PHP - Adding Query String To Any Url

Hi,

This is probably very easy for someone who knows how but I cant seem to get it working.

Can someone please tell me how I can alter this query to select a 'WHERE gender = female' so instead of picking a random user it picks a random female user ?

Code: [Select]
$offset_result = mysql_query( " SELECT FLOOR(RAND() * COUNT(*)) AS `offset` FROM `users` ");
$offset_row = mysql_fetch_object( $offset_result );
$offset = $offset_row->offset;
$user1b = mysql_query( " SELECT * FROM `users` LIMIT $offset, 1 " );
$user1= mysql_fetch_array($user1b);

Also, would it be easy to modify it so it picks a second random user that isnt the same as the first user ? so I end up with 2 random females ( $user1 and $user2 )

Many thanks,

Scott

Hi

I have an array of items that I am currently displaying a foreach loop

What I'm looking to do, is add a string, to all of items, a part from the last 4:


foreach ($pager->getResults() as $items => $item)
{
echo $item->getName(); 
//need to add a string to this for the all but the last 4
}


Thanks

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

Greetings,

I need a bit of php coding help.  I have a query that gives me results.  What I need to do is take those results and, I think, use php to compare the results and add a 1 to the correct section.

Example:

3 tables
table name- retail_sales
id
sales_id
amount
week

table name- online_sales
id
sales_id
amount
week

table name- sales_person
sales_id
name

I have the query that breaks down results by week:

week 1 | retail_sales.amount | online_sales.amount

What I need is some php code that will then compare the two together and:
 if retail_sales.amount > online_sales.amount +1 to retail
OR if retail_sales.amount < online_sales.amount +1 to online

so you would end up with a result like a sports record (6-2)

Any ideas

Thank you for any help

I need to add likes_username to this query and use a DISTINCT on it.   It currently counts how many likes a status has but in the table there are some statuses with multiple likes from the same username. 

SELECT s.*, COUNT(l.likes_location_id) AS likeCount
 FROM stream AS s
  LEFT JOIN likes AS l ON ( l.likes_location_id = s.stream_id )
 GROUP BY s.stream_id
 ORDER BY s.stream_id DESC LIMIT 50

Many thanks,  

Hello,

The query below works well. It pulls information from 3 MySQL tables: login, submission, and comment.

It creates a value called totalScore2 based on calculation of values pulled from these three tables.

The MySQL tables "comment" and "submission" both have the following fields:

Code: [Select]
loginid submissionid
In the table "submission," each "submissionid" has only one entry/row, and thus only one "loginid" associated with it.

In the table "comment," the field "submissionid" could have several entries/rows, and could be associated with multiple "loginid"s.

Each time one of the "submissionid"s in "comment" is associated with the same "loginid" that it has in the table "submission," I would like to add this as a factor to the equation below. I would like to multiple instances like this times (-10).

How could I do this?

Thanks in advance,

John

$sqlStr2 = "SELECT 
    l.loginid, 
    l.username, 
    l.created,
    DATEDIFF(NOW(), l.created) + COALESCE(s.total, 0) * 5 + COALESCE(scs.total, 0) * 10 + COALESCE(c.total, 0) AS totalScore2
FROM login l    
LEFT JOIN (
    SELECT loginid, COUNT(1) AS total 
    FROM submission 
    GROUP BY loginid
) s ON l.loginid = s.loginid
LEFT JOIN (
    SELECT loginid, COUNT(1) AS total 
    FROM comment 
    GROUP BY loginid
) c ON l.loginid = c.loginid
LEFT JOIN (
    SELECT S2.loginid, COUNT(1) AS total 
    FROM submission S2
    INNER JOIN comment C2
    ON C2.submissionid = S2.submissionid
    GROUP BY S2.loginid
) scs ON scs.loginid = l.loginid
GROUP BY l.loginid
ORDER BY totalScore2 DESC 
LIMIT 25";

