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emediastudios

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Posts posted by emediastudios

    [SOLVED] Upload More images

    in PHP Coding Help

    Posted

    I pasted your code over mine and now i get this.

    Warning: move_uploaded_file(../news/) [function.move-uploaded-file]: failed to open stream: Permission denied in C:\Program Files\Apache Group\Apache2\htdocs\Prepare2win\admin\upload_news.php on line 18

Warning: move_uploaded_file() [function.move-uploaded-file]: Unable to move 'C:\Windows\TEMP\php1CB6.tmp' to '../news/' in C:\Program Files\Apache Group\Apache2\htdocs\Prepare2win\admin\upload_news.php on line 18
Sorry, there was a problem uploading your file.
Warning: move_uploaded_file(../news/) [function.move-uploaded-file]: failed to open stream: Permission denied in C:\Program Files\Apache Group\Apache2\htdocs\Prepare2win\admin\upload_news.php on line 18

Warning: move_uploaded_file() [function.move-uploaded-file]: Unable to move 'C:\Windows\TEMP\php1CC7.tmp' to '../news/' in C:\Program Files\Apache Group\Apache2\htdocs\Prepare2win\admin\upload_news.php on line 18
Sorry, there was a problem uploading your file.
Warning: move_uploaded_file(../news/) [function.move-uploaded-file]: failed to open stream: Permission denied in C:\Program Files\Apache Group\Apache2\htdocs\Prepare2win\admin\upload_news.php on line 18

Warning: move_uploaded_file() [function.move-uploaded-file]: Unable to move 'C:\Windows\TEMP\php1CC8.tmp' to '../news/' in C:\Program Files\Apache Group\Apache2\htdocs\Prepare2win\admin\upload_news.php on line 18
Sorry, there was a problem uploading your file.

    ???

     

    Thanks for your help

    [SOLVED] Upload More images

    in PHP Coding Help

    Posted

    Thanks for your help.

    I did as you said and get this error

    Parse error: syntax error, unexpected $end in C:\Program Files\Apache Group\Apache2\htdocs\Prepare2win\admin\upload_news.php on line 39

     

    My code looks like this now

    <?php require_once('../Connections/p2w.php'); ?>
<?php 

//This is the directory where images will be saved 
$target = "../news/"; 
$target = $target . basename( $_FILES['photo']['name']); 

//This gets all the other information from the form 
$headline=$_POST['headline']; 
$contents=$_POST['contents']; 
$photo1=($_FILES['photo1']['name']); 
$photo2=($_FILES['photo2']['name']); 
$photo3=($_FILES['photo3']['name']);
$title1=$_POST['title1']; 
$title2=$_POST['title2'];
$title3=$_POST['title3'];  
$link=$_POST['link']; 





//Writes the photo to the server 
for ($x=1;$x<=3;$x++){
    $photo ='photo'.$x;
    if(move_uploaded_file($_FILES[$photo]['tmp_name'], $target)){


//Tells you if its all ok 
echo "Success."; 
//Writes the information to the database 
mysql_query("INSERT INTO `news` VALUES (NULL,'$headline','$contents','$photo1','$photo2','$photo3','$title1','$title2','$title3','$link')") ; 
} 
else { 

//Gives and error if its not 
echo "Sorry, there was a problem uploading your file."; 
} 
?>

     

    [SOLVED] Upload More images

    in PHP Coding Help

    Posted

    i changed my code to this.

    <?php require_once('../Connections/p2w.php'); ?>
<?php 

//This is the directory where images will be saved 
$target = "../news/"; 
$target = $target . basename( $_FILES['photo1']['name']); 

//This gets all the other information from the form 
$headline=$_POST['headline']; 
$contents=$_POST['contents']; 
$photo1=($_FILES['photo1']['name']); 
$photo2=($_FILES['photo2']['name']); 
$photo3=($_FILES['photo3']['name']);
$title1=$_POST['title1']; 
$title2=$_POST['title2'];
$title3=$_POST['title3'];  
$link=$_POST['link']; 





//Writes the photo to the server 
if(move_uploaded_file($_FILES['photo1']['tmp_name'], $target)) 
{ 


//Tells you if its all ok 
echo "Success."; 
//Writes the information to the database 
mysql_query("INSERT INTO `news` VALUES (NULL,'$headline','$contents','$photo1','$photo2','$photo3','$title1','$title2','$title3','$link')") ; 
} 
else { 

//Gives and error if its not 
echo "Sorry, there was a problem uploading your file."; 
} 
?>

    Only uploads the one file, i know why but how i get it to do the 3?

