALUES ('','$username','$image_name','','','','','uploads/$username/images/$image_name','uploads/$username/images/thumbs/thumb_$image_name','$type','$size')");

}

else

{
    

echo implode($error);

}
}
?>
<form method="post" enctype="multipart/form-data" action="upload_images.php">
<input type="file" name="image[]" /><br />
<input type="file" name="image[]" /><br />
<input type="file" name="image[]" /><br />
<input type="file" name="image[]" /><br />
<input type="file" name="image[]" /><br />
<input type="file" name="image[]" /><br />
<input type="submit" name="submit" value="Upload">
</form>


Thanks

I have a script that uploads my file, but this script requires apache. I want to get this file onto my other server that's running nginx. Is there a way for PHP to write a file to nginx?

Alternatively, is there a way for PHP to copy a file from one server to another? Because that will work as well.

How do I do this?

i am making song site
i want to upload songs files on another instread on which m running my script for uploading songs and adding information in data...
mean want to know is there any function or method that i upload files through the panel of www.site1.com
and my files upload on an other server www.site2.com.
which code i have to write for doing this

Hi

I have a form which is used to upload files to my server, now i need to move files to another server as soon they are uploaded, but move_uploaded_file will work with local server where form has been submitted.

 i was thinking to add some ftp functions to upload that file to remote server but is there any better way?  that user dont notice ftp connection timing. i wants to use less ftp connections and user should not notice delay timing for image upload.

Thanks  for any suggestion.

Hi there! im new to php , im having problem when upload image to server. Hope somebody can help me.

i upload picture in admin CMS, the pic able to save in mysql but is not save in my server, the script below :
Code: [Select]
$catImage = uploadImage('fleImage', SRV_ROOT . 'images/category/'); 
i have check the folder/directory name all correct but the picture wont save in my server images folder & i try this script but not working as well:
Code: [Select]
$target_path = "images/category/";
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']); 
i got this tutorial from :
http://www.phpwebcommerce.com/

Thank You In Advance

I have a website where users can add their friends. What I am trying to achieve is to show every user that who are their friends that are currently online. If any one knows a link to any script or tutorial that explains this, Please help.

Thanks,

Faisal

I looked for a script that shows the number of users online on the site i am making, and found this. I don't know if it is for just showing if the user is online on that one page like a profile page, or if it is for all users in the database. Also, when using the code, it shows "Users Online: 1" then when I reload the page, it says "2' and again, then "3" even though no users have logged in. Thank you.

the script

Code: [Select]
<?php /////////////////////////////////////////////////////////////////////////////////////////


////IS USER ONLINE SCRIPT PART////

 

$session=session_id();
$time=time();
$time_check=$time-600; //SET TIME 10 Minute

 //CONNECT TO DB
//connect info here


// Connect to server and select databse
mysql_connect("$host1", "$username1", "$password1")or die("cannot connect to server");
mysql_select_db("$db_name1")or die("cannot select DB");

$sql="SELECT * FROM $tbl_name1 WHERE session='$session'";
$result=mysql_query($sql)or die(mysql_error());

$count=mysql_num_rows($result);

if($count=="0"){
$sql1="INSERT INTO $tbl_name1(session, time)VALUES('$session', '$time')";
$result1=mysql_query($sql1)or die(mysql_error());
}
else {
"$sql2=UPDATE $tbl_name1 SET time='$time' WHERE session = '$session'";
$result2=mysql_query($sql2)or die(mysql_error());
}

$sql3="SELECT * FROM $tbl_name1";
$result3=mysql_query($sql3)or die(mysql_error());

$count_user_online=mysql_num_rows($result3);

echo "User online : $count_user_online ";

// if over 10 minute, delete session
$sql4="DELETE FROM $tbl_name WHERE time<$time_check";
$result4=mysql_query($sql4);

mysql_close();

// Open multiple browser page for result
/////////////////////////////////////////////////////////////////////////////////////////////



?>

I need to upload an image using php and store the image as blob in mysql. I understand the standard way is to create a form for the user, let the user choose the image to upload, click submit.

But I need to create a php, for example, uploadImage.php?image="c:\myPhoto.jpg"
and when I type http:\\www.myhome.com\uploadImage.php?image="c:\myPhoto.jpg"
from a web browser in my pc, I can automatically upload myPhoto.jpg to the myhome.com server?

