PHP - Query Expects Paramater To Be String?

I have a strange error from the code below. I have run the query with the resulting values in PHPMyAdmin and it works fine.

The code is:
$query = "SELECT * FROM sn_matches WHERE (P1_ID='".$P1_ID."' AND P2_ID='".$P2_ID."') OR (P1_ID='".$P2_ID."' AND P2_ID='".$P1_ID."')";
while($row = mysql_fetch_assoc($query)){

The error is:
Quote

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, string given in /customers/ronflorax.com/ronflorax.com/httpd.www/snookerstats/h2h.php on line 21


Line 21 is the one which has mysql_fetch_assoc on it.

Any help would be much appreciated!

Here is the error:
Code: [Select]
Warning: image_type_to_extension() expects parameter 1 to be long, string given in C:\wamp\www\Login\inc\edit-profile.php on line 23
Here is the code for line 23:
Code: [Select]
$type = image_type_to_extension($_FILES['pic']['type'], true);
Here is the code for the whole page:
Code: [Select]
<?php
if(!$userb)
{
login('?p=edit-profile');
die();
}
if(isset($_POST['submit']))
{
$mottof = mysql_real_escape_string(bb($_POST['motto']));
$aboutf = mysql_real_escape_string(bb($_POST['about']));
$interesrsf = mysql_real_escape_string(bb($_POST['interests']));
$booksf = mysql_real_escape_string(bb($_POST['books']));
$authorsf = mysql_real_escape_string(bb($_POST['author']));
$randf = mysql_real_escape_string(bb($_POST['random']));
if($_FILES['pic']['type'] == 'image/gif' || $_FILES['pic']['type'] == 'image/jpeg' || $_FILES['pic']['type'] =-'image/pjpeg' || $_FILES['pic']['type'] == 'image/png')
{
if($_FILES['pic']['size'] < 8388608)
{
if($_FILES['pic']['name'] != '')
{

//mkdir('/images/user_images/' .$user_log);
$type = image_type_to_extension($_FILES['pic']['type'], true);
move_uploaded_file($_FILES['pic']['tmp_name'], 'images/user_images/' .$user_log .'pp' .$type);
$path = 'images/user_images/' .$user_log .'pp' .$type;
}
else 
{
$path = 'images/user_images/user.jpg';}
}
$q2 = "UPDATE login_info
SET profile_picture = '$path'
WHERE user='$user_log'

";
mysql_query($q2) or die(mysql_error());


}


$query = "
UPDATE login_info
SET motto = '$mottof', about = '$aboutf', fav_books = '$booksf', fav_authors = '$authorsf', interests = '$interesrsf', random = '$randf'
WHERE user='$user_log'
";
mysql_query($query);






echo '<h3> Profile Updated </h3>';
}
$q = "SELECT * FROM login_info WHERE user='$user_log' ";
$s = mysql_query($q) or die(mysql_error());
$r = mysql_fetch_assoc($s);
$motto = $r['motto'];
$about = $r['about'];
$books= $r['fav_books'];
$authors = $r['fav_authors'];
$rand = $r['random'];
$interests = $r['Interests'];
if(isset($_GET['act'])){$red = '?p=' .$rpage ;} else {$red='?p=edit_profile';}
echo"

<form action='?p=edit-profile' enctype='multipart/form-data' method='post' >
<label>Motto: </label>
<input type='text' name='motto' value='$motto' placeholder='Motto' />
<label> About :</label>
<textarea name='about' cols='40' rows='10'> $about</textarea>
<label> Interests: </label>
<textarea name='interests' cols='40' rows='10'> $interests </textarea>
<label> Favorite Books: </label>
<textarea name='books' cols='40' rows='10'> $books </textarea>
<label> Favorite Authors </label>
<textarea name='author' cols='40' rows='10'> $authors </textarea>
<label> Random: </label>
<textarea name='random' cols='40' rows='10'> $rand </textarea>
";
?>
<script type='text/javascript'>
function if_checked()
{
if($('#cur').is(':checked'))
{
$('#pic').attr('disabled', true);
}
else
{
$('#pic').removeAttr('disabled');
}
}
</script>
<label> Profile Pictu </label>
<input type='checkbox' name= 'current' id='cur' onchange="if_checked();" checked /> <label for='current' style='display:inline;'> Current profile picture </label> <br />
<input type='file' id='pic' name='pic' disabled="disabled" />
<br />
<input type='submit' value ='Submit' name='submit' />
</form>
Ive used the function before and I havent encountered this error. Im not sure what it means.