    [SOLVED] Upload More images

    in PHP Coding Help

    Posted

    I have this code that works good, just have to implement some filters.

    How would i change this code if i wanted to add 3 images instead of just the 1.

    I have input fields and fields in my database called photo1, phot02 and photo3.

     

    <?php require_once('../Connections/gcproperty.php'); ?>
<?php 

//This is the directory where images will be saved 
$target = "../images/news/"; 
$target = $target . basename( $_FILES['photo']['name']); 

//This gets all the other information from the form 
$headline=$_POST['headline']; 
$link=$_POST['link']; 
$content=$_POST['content']; 
$photo1=($_FILES['photo'1]['name']); 





//Writes the photo to the server 
if(move_uploaded_file($_FILES['photo1']['tmp_name'], $target)) 
{ 


//Tells you if its all ok 
print "<meta http-equiv=\"refresh\" content=\"0;URL=news_added_successfully.php\">"; 
//Writes the information to the database 
mysql_query("INSERT INTO `news` VALUES ('$headline','$photo1','$content','$link')") ; 
} 
else { 

//Gives and error if its not 
echo "Sorry, there was a problem uploading your file."; 
} 
?>

    [SOLVED] Delete record

    in PHP Coding Help

    Posted

    thanks, i changed it and it gave me this error

    You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '` WHERE id='1'' at line 1

     

    thats with your code,  ??? and when i have no (') i get what i had first?

    Weird

    [SOLVED] Delete record

    in PHP Coding Help

    Posted

    Im some having  trouble with deleting a record, surely it isnt that hard, dont know what im doing wrong.

    this is the error i am gettin

     

    You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '` WHERE id=1' at line 1

    Here is my delete code

     

    <?php
if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") 
{
  $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;

  $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue);

  switch ($theType) {
    case "text":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;    
    case "long":
    case "int":
      $theValue = ($theValue != "") ? intval($theValue) : "NULL";
      break;
    case "double":
      $theValue = ($theValue != "") ? "'" . doubleval($theValue) . "'" : "NULL";
      break;
    case "date":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "defined":
      $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
      break;
  }
  return $theValue;
}
}

$currentPage = $_SERVER["PHP_SELF"];

if ((isset($_POST['id'])) && ($_POST['id'] != "")) {
  $deleteSQL = sprintf("DELETE FROM ``application`` WHERE id=%s",
                       GetSQLValueString($_POST['id'], "int"));

  mysql_select_db($database_p2w, $p2w);
  $Result1 = mysql_query($deleteSQL, $p2w) or die(mysql_error());

  $deleteGoTo = "application_deleted.php";
  if (isset($_SERVER['QUERY_STRING'])) {
    $deleteGoTo .= (strpos($deleteGoTo, '?')) ? "&" : "?";
    $deleteGoTo .= $_SERVER['QUERY_STRING'];
  }
  header(sprintf("Location: %s", $deleteGoTo));
}

$maxRows_Recordset1 = 2;
$pageNum_Recordset1 = 0;
if (isset($_GET['pageNum_Recordset1'])) {
  $pageNum_Recordset1 = $_GET['pageNum_Recordset1'];
}
$startRow_Recordset1 = $pageNum_Recordset1 * $maxRows_Recordset1;

mysql_select_db($database_p2w, $p2w);
$query_Recordset1 = "SELECT * FROM `application`";
$query_limit_Recordset1 = sprintf("%s LIMIT %d, %d", $query_Recordset1, $startRow_Recordset1, $maxRows_Recordset1);
$Recordset1 = mysql_query($query_limit_Recordset1, $p2w) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);

if (isset($_GET['totalRows_Recordset1'])) {
  $totalRows_Recordset1 = $_GET['totalRows_Recordset1'];
} else {
  $all_Recordset1 = mysql_query($query_Recordset1);
  $totalRows_Recordset1 = mysql_num_rows($all_Recordset1);
}
$totalPages_Recordset1 = ceil($totalRows_Recordset1/$maxRows_Recordset1)-1;

