PHP - Moved: Php Sql String Query Using Where And Commands - Going Nuts!

This topic has been moved to PHP Regex.

http://www.phpfreaks.com/forums/index.php?topic=326004.0

This topic has been moved to Apache HTTP Server.

http://www.phpfreaks.com/forums/index.php?topic=349296.0

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

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

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 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.

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

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???

i want to change this code :

$result = mysql_query("select count(*) from 3gp where category='Bollywood'");

with this

$result = mysql_query("select count(*) from 3gp where category='$category'");

but i am getting error message "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/funxy/public_html/salmobile.info/3gp-videos.php on line 151". I am calling the page using http://abc.com/3gp-videos.php?category=Bollywood

Please help.


Code: [Select]
<?php
 

$db="funxy_db";

$conn = mysql_connect("localhost","funxy_saba","myasco001");
@mysql_select_db($db) or die( "Unable to select database, Please contact your administrator");









class Pager 
   { 
       function getPagerData($numHits, $limit, $page) 
       { 
           $numHits  = (int) $numHits; 
           $limit    = max((int) $limit, 1); 
           $page     = (int) $page; 
           $numPages = ceil($numHits / $limit); 

           $page = max($page, 1); 
           $page = min($page, $numPages); 

           $offset = ($page - 1) * $limit; 

           $ret = new stdClass; 

           $ret->offset   = $offset; 
           $ret->limit    = $limit; 
           $ret->numPages = $numPages; 
           $ret->page     = $page; 

           return $ret; 
       } 
   }  


 
    // get the pager input values 
    $page = $_GET['page']; 
    $limit = 3; 
    $result = mysql_query("select count(*) from 3gp where category='Bollywood'"); 
    $total = mysql_result($result, 0, 0); 

    // work out the pager values 
    $pager  = Pager::getPagerData($total, $limit, $page); 
    $offset = $pager->offset; 
    $limit  = $pager->limit; 
    $page   = $pager->page; 

    // use pager values to fetch data 
    $query = "select * from 3gp where category='Bollywood' order by id DESC limit $offset, $limit"; 
    $result = mysql_query($query); 

    // use $result here to output page content 


//my addition

//grab all the content

   //Custom Table Stsrt//
        $cols = 4; //number of coloms
        $i =1;
        echo "<table border=\"0\" cellpadding=\"2\" cellspacing=\"2\" width=\"100%\" id=\"table1\" bordercolor=\"#FFFFFF\" bgcolor=\"#FFFFFF\">"
             ."<tr>";

while($r=mysql_fetch_array($result))
{
   //the format is $variable = $r["nameofmysqlcolumn"];
   //modify these to match your mysql table columns
  
   $id=$r["id"];
   $name=$r["name"];
   $views=$r["views"];
   $url=$r["url"];
   $image=$r["image"];
   $category=$r["category"];
  
   
   
   //display the row
   
   
   
   $mybox = "
   
   <br>
   <a href='http://salmobile.info/3gp-videos-download.php?id=$id' class=\"classd\"><b><u>&nbsp;$name</u></b></a>
   <BR>
   <span class=\"text\"><img src=\"$image\" width=\"60\" height=\"60\"></span>
   <BR>
   <span class=\"text2\">Download This 3GP Video</span>
   <br>";


                if (is_int($i / $cols)){
                    echo "<td width='336' align='center' style=\"border-style: dotted; border-width: 0\">$mybox</td></tr><tr>";
                }else{
                    echo "<td width='336' align='center' style=\"border-style: dotted; border-width: 0\">$mybox</td>";
                }
             if ( $i / $cols == 3) echo "<td colspan='3' align=\"center\">









</td></tr><tr>";



if ( $i / $cols == 8) echo "<td colspan='3' align=\"left\">



 

</td></tr><tr>";


             $i++;
          //end if
       }//end while
       echo "</tr></table>";
       //Custom Table End//
   


//ends my addition



    // output paging system (could also do it before we output the page content) 
    if ($page == 1) // this is the first page - there is no previous page 
        echo "<font color=\"#FFB300\">Previous</font>"; 
    else            // not the first page, link to the previous page 
        echo "<a href=\"http://salmobile.info/index-" . ($page - 1) . ".html\" id=\"navigationURL\"><font color=\"#FFB300\">Previous</font></a>"; 

    for ($i = 1; $i <= $pager->numPages; $i++) { 
        echo " <font color=\"#FFFFFF\">|</font> "; 
        if ($i == $pager->page) 
            echo "<font color=\"#FFB300\"> $i</font>"; 
        else 
            echo "<a href=\"http://salmobile.info/index-$i.html\" id=\"navigationURL\"> <font color=\"#FFB300\">$i</font></a>"; 
    } 

    if ($page == $pager->numPages) // this is the last page - there is no next page 
        echo "Next"; 
    else            // not the last page, link to the next page 
        echo "&nbsp;&nbsp;&nbsp;<a href=\"http://salmobile.info/index-" . ($page + 1) . ".html\" id=\"navigationURL\"><font color=\"#FFB300\">Next</font></a>"; 
?>

Hi, how do I store as a string (in a variable) a mysql query b/c what I'm doing below outputs Resource id in client browser:

Code: [Select]
<?php
//database connection set up etc
$show=mysql_query("SELECT file_Name FROM xdocument WHERE doc_id=95");
print $show;
?>

Any help much appreciated, thanks.

I want the page to refresh on with the query string, but this isn't working for me...

Code: [Select]
header("Location: " . $_SERVER['php_self'] . "?" . $_SERVER['query_string'] );

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

I have this form

<form action="" method="get" name="size"> Størrelse <hr><label><input type="radio" value="9" name="size" onchange="this.form.submit()">92 (2år)</label>


<label><input type="checkbox" value="147" name="size" onchange="this.form.submit()">68 (6-9 mdr.)
</label><label><input type="checkbox" value="150" name="size" onchange="this.form.submit()">86 (18-24 mdr.)</label>
<input type="checkbox" value="149" name="size" onchange="this.form.submit()">80 (12-18 mdr.)</label>
<label><input type="checkbox" value="148" name="size" onchange="this.form.submit()">74 (9-12 mdr.)</label>


</form>

I want to make it so that when the user clicks one checkbox for example size 68 and then checks size 80 I want both sizes to be in the URL query string. But every time the user clicks a new checkbox, the query string changes accordingly. Maybe with a separator like ?size=4:23 and then I will handle it with the DB stuff.   Please tell me how to add multiple elements in the query string with this form, thank you!
Edited by Stefany93, 21 May 2014 - 08:03 AM.