PHP - Skipping An Entry In Foreach

Hello, i have an array like:
Code: [Select]
Array
(
    [0] => http://www.google.com/asdasfsdafas
    [1] => http://www.google.com/join.html
    [2] => http://www.google.com/dfsgjkd
    [3] => http://www.google.com/join.html
    [4] => http://www.google.com/ddfgyyhujh
    [5] => http://www.google.com/join.html
    [6] => http://www.google.com/ghklkl
)I'm trying to  transform it like:
Code: [Select]
Array
(
    [0] => http://www.google.com/asdasfsdafas
    [1] => http://www.google.com/dfsgjkd
    [2] => http://www.google.com/ddfgyyhujh
    [3] => http://www.google.com/ghklkl
)I used array_fliter but with no success any advise?

Thanks

Hey Guys,

i have now managed to upload my site, products are listed from the database-great!
However, when checkout is clicked you cannot update the cart with quantities and then click pay, it instead skips this and just empties the cart. Spent hours and getting nowhere!

Can sum1 please help?

My site: http://www.afebmunn.webhostingdeluxe.com/inkcrazy/index.php

I have three records in this table, yet the script below only displays two records, skipping the first item.  I have tried it in other fields also, and it is always skipping the first record?
Code: [Select]
$result = mysql_query("SELECT * FROM `content_type_ads`");

$row = mysql_fetch_array( $result );

while($row = mysql_fetch_array($result)){
echo $row['nid']. " - ". $row['field_item_id_value'];
echo "<br />";
}

Hey guy's, have been stuck at this hurdle for a little whole now and I can't seem to find the right solution and would very much appreciate anyones help on this.

Ok, so I have a table of votes and each time a user votes, they're added to this table. Now, I have a 'pool' table which will display certain items. What I'm trying to do is display the 'pool' items, but if this specific user has already voted on this particular item, then I want it to skip the item in the 'pool' and go to the next item, until the argument of has that user voted is false.


The current code I'm pasting will obviously not work, it isn't even in the right position, but I've separated it so that you can hopefully understand what I'm trying to do in order to help.



Below is the code I use to check if a user has already voted on a particular photo:

<?php
$result=mysql_query("SELECT * FROM votes WHERE vVoter='$_SESSION[userID]' AND vPhoto='$_SESSION[currentphoto]'");
$rowCheck = mysql_num_rows($result);
$alreadyv = 0;
if($rowCheck > 0){
   $alreadyv = 1;
        //user has already voted on this photo, need to use this info to be able to skip it in the 'pool'
} else {
   //do nothing
}
?>

totalVoted is basically how many times a user has voted, so that people who vote more, get more votes:

<?php
$result=mysql_query("SELECT * FROM pool ORDER BY totalVoted DESC LIMIT 1");
$rowCheck = mysql_num_rows($result);

if($rowCheck > 0){
   while($row = mysql_fetch_array($result)){
      $photoid = $row['photoInPool'];
      
         $result2=mysql_query("SELECT * FROM photos WHERE pID='$photoid'");
         $rowCheck2 = mysql_num_rows($result2);

         if($rowCheck2 > 0){
            while($row2 = mysql_fetch_array($result2)){
               if($alreadyv == 1) {
                  //I NEED THIS TO SHOW THE NEXT PHOTO IN THE ARRAY, HOWEVER, IF THEY HAVE ALREADY VOTED ON THE NEXT PHOTO, IT WILL STILL DISPLAY IT, THIS IS MY PROBLEM I NEED HELP WITH PLEASE
               } elseif ($alreadyv != 1) {
                  $fname = $row2['pFileName'];
               }
            }
         }
   
      $_SESSION[currentphoto] = $row['photoInPool'];
      echo "<center><img src='user_images/$fname'></center>";
   }
}

?>

I hope I've explained this well enough and hope somebody can help. Many thanks in advance

I have a script that uploads and images, creates 2 versions of it, re-sizes the images and inserts all the info into a DB.

The odd thing is that if I don't have an image to upload, it will still run the script but it will completely skip over the description.  If I upload an image, it works fine.  Not too sure what the issue is.
Is it because my variable of $desc causing a conflict?