$queryString_Recordset1 = "";
if (!empty($_SERVER['QUERY_STRING'])) {
  $params = explode("&", $_SERVER['QUERY_STRING']);
  $newParams = array();
  foreach ($params as $param) {
    if (stristr($param, "pageNum_Recordset1") == false && 
        stristr($param, "totalRows_Recordset1") == false) {
      array_push($newParams, $param);
    }
  }
  if (count($newParams) != 0) {
    $queryString_Recordset1 = "&" . htmlentities(implode("&", $newParams));
  }
}
$queryString_Recordset1 = sprintf("&totalRows_Recordset1=%d%s", $totalRows_Recordset1, $queryString_Recordset1);
?>

     

    I have a record in my database with a incrementing primary key called id.

    and a hidden field in the form with id posted within it.

     

    Thanks for any help :)

    Show images only

    in PHP Coding Help

    Posted

    I have no idea, i dont get any errors, just a blank page.

    I added a filter  and array butim sure its all wrong.

    Heres my new code.

    <?php
//filter extensions
$file_name = $file;
function valid_ext($file_name)
{
$valid = array("jpeg","jpg","png","gif");
$extension = strtolower(substr($file_name,-3,3));
return (in_array($extension, $valid));
}
///Directory Settings
$folder = "../images/"; 
$handle = opendir($folder);

# Making an array containing the files in the current directory: 
while (false !== ($file = readdir($handle)))
{
    if ($file != '.' && $file != '..')
       $files[] = $file; 
} 
closedir($handle); 

//Test Check
$Valid = false;
if(valid_ext($file_name))

#echo the files 
foreach ($files as $file) { 
   echo "<img src='$folder$file' alt='image' width='150' /><br />";
   echo "<a href='$folder$file'>$file</a> - ";
   echo "<a href=filedelete.php?file=$file>Delete this Image</a><br /><hr />";
} 
?>

    Show images only

    in PHP Coding Help

    Posted

    I have imported a function into the code but am unsure how to call upon it

     

    <?PHP 
//filter extensions
function valid_ext($file_name)
{
$valid = array("jpeg","jpg","png","gif");
$extension = strtolower(substr($file_name,-3,3));
return (in_array($extension, $valid));
}
///Directory Settings
$folder = "../images/"; 
$handle = opendir($folder);

# Making an array containing the files in the current directory: 
while (false !== ($file = readdir($handle)))
{
    if ($file != '.' && $file != '..')
       $files[] = $file; 
} 
closedir($handle); 

#echo the files 
foreach ($files as $file) { 
   echo "<img src='$folder$file' alt='image' width='150' /><br />";
   echo "<a href='$folder$file'>$file</a> - ";
   echo "<a href=filedelete.php?file=$file>Delete this Image</a><br /><hr />";





} 
?>

    Show images only

    in PHP Coding Help

    Posted

    I have a code that displays all the files in a folder as well as a thumbnail image.

    This is a page that allows my client the option to delete a image from his dir if it is no longer needed.

    I was hoping that i could alter my code so that it only displayed images in jpg, jpeg, gif or png.

    and if there is files or folders that are not of this format in the selected(../images) directory, not echo them.

    Here is my code.

    <?PHP 
$folder = "../images/"; 
$handle = opendir($folder);

# Making an array containing the files in the current directory: 
while (false !== ($file = readdir($handle)))
{
    if ($file != '.' && $file != '..')
       $files[] = $file; 
} 
closedir($handle); 

#echo the files 
foreach ($files as $file) { 
   echo "<img src='$folder$file' alt='image' width='150' /><br />";
   echo "<a href='$folder$file'>$file</a> - ";
   echo "<a href=filedelete.php?file=$file>Delete this Image</a><br /><hr />";



} 
?>

    MAny thanks.