PHP Warning:  file_exists() expects parameter 1 to be string, array given in /home/mysite/public_html/display.mysite.com/wp-content/themes/mytheme/form.php on line 9

PHP Warning:  file_exists() expects parameter 1 to be string, array given in /home/mysite/public_html/display.mysite.com/wp-content/themes/mytheme/form.php on line 13

I have a form that is having some issues and in the error log I see a bunch of these warnings. What is wrong with this??   I numbered the offending lines below (lines 9 & 13)   Thanks for any help on this.

function CheckExistance($VUrl)

{

/*9*/    if ( file_exists($VUrl) )

        echo $VUrl;

/*13*/    elseif ( file_exists(str_replace('www.display.mysite.com','www.display.com', $VUrl)) )

        echo str_replace('www.display.mysite.com','www.mysite.com', $VUrl);

    else

        echo str_replace('www.mysite.com','www.display.mysite.com', $VUrl);

}

Edited by damion, 02 July 2014 - 06:01 PM.

if($pid != "") {
    $bname = $_REQUEST['bname'];
    $btitle = $_REQUEST['btitle'];
    $btags = $_REQUEST['btags'];
    $bdesc = $_REQUEST['bdesc'];
    $btext = $_REQUEST['btext'];
    $bimg = $_REQUEST['bimg'];
    $bimgalt = $_REQUEST['bimgalt'];
            
            $data = "";
                if($bname!="") { $data = $data." 'bname' => ".$bname.", "; }
                if($btitle!="") { $data = $data."'btitle' => ".$btitle.", "; }
                if($btags!="") { $data = $data."'btags' => ".$btags.", "; }    
                if($bdesc!="") { $data = $data."'bdesc' => ".$bdesc.", "; }
                if($btext!="") { $data = $data."'btext' => ".$btext.", "; }    
                if($bimg!="") { $data = $data."'bimg' => ".$bimg.", "; }    
                if($bimgalt!="") { $data = $data."'bimgalt' => ".$bimgalt.", "; }
                $data = $data."'pid ' =>". $pid.", " ;
                $data = "[".$data."]";
                
            
            $build = "";
                if($bname!="") { $build = $build." page_name = :bname,"; }
                if($btitle!="") { $build = $build." page_title = :btitle,"; }
                if($btags!="") { $build = $build." page_tags = :btags,"; }
                if($bdesc!="") { $build = $build." page_desc = :bdesc,"; }
                if($btext!="") { $build = $build." page_text = :btext,"; }
                if($bimg!="") { $build = $build." page_img = :bimg,"; }
                if($bimgalt!="") { $build = $build." page_imgalt = :bimgalt,"; }
                $build = $build." page_id = :pid";
                
                
                    
            $sql = "UPDATE pages SET ".$build." WHERE page_id=:pid";
            echo $sql."<br /><br />";            
            echo $data."<br /><br />";    
            
            $stmt= $pdo->prepare($sql);
            $stmt->execute($data);

 Result of echo SQL and Data

UPDATE pages SET page_name = :bname, page_title = :btitle, page_tags = :btags, page_desc = :bdesc, page_text = :btext, page_img = :bimg, page_imgalt = :bimgalt, page_id = :pid WHERE page_id=:pid

[ 'bname' => Home, 'btitle' => Market Net, 'btags' => full stack development, search engine ready, professional web development, web design, 'bdesc' => Market net gives you online access to market traders country wide, Support Local businesses today, 'btext' =>

Market Net is the place for you to find those goodies you saw on the markets, buy them online and support small local businesses. Watch this space for more information.

Test

#D hvac

, 'bimg' => supportlocal.jpg, 'bimgalt' => Support Local Small Businesses, 'pid ' =>106, ]

Yet I am getting said error message.  What am I doing wrong?

I'm going to create a web page with <iframe> tag & want that the src of <iframe> change as 

I've made a few attempts at this, but I'm likely not making much sense.... I'm working on how to phrase the question.

I had a MySQL database where I used to combine user first AND last names in a single string... I called that string "Member"

So if I filtered:
www.mysite.com/members.php?Member=Peter

It would find and display all Peters: Peter Rabbit and Peter Griffith and Pope Peter.  This is what I want.

I'm now keeping each name (first and last) in separate strings (as separate variables?), and I am combining them in an array to display on the website:

Code: [Select]
$First = $First;
$Last = $Last;

$member = array($Last, $First,);
foreach ($Member as $key => $v ) if (!$v) unset ($Member[$key]);
$Member = implode(', ', $Member);
echo $Member;

Now I've lost the ability to search all the members for Peter.

So if I filtered:
www.mysite.com/members.php?First=Peter
...it would work.... for Peter Rabbit and Peter Griffith, but not Pope Peter

I cannot query the entire array?  I can no longer check members (first and last names) for Wayne?
www.mysite.com/members.php?Member=Peter

I'm betting I can, but there's a lot more to it... and 8 hours of Googling haven't helped much.

Thanks for listening.