Thank you for your kind help.

Hi Guys

I don't why it happen. probably i missed out something.

Here is my code.

Code: [Select]
<?php
if($_REQUEST["xtvt"]=="addEdit")
{
$sql02="INSERT INTO attachment (remarks, attached_by, attached_date)
      VALUES('".$_REQUEST["remarks"]."', '".$_SESSION["SESSION_ID"]."', now())";
$res02=mysql_query($sql02);

//--------- attach file ----------------------------
$name = $_FILES["file"]["name"]; 
if ($name != "")
{
$selectDB="SELECT max(id) AS id FROM attachment";
$qryDB= @$res1=mysql_query($selectDB);
$fetchDB=mysql_fetch_array($qryDB);

$maxid=$fetchDB[0];
$result=move_uploaded_file($_FILES["file"]["tmp_name"], "upload//" . $_FILES["file"]["name"]);
if (!$result) {
echo "Error uploading file";
}
$update_faultLog = "UPDATE attachment SET attachment='$name' WHERE id='$maxid'";
$res1=mysql_query($update_faultLog);
}
}
?>

<table width="625" border="0">
  <tr>
    <td width="78">Select File <span class="style1">*</span></td>
    <td width="5">:</td>
    <td width="539"><input type="file" name="file" id="file"/></td>
  </tr>
  <tr>
    <td>Remarks <span class="style1">*</span></td>
    <td>:</td>
    <td><pre><textarea name="remarks" id="remarks" cols="45" rows="3"></textarea></pre></td>
  </tr>
  <tr>
    <td>&nbsp;</td>
    <td>&nbsp;</td>
    <td><input type="reset" id="bttnReset" name="bttnReset" value="Reset" onclick="Reset()"/>
<input type="button" id="bttnAdd" name="bttnAdd" value="Add" onclick="addEdit()"/></td>
  </tr>
</table>

Its work at localhost but not when i put at server...

please help me...

Thank You

I designed a web form that takes as input a file and that should upload that file on the server. The file is located on the client machine. The page containing the web form is handled by PHP script.

I found out (reading 9483658954 forum topics so far) that there are 2 methods for getting a file uploaded. The first one is using FTP, and the second one using $_FILES variable. Unfortunately i have complains regarding both of the methods.

Firstly, when attempting to upload the file using FTP it seems that rather the FTP server is calling "upload" a simple copy action (because the only successful thing i could do with FTP was to copy a file from location A to location B, both A and B being on the server) or I don't know to use FTP (which is probably the right answer). Then, when using $_FILES variable, I was forced to include ' enctype="multipart/form-data" ' in the form tag. After a long documentation I finally figured out that by including the enctype blah blah, the $_POST and $_FILES variables were NULL after the form was submitted.Still can't understand why!



Are there any other methods for uploading a file from the client machine to the server? Can I get a competent script for this?

Hey everyone,

I'm currently working on a friends online script and i have a slight problem that i need help with.

Basically the code first searches "TBL_Friends" to see if you have any friends added. If it returns results it then turns your friends ID's into a variable. It then searches "TBL_Users_Online" to see if any body is logged based on the friend's ID it returned before. The first bit of the code works and it retrieves all the friends i got added. The second half is odd, if i have one or two friends added it will show that one is online. If i have more then three friends added it returns no results. I know my code is a bit sloppy and probably not the best way of writing it, im still learning PHP.

Anyways this is the code, any help is appreciated.

Code: [Select]
<?php
$FriendsOnline = mysql_query("SELECT Sender_ID FROM TBL_User_Friends WHERE Reciever_ID = $UserID");
while($fo=mysql_fetch_array($FriendsOnline))
{
     $FriendsOnlineID = $fo[Sender_ID];

     $FriendsOnlineNumber = mysql_query("SELECT * FROM TBL_Users_Online WHERE User_ID = $FriendsOnlineID");
     $FriendsNumber = mysql_num_rows($FriendsOnlineNumber);

     echo $FriendsNumber;
}
?>
$SenderID = Friends ID
$Reciever_ID = User ID
$UserID = User ID