If ($_GET['CODE'] == '1') {
    $ThreadID = mysql_escape_string($_GET['threadid']);
    $ForumID = mysql_result(mysql_query("SELECT forum_id FROM ".FORUM_THREADS." WHERE thread_id='{$ThreadID}'", $db), 0, 'forum_id');
    $PostText = mysql_escape_string($_POST['replytext']);
    $IP = $_SERVER['REMOTE_ADDR'];
    $PostQuery = "INSERT INTO ".FORUM_POSTS." (`user_id`, `thread_id`, `post_text`, `forum_id`, `ip_address`, `timestamp`) VALUES ('{$memid}', '{$ThreadID}', '{$PostText}', '{$ForumID}', '{$IP}' , CURRENT_TIMESTAMP)";
    $PostRes = mysql_query($PostQuery, $db);


For some reason whenever this query goes through, it also adds a blank result using all the same information but without any $PostTest so essentially it does query:

Code: [Select]
INSERT INTO ".FORUM_POSTS." (`user_id`, `thread_id`, `post_text`, `forum_id`, `ip_address`, `timestamp`) VALUES ('{$memid}', '{$ThreadID}', '', '{$ForumID}', '{$IP}' , CURRENT_TIMESTAMP)

This is the only query that does this part of the code and I can't see why it is adding 2 queries.

FORUM_POSTS = constant containing table name.

So I have this snippet of code:

Code: [Select]
             

                $row['ratings'] = "";


                        $ratingsSQL = $this->ipsclass->DB->query(
                                                                "SELECT `pid` , `rating` , COUNT(`rating`) as ratings
                                                                FROM `ibf_ratings`
                                                                WHERE `pid` = '" . $row['pid'] . "' GROUP BY `rating`
                                                                ");

                        $i = 0;
                        $ratingshtml = "";

                        if( $this->ipsclass->DB->get_num_rows( $ratingsSQL ) )
                        {
                                while( $rrow = $this->ipsclass->DB->fetch_row( $ratingsSQL ) )
                                {
                                        $ratingshtml .= "<div class=\"rate_it_display\" style=\"background-image: url( 'style_images/rate/" . str_replace( ' ' , '' , strtolower( $this->ratings[$rrow['rating']] ) ) . ".png' ) \"> " . $rrow['ratings']. " x " . $this->ratings[$rrow['rating']] . "!</div>";
                                        $i++;
                                }

                                $ratingshtml .= "<a href=\"#\" onclick=\"return RateItExpand( {$row['pid']} );\" style=\"color: #999;\">(list)</a>";

                        }

                        $row['ratings'] = $this->ipsclass->compiled_templates['skin_topic']->ratings( $row['pid'] , $row['author_id'] , $ratingshtml );
I want to add an if statement to the code so that the ratings won't load unless a user clicks a button at the top of the screen.  Can anyone help me with this?  I'd be very appreciative.  Thanks in advance.

Hello everyone,

I have this very short script I wrote with help from a book using arrays and the rand function.
Basically it changes the images randomly on browser load or refresh.
However, I am trying to add an additional, not sure what to call it, a string or another caption for each images.

Instead of this being a caption, it will be a link, so a link will appear under each caption that goes to another page
when it is clicked on.

I tried adding another caption and adding a like to it but getting syntax error.

Thanks everyone!

Here is the code: Code: [Select]
<?php
$images = array(
array('file' => 'image1',
'caption' => 'Caption 1'),

array('file' => 'image2',
'caption' => 'Caption 2'),
);


$i = rand(0, count($images)-1);
$selectedImage = "graphics/{$images[$i]['file']}.png";
$caption = $images[$i]['caption'];
?>
And here is where the images are written to the page:
Code: [Select]
<div id="stage">
<img src="<?php echo $selectedImage; ?>" alt="Random image" />
<p id="caption"><?php echo $caption; ?></p>
</div>
IC

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

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???

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.

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..

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