~Wayne

Hi guys,   I am using this code to open and close a pop up window, but as soon as i click the close button this   http://localhost/popup.php?random=&button=   automatically adds in the url, Please tell me what is wrong with the script

<script type="text/javascript">

 $(document).ready(function(){
    $('a.popup-window').click(function(){
       
       var popupBox = $(this).attr('href');
       $(popupBox).fadeIn(400);
	   
	   var popMargTop = ($(popupBox).height() + 24)/2;
	   var popMargLeft = ($(popupBox).width() + 24)/2;
	   
	   $(popupBox).css({
		   'margin-top' : -popMargTop,
		   'margin-left' : -popMargLeft
	   });
	   
	   $('body').append('<div id="mask"></div>');
	   $('#mask').fadeIn(400);   
	   return false;
	});
    
	$('button.close,#mask').live('click', function(){
		   $('#mask,.popupInfo').fadeOut(400,function(){
			   $('#mask').remove();
			   });	
			   return false;
			});	  	
    
       
 
 });
 $(document).keyup(function(e){
	 if(e.keyCode ==27){
		 $('#mask,.popupInfo, #popup-box').fadeOut(400);
		 return false;
		 }
	 });

 </script>
</head>

<body>
<a href="#popup-box" class="popup-window">Click</a>
  <div id="popup-box" class="popupInfo">
    <form>
      <label>ANYTHING</label></br>
      <input type="text" name="random"/></br>
      <button type="submit" name="button" class ="close">close</button>
    </form>
   </div> 

</body>
</html>

Edited by chauhanRohit, 27 June 2014 - 09:46 AM.

I have seen forums highlight what we searched on google to get to their site.

is there a script that would allow me do the same???

Is it possible to add a query string for example some_var=jk84 to any sort of link be it, http://www.website.com , http://www.website.com/?some_id=4, or http://www.website.com/post=45&category=9 or http://www.website.com/somepost/ ?

While adding that extra query string how can I make sure I'm not affecting the website's content or causing some script error?

Hi there,
im trying to have a form show up when user clicks "add joke". I need the variable to be retrieved from the url query string. I cant get the form to show up.  I think its either an issue with the GET function at the top or the link down at the bottom. Please help!


<?php
  // If the user wants to add a joke
  
    $_GET['addjoke'] = $addjoke;

  if (isset($addjoke)):
?>

<FORM ACTION="<?php echo($PHP_SELF); ?>" METHOD=POST>
<P>Type your joke he <BR>
<TEXTAREA NAME="joketext" ROWS=10 COLS=40 WRAP>
</TEXTAREA><BR>
<INPUT TYPE=SUBMIT NAME="submitjoke" VALUE="SUBMIT">
</FORM>

<?php
  else:

    // Connect to the database server
    $dbcnx = @mysql_connect("servername", "username", "password");
    if (!$dbcnx) {
      echo( "<P>Unable to connect to the " .
            "database server at this time.</P>" );
      exit();
    }

    // Select the jokes database
    if (! @mysql_select_db("jhodara2") ) {
      echo( "<P>Unable to locate the joke " .
            "database at this time.</P>" );
      exit();
    }

    // If a joke has been submitted,
    // add it to the database.

$joketext = $_POST['joketext']; 
$submitjoke = $_POST['submitjoke']; 

    if ("SUBMIT" == $submitjoke) {
      $sql = "INSERT INTO jokes SET " .
             "JokeText='$joketext', " .
             "JokeDate=CURDATE()";
      if (mysql_query($sql)) {
        echo("<P>Your joke has been added.</P>");
      } else {
        echo("<P>Error adding submitted joke: " .
             mysql_error() . "</P>");
      }
    }
  
    echo("<P> Here are all the jokes " .
         "in our database: </P>");
  
    // Request the text of all the jokes
    $result = mysql_query(
              "SELECT JokeText FROM jokes");
    if (!$result) {
      echo("<P>Error performing query: " .
           mysql_error() . "</P>");
      exit();
    }
  
    // Display the text of each joke in a paragraph
    while ( $row = mysql_fetch_array($result) ) {
      echo("<P>" . $row["JokeText"] . "</P>");
    }
  
    // When clicked, this link will load this page
    // with the joke submission form displayed.
    echo("<P><A HREF='$PHP_SELF?addjoke=1'>Add a Joke!</A></P>");
  
  endif;
  
?>


see the problem live at http://www.freewaycreative.com/insert2.php

Hello everyone.

I have a small problem.  I might receive a ?aff=## or not on the end of my url when I get a visitor to my website.  This depends on if they are sent from a affiliate website or not.  If they are it shows fine on the home page but I loss it if they go to another page on my website.  I need to keep this ?aff=## information while they look at the other pages.  How would I capture and pass this information to my other pages as they surf my site? 
I tried this to no avail.
$aff=$_GET['aff'];
I have no clue no adding it back to the next page they go to.  Any ideas would be helpful..