Here's my code:

Code: [Select]

<?php 
ini_set("display_errors",1);
include 'dbconn.php';

$change="";
$abc="";


 define ("MAX_SIZE"," 10000");
 function getExtension($str) {
         $i = strrpos($str,".");
         if (!$i) { return ""; }
         $l = strlen($str) - $i;
         $ext = substr($str,$i+1,$l);
         return $ext;
 }

 $errors=0;
  
 if($_SERVER["REQUEST_METHOD"] == "POST")
 {
  $image =$_FILES["file"]["name"];
$uploadedfile = $_FILES['file']['tmp_name'];
     
 
  if ($image) 
  {
 
  $filename = stripslashes($_FILES['file']['name']);
 
   $extension = getExtension($filename);
  $extension = strtolower($extension);


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

  $change='<div class="msgdiv">Unknown Image extension </div> ';
  $errors=1;
  }
  else
  {

 $size=filesize($_FILES['file']['tmp_name']);


if ($size > MAX_SIZE*1024)
{
$change='<div class="msgdiv">You have exceeded the size limit!</div> ';
$errors=1;
}


if($extension=="jpg" || $extension=="jpeg" )
{
$uploadedfile = $_FILES['file']['tmp_name'];
$src = imagecreatefromjpeg($uploadedfile);

}
else if($extension=="png")
{
$uploadedfile = $_FILES['file']['tmp_name'];
$src = imagecreatefrompng($uploadedfile);

}
else 
{
$src = imagecreatefromgif($uploadedfile);
}

echo $scr;

list($width,$height)=getimagesize($uploadedfile);


$newwidth=150;
$newheight=($height/$width)*$newwidth;
$tmp=imagecreatetruecolor($newwidth,$newheight);


$newwidth1=50;
$newheight1=($height/$width)*$newwidth1;
$tmp1=imagecreatetruecolor($newwidth1,$newheight1);

imagecopyresampled($tmp,$src,0,0,0,0,$newwidth,$newheight,$width,$height);

imagecopyresampled($tmp1,$src,0,0,0,0,$newwidth1,$newheight1,$width,$height);

$famname = $_POST['famname'];
$desc =  $_POST['description'];
$petname = $_POST['petname'];
$letter = $_POST['letter'];
$filename = "images/". $_FILES['file']['name'];
$filename1 = "images/small". $_FILES['file']['name'];



imagejpeg($tmp,$filename,100);

imagejpeg($tmp1,$filename1,100);

imagedestroy($src);
imagedestroy($tmp);
imagedestroy($tmp1);
}}

}

//If no errors registred, print the success message
 if(isset($_POST['Submit']) && !$errors) 
 {
 
 $insert="insert ignore into tbl_tribue (petname,familyname,letter,description,imgpath,thumbpath) values ('$petname','$famname','$letter','$desc','$filename','$filename1')";
   mysql_query($insert);
   //mysql_query("update gallery set imgpath='$filename'");
  $change=' <div class="msgdiv">Pet Added Successfully!</div>';

print $insert;

 }
 
?>


I had turned on the error and the print to see the output.  Anyone have any ideas?  Thanks!

This is new. For some reason, the PHP errors are not counting my //commented lines.   Example:

<?php
//my comment
:
?>

Parse error: syntax error, unexpected ':' in D:\www\myscript.php on line 2   The error is on line 3. If I add more comments, the error is the same:

<?php
//my comment
//my comment
//my comment
//my comment
//my comment
:
?>

Parse error: syntax error, unexpected ':' in D:\www\myscript.php on line 2   Where is this setting? Thanks.

We have a php application that is just a form, that takes in data to a mysql database and then displays that data in queue(php web page) for the client. It has been working for the past 5  or more years. Six weeks ago, the client noticed that the numbering started skipping a number each time a request was entered.  It then seemed to stop, now it has started again.  See attached image.

Some detail he   after user has completed web form and hits submit, three fields are populated in the mysql "REQUEST" table, then the actually data from the submitted web form is sent to the ANSWER table.  The REQUEST table is in sync and is always populated, and numbering is in sequence but upon submit, the actual data from the request does NOT always go to the ANSWER table hence the "skipping of numbers."
     
What do I even begin looking at/for to determine what happened?  Reminder: this is a 5+ year old app that worked fine until maybe 6 weeks ago.  No changes have been made to the database or the code base since 2008.  The code that adds the data to the ANSWER table [the one that appears to skip] is below.  The applicate is in php 4.3.1, mysql 5, linux server.