     

    Oh i have this code too, maybe i could use some code out of it and combine it with the upper code?

    <?php
echo "<select name=\"file\">\n";
foreach (new DirectoryIterator('.') as $file) {
   // if the file is not this file, and does not start with a '.' or '..',
   // then store it for later display
   if ( (!$file->isDot()) && ($file->getFilename() != basename($_SERVER['PHP_SELF'])) ) {
      echo "<option>";
      // if the element is a directory add to the file name "(Dir)"
      echo ($file->isDir()) ? "(Dir) ".$file->getFilename() : $file->getFilename();
      echo "</option>\n";
   }
}
echo "</select>\n";
?>

     

    Delete files from directory

    in PHP Coding Help

    Posted

    now all i got to do is sort out the filedelete.php

    I was given this code below but nothing, justs timesout

     

    <?php
$filename = 'gcproperty/images'.$_GET['file'];
$file = $_GET['file'];

// set up basic connection
$ftp_server = "localhost";
$ftp_user_name = "root";
$ftp_user_pass = "5050888202";
$conn_id = ftp_connect($ftp_server);


// login with username and password
$login_result = ftp_login($conn_id, $ftp_user_name, $ftp_user_pass);

// try to delete $file
if (ftp_delete($conn_id, $filename)) {
echo "<p><center>The file $filen was deleted successfully.<br />";
echo "<center><a href=admin.php>Back</a>";
} else {
echo "<p><center>Could not delete $file.<br />";
echo "<center><a href=admin.php>Back</a>";
}

// close the connection
ftp_close($conn_id);

?>

     

     

    Delete files from directory

    in PHP Coding Help

    Posted

    i have this code

    <?php
echo "<img src='$folder$file' alt="image" width="150" /><br />";?>

    and i get this error

    Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in C:\Program Files\Apache Group\Apache2\htdocs\gcproperty\admin\filelist.php on line 17

    Delete files from directory

    in PHP Coding Help

    Posted

    loaded the page and read the souce code. its here

     

    <?php
<a href=../images/Conventoncntr.jpg><label title=Conventoncntr.jpg>Conventoncntr.jpg</a><br /></label><img src="$folder$file" alt="image" width="150" /><br /><a href=filedelete.php?file=Conventoncntr.jpg>Delete</a><br />
?>

     

    something not right there.

    this is the code generated from

     

    <?php
#echo the files 
foreach ($files as $file) { 
    echo "<a href=$folder$file><label title=$file>$file</a><br /></label>";
echo '<img src="$folder$file" alt="image" width="150" /><br />';
echo "<a href=filedelete.php?file=$file>Delete</a><br /><br />";

} 
?>

    Delete files from directory

    in PHP Coding Help

    Posted

    seems so strange as the file names echo correctly, why doesnt the link or thumnail work.

     

    Code again with a couple of changes

    <?PHP 

$folder = "../images/"; 
$handle = opendir($folder);

# Making an array containing the files in the current directory: 
while (false !== ($file = readdir($handle)))
{
    if ($file != '.' && $file != '..')
       $files[] = $file; 
} 
closedir($handle); 

#echo the files 
foreach ($files as $file) { 
    echo "<a href=$folder$file><label title=$file>$file</a><br /></label>";
echo '<img src="$folder$file" alt="image" width="150" /><br />';
echo "<a href=filedelete.php?file=$file>Delete</a><br /><br />";

} 
?>

    Delete files from directory

    in PHP Coding Help

    Posted

    As i am a newbie i think i may of found a little problem. When you are echoing the file location within the html ''img scr'' you need to include the full path details and not just $file, which is the name of the file. You need to set up a variable that contains both $folder and $file

    Gave that a go before, it didnt work either.

    ;)

    thanks for trying champ, and yes that makes sense

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