Greetings,

I'm looking for a way to pass a query string (from page1) as part of a query string (to page2) as a single key=>value pair. The idea is the use the query string to return the user to the previous page after the action has been completed.

query results[page1]->view record/action selection[page2]->back to results[page1]

I'm sure someone has been down this path before. P.S. the script is all contained within one file, thus the filename.ext is already known.

Thanks

Hi,
My issue here is that I cant get my query string variables to ONLY feed into if/else statement  on my secondary page. I include my secondary page (fine.php) from my index page. My query string variables keep being fed back into my original if/ese statement on index.php.

here is the if/else on index.php (these links work fine):

<?php
if($_SERVER['QUERY_STRING']=='/index.php' || $_SERVER['QUERY_STRING']=='')
{
include 'port.php';
}

elseif (isset($_GET['pos'])){
include 'pos.php';
}

elseif (isset($_GET['web'])){
include 'web.php';
}

elseif (isset($_GET['fine'])){
include 'fine.php';
}

        else {include '404.php';}






here is the if/else on my secondary page (fine.php). These links are supposed to alert the if/else in the next table cell. However, they instead alert the if/else in index.php.


<td><br/>
<a href="?backset"><img src="fine/thumbs/x-backset.jpg" border="0"></a><br/><br/> 
<a href="?backside"><img src="fine/thumbs/x-backside.jpg" border="0"></a><br/><br/> 
<a href="?bannerprint"><img src="fine/thumbs/x-bannerprint.jpg" border="0"></a><br/><br/> 
<a href="?chopu"><img src="fine/thumbs/x-chopu.jpg" border="0"></a><br/><br/> 
</td>
<td><br/>
<div id="DivPiece" align="left">
<?PHP 

if (isset($_GET['backset'])){
       include 'fine/backset.php';
}

elseif (isset($_GET['backside'])){
     include 'fine/backside.php';
}

elseif (isset($_GET['bannerprint'])){
     include 'fine/bannerprint.php';
}

elseif (isset($_GET['chopu'])){
include 'fine/chopu.php';
}

?>
</div>
</td>



How can I get the links on the secondary page to only alert the if/else statement on that page, and BLOCK the if/else statement on index.php from seeing them? I still want to use the query string though.

Thanks!
[/quote]

Hi All
Thanks in advance for your help.

I want to have to following query string Type=myparam&Username=dazd&Password=nk98830&id=0&Cols_Returned=numfrom,sentdata

But my code returns the following
Type=myparam&Username=dazd&Password=nk98830&id=0&Cols_Returned=%2F%22numfrom%2F%22%2C%2F%22sentdata%2F%22

Below is the code:
$data= array(
"Type"=> "myparam",
"Username" => "dazd",
"Password" => "nk98830",
"id" => "0",
"Cols_Returned" => '/"numfrom/",/"sentdata/"'

) ; //This contains data that you will send to the server.
$data = http_build_query($data); //builds the post string ready for posting
echo "The Query String is                   ";
echo $data;


Regards

hey @requinix, I know you and I have talked about this before but I just want to clarify this issue.  Sadly, I have produced very few websites that have made use of dynamics in the form of database querying.  But, the next project that I'm looking at will require it.  The last thing I did with a query string involved the recording of downloads of file on a website.  and on the page, I had this code:

The archive is located he    <a href="dl.php?f=archive.zip">archive.zip</a>

and the ''dl'' page had this code:

if(!$_GET['f']) error('Missing parameter!');

        $stmt = $conn->prepare("INSERT INTO tblDownloads (ip, file, date, time) 
        VALUES (?, ?, ?, ?)");
        $stmt->bind_param("ssss", $ip, $file, $date, $time);
        $stmt->execute();
        $stmt->close();
        $conn->close();
	
	    header("Location: archive.zip");
	    exit;

and all the variables in the above ''dl'' page are declared in a file that is required by it, called ""conn.php"".

My question for anyone here is:

I am going to be producing a website similar to the testing site called ""jsFiddle"":  https://jsfiddle.net/ .  And it is slated to have a huge amount of content on it.  for example, a main page might have 100 links on it whereby, if a user clicks on any of them, it should take them to another page (or load async content) that illustrates and example of any given issue being asked about.  So, I know we've talked about $_GET before Requinix, and how most websites use it to produce dynamic content.  And we've also talked about the concept of markdown.  But once again, can someone here remind me....what's the best way to go about doing this?  Storing the actual ""answer to the issue"" content in a backend database and pulling/displaying it appropriately, kind of like what I am showing I did with the file in the above code?  thanks.

Adam

Edited September 6, 2020 by ajetrumpet