This topic has been moved to Application Frameworks.

http://www.phpfreaks.com/forums/index.php?topic=355772.0

Hey, So what im trying to do is put my database variables into a session array. So this is what im trying to accomplish...

$_SESSION['Name_of_Row'] = $value

 This is the script I wrote:

Code: [Select]
Function setupSession(){
session_start();
$query = "SELECT * FROM users WHERE u_id ='{$this->u_id}'";
    $result = mysql_query($query);
        $row = mysql_fetch_array($result);

foreach($row as $key => $value){
    if(!empty($value)){
        $_SESSION[$key] = $value;   
            }
                }
}

When that runs I get the following warning. Can anyone tell me what this means and how to fix it?

Error: Notice: Unknown: Skipping numeric key 0 in Unknown on line 0

I'm not sure if the title explains it very well, but here is an image of it. I am trying to figure out how to access the option selected (equal, not equal, etc.), along with its input box, for each column (the checkboxes on the right) selected. The purpose of this is to allow a non-programming administrator user to create a custom query to access information. I can find out which checkboxes are selected no problem, but finding the accompanying drop-down option and input box is difficult for me. How do I do this?

If you have a suggestion on how to change it, I'm very open. Thank you in advance!



Code: [Select]
if(!isset($_POST['submit']))
{
echo "
<form method='post'>
<table>
 <tr>
  <td valign='top'>SELECT</td>
  <td valign='top'>
   <table>";
 
// LIST ALL COLUMNS IN A TABLE
$result = mysql_query("SELECT * FROM table_name") or die(mysql_error());
$rowcount=mysql_num_rows($result);
$y=mysql_num_fields($result);
for ($x=0; $x<$y; $x++)
{ echo "
    <tr>
<td>
  <input type='checkbox' name='fields[]' value='".mysql_field_name($result, $x)."'>".mysql_field_name($result, $x)."
     </td>
</tr>"; }
 
   echo "
   </table>
  </td>
  <td valign='top'>FROM allocations</td>
  <td valign='top'>WHERE</td>
  <td>
   <table>";
   
// LIST ALL COLUMNS IN A TABLE
$result = mysql_query("SELECT * FROM table_name") or die(mysql_error());
$rowcount=mysql_num_rows($result);
$y=mysql_num_fields($result);
for ($x=0; $x<$y; $x++)
{ echo "
    <tr>
<td>
  <input type='checkbox' name='column[]' value='".mysql_field_name($result, $x)."'>".mysql_field_name($result, $x)."
</td>
<td>
  <select name='operator[]'>
          <option value='='>equals</option>
          <option value='<>'>does not equal</option>
          <option value='>'>greater than</option>
          <option value='<'>less than</option>
          <option value='>='>greater than or equal to</option>
          <option value='<='>less than or equal to</option>
          <option value='LIKE'>is like</option>
         </select>
</td>
     <td><input type='text' name='criteria[]'></td>
</tr>"; }

   echo "
   </table>
  </td>
  <td valign='top'><input type='submit' value='Process Query' name='submit' class='submit'></td>
 </tr>
</table>
</form>";
}
else
{
 echo "<b>Fields to edit:</b> ";
foreach ($_POST['fields'] as $fields) {
echo $fields,", ";
}

 echo "<br><br>";
 echo "<b>Columns to query:</b> ";
foreach ($_POST['column'] as $columns) {
echo $columns," HERE IS THE SPOT, ";
}
}

Hey folks,

Sorry for being a pain in the ass.

I am trying to submit data to my database via a form and when I click Submit, I get:

Duplicate entry '' for key 1

I understand that it means I have a duplicate entry with the ID of 1 or something like that. I can't find where the issue is.

Here is the form:


<form actin="" id="settings" name="settings">
                    <table class="listing form" cellpadding="0" cellspacing="0">
                        <tr>
                            <th class="full" colspan="2"><?php echo $lang_settings; ?></th>
                        </tr>
                        <tr>
                            <th colspan="2"><?php echo $lang_settings_description; ?></th>
                        </tr>
                        <tr>
                           <td><?php echo $lang_sitename; ?>: </td> <td><input type="text" name="sitename" value="<?php echo $site_name; ?>" width="172" /> <em>Site name for logo</em></td>
                        </tr>
                        <tr>
                           <td><?php echo $lang_email; ?>: </td> <td><input type="text" name="email" value="<?php echo $site_email; ?>" width="172" /> <em>Your email address</em></td>
                        </tr>
                        <tr>
                           <td><?php echo $lang_yourname; ?>: </td> <td><input type="text" name="name" value="<?php echo $your_name; ?>" width="172" /> <em>Your own name</em></td>
                        </tr>
                        <tr>
                           <td><?php echo $lang_meta_description; ?>: </td> <td><input type="text" name="meta-description" value="<?php echo $description; ?>" width="172" /> <em>SEO</em></td>
                        </tr>
                        <tr>
                           <td><?php echo $lang_keywords; ?>: </td> <td><input type="text" name="meta-keywords" value="<?php echo $keywords; ?>" width="172" /> <em>Separate with Commas</em></td>
                        </tr>
                       <tr>
                       <td><input type="submit" class="button" name="submit" value="<?php echo $lang_button_savesettings; ?>"></td>
                       </tr>
                    </table>
</form>


Here is the Insert code:

$insert = "INSERT INTO settings (site_name, description, keywords, email, name) 
VALUES ('$sitename', '$meta_description', '$meta_keywords', '$site_email', '$your_name')"; 
mysql_query($insert) or die(mysql_error());


Can anyone please tell me where I am going wrong here?

Much appreciated.

Hey guys!

I know, I know this problem is EVERYWHERE but i just dont understand! I have a solid knowlage of php but my SQL skills are low, so i dont know too much about Keys and stuff. But my error is: Duplicate entry '' for key 2.

The thing that im working on at this section is logging in with facebook. The code that presents my error is:

    $sql = "SELECT * FROM users WHERE uid=".$uid;
    $fbid = mysql_query($sql);
$num_rows = mysql_num_rows($fbid);

if(mysql_num_rows($fbid) < 1) {
echo "You are not logged in.    ";
mysql_query("INSERT INTO `users` (`uid`) VALUES ('".$uid."')")
or die(mysql_error());  

} else {
mysql_query("UPDATE users SET logged = '1' WHERE uid=".$uid);
//mysql_query("UPDATE users SET full_name = $me WHERE uid=".$uid);

echo "Your Logged in ";
echo $me['name'];
?> Continue to <a href="removed :)"> My Settings </a>. <?
}


Any help is welcome

Hello there,

I have some code here which sends a number of variables from flash to SQL...

I would simply like to add the functionality to overwrite records which have the same 'name' or 'pseudo'... can anyone help me please ? Thanks in advance Martin


<?php

$pseudo=$_POST['var1'];
$score=$_POST['var2'];
$table = $_POST['tab'];

$dategame = $_POST['tempjoueur'];
//$micro = microtime();
//$dategame = time()."".substr($micro, 2, 6);

$_COOKIE['User'] = $_SERVER['REMOTE_ADDR'];
$envoie = InsertDatas($table, "name, score, dategame", "'".$pseudo."','".$score."','".$dategame."'");
if ($envoie) {
print_r("OK, $pseudo, $score, $dategame,$ipclient");
}
else { echo "BAD, $pseudo, $score, $dategame,$ipclient"; }
?>

For some reason this only allows one SQL to be added...

// SQL Connection
$username="monstert_admin"; $password="admin"; $database="monstert_admin";
$connection = mysql_connect("localhost", $username, $password) 
or die("Connection Failure to Database");
// Select Database
mysql_select_db($database, $connection) or die ($database . "No Database" . $username);

//Select everything from the the table
$MyQuery = "SELECT * FROM photos";
$retrieve = mysql_query($MyQuery) or die(mysql_error());
if(mysql_num_rows($retrieve) != 0):
$row = mysql_fetch_assoc($retrieve);
else:
echo '';
endif;

if(isset($_POST['Submit']) && !$errors)
{
$url = $newname;

   include('img.php');
   $image = new SimpleImage();
   $image->load($url);
   $image->resize(500,315);
   $image->save($newname);

mysql_query("INSERT INTO photos (url) VALUES ('$url')");

echo "File Uploaded Successfully as <i> ";
echo $newname;
echo "</i>";
}

What would the issue be? I only have two columns - ID and url

Thanks in advance!