Jump to content

ichversuchte

Members
  • Posts

    24
  • Joined

  • Last visited

    Never

Contact Methods

  • Website URL
    http://www.forthestudentsbookstore.com

Profile Information

  • Gender
    Not Telling

ichversuchte's Achievements

Newbie

Newbie (1/5)

0

Reputation

  1. Thank you, This worked great! <?php echo $price; ?>
  2. Below there is a table that has a image that has on click event that opens a amorizationcalculator. I am trying to pass the $price from a queried row.  The data is there, i echoed it.  But it will not pass to the URL, any ideas? [code] <table width="555" height="144" border="0">   <tr>     <td width="177" rowspan="3"><?php  print "<img src=\"image.php?iid=$iid\" width='213' height='153'>"; ?></td>     <td colspan="2">&nbsp;<?php print $title ?></td>   </tr>   <tr>     <td width="73"><div align="center"><img src="images/car.gif" width="29" height="9"></div></td>     <td> Only <?php print $mileage ?> miles</td>   </tr>   <tr>     <td width="73"><div align="center"><A ONCLICK="window.self.open('AmorizationCalc.php?price=<?php $price ?>','','width=550,height=400,screenX=400,screenY=400,top=400,left=400,scrollbars=yes');"=""><img src="images/tigracalc.gif" width="15" height="13" alt="Calculate Payment" alt="Calculate Payment" title="Calculate Payment"></div></A> </td>     <td>Asking <?php print convert_to_money($price) ?></td>   </tr> </table> [/code]
  3. I need to pull a value from the url that is passed.  Say that the url is this: http://www.forthestudentsbooks.com/usersendinginfo.php?lid=26 I need to get the 26 from the url so that I can use it on the page im working with.  What would the code look like to do this? Thank you
  4. I am new to javascript and for some reason i am having trouble implementing this. I tried using the code above and i created the three documents script1.html, script2.html, script3.html. I need to use this code which fills the combobox : [code] <?php include('../Scripts/conn/cid.php'); $sql = "SELECT ra_name FROM staff_elkin ORDER BY ra_name ASC"; $result = mysql_query($sql)           or die('Query failed. ' . mysql_error()); if (mysql_num_rows($result) > 0) {     '<select name="raname" id="raname" onchange=Change_Iframe_Contents(this)>'; while ($row = mysql_fetch_array($result))     {         $title = $row['ra_name'];     echo("<option value='$title'>$title</option>");               }     echo '</select>';    }     include('../Scripts/close/end.php'); ?> [/code] I need to take the combobox on a onchange and update the iframe. Depending on the choice which is [code] $title [/code] i need to update the iframe after running that varible through this sql statement - [code] <?php include_once('conn/conn.php'); $query = "SELECT raname1, raname2, date1, raname3, date2           FROM dutyswitch_elkin where raname1 = '".$title."'";            $result = mysql_query($query) or die('Error, query failed'); echo '<table border="0">'; while(list($raname1, $raname2, $date1, $raname3, $date2, ) = mysql_fetch_array($result)) { echo "<tr><td>$raname1 <em>made the request:</em> <br>$raname2 <em>will sit on</em> $date1 <em>and</em> </br>$raname3 <em>will sit on</em> $date2</td></tr>"; } echo '</table>'; echo '<br>'; ?> [/code] I have been trying to get this work for a while. I have a couple of books but I can't seem to figure out. Can someone please help me. I don't need it done for me, but can someone please direct me. Thank you for your help in advance....
  5. [!--quoteo(post=370957:date=May 3 2006, 01:14 PM:name=ober)--][div class=\'quotetop\']QUOTE(ober @ May 3 2006, 01:14 PM) [snapback]370957[/snapback][/div][div class=\'quotemain\'][!--quotec--] This has turned into a Javascript question. You can use the onchange event instead of onclick to fire a function from the selection in a select box. I'm moving this to Javascript. [/quote] Can someone help me with this javascript question?
  6. The only think that I can see is that you are trying to us mysql connect to connect to $dbname - test before you call that varible. That might be why you stop at check1 and don't go any futher. Try [code] <?php $dbhost = 'localhost'; $dbuser = 'root'; $dbpass = 'cze00345'; $dbname = 'test'; echo 'check 1'; $conn = mysql_connect($dbhost, $dbuser, $dbpass); if (!$conn) { echo("ERROR: " . mysql_error() . "\n");    } echo 'check 2'; mysql_select_db($dbname); ?> [/code]
  7. If no one can help me, can someone at least point me to a site or reference a book that can help me. Thank you
  8. I have been reading and figured out iframes. Is it possible to to use the onclick with a combobox withouta button. I want to post with the click on the mouse on the comboox not a button. Please help
  9. you first need to connect to the database.... which can look like this [code] <?php     $usr = "username";     $pwd = "password";     $db = "database name";     $host = "localhost";     $cid = mysql_connect($host,$usr,$pwd);     mysql_select_db($db);     if (!$cid) { echo("ERROR: " . mysql_error() . "\n");    } ?> [/code] Then need to the form and this code...you might have to tweak it a bit [code] <?php        # back on to this page (POST METHOD)     if ($REQUEST_METHOD=="POST") {                  $SQL = " INSERT INTO email_test ";         $SQL = $SQL . " (address, ) VALUES ";         $SQL = $SQL . " ('$address') ";                          #execute SQL statement         $result = mysql_db_query($db,"$SQL",$cid);                  # check for error         if (!$result) { echo("ERROR: " . mysql_error() . "\n$SQL\n");    }               echo ("<P><B>Your Request Was Submitted Successfully!!!</B></P>\n");     }     mysql_close($cid); ?> [/code] should get you started.....
  10. trying adding the this [code] <?php .....  ?> [/code] around the code and your code shouldn't be shown like it has been. and does westminster stand for westminster college in pa
  11. I am trying to update a I frame when a person chooses a name from a combobox. I have this so far, which populates the combobox : [code] <?php include('./Scripts/conn/cid.php'); $sql = "SELECT ra_name FROM staff_elkin ORDER BY ra_name ASC"; $result = mysql_query($sql)           or die('Query failed. ' . mysql_error()); if (mysql_num_rows($result) > 0) {     echo '<select name="raname" id="raname">'; while ($row = mysql_fetch_array($result))     {         $title = $row['ra_name'];     echo("<option value='$title'>$title</option>");     }     echo '</select>';    }     include('./Scripts/close/end.php'); ?> [/code] From here I want to update a iframe, and i have no experience with iframes. I want to use one because I don't want to refresh the page. When they select a name, I want the I frame to be updated. I am goign to run a simple sql statement that pulls the names from that query. Can someone help me get started ? :D please
  12. When i used that code, it did the same thing. only filled one combobox....
  13. Ok i did that and the code looks like this revised... That fixed the double combobox problem, but now its only filling one name into the box. [code] <?php include_once('Scripts/conn/cid.php'); $sql = "SELECT ra_name FROM staff_elkin"; $result = mysql_query($sql)           or die('Query failed. ' . mysql_error()); if (mysql_num_rows($result) > 0) {     while ($row = mysql_fetch_array($result))     {             $title = $row['ra_name'];          }     ?>     <select name="raname" id="raname">     <?php echo("<option value='$title'>$title</option>"); ?>     </select>     <?php      }     include_once('Scripts/close/end.php'); ?> [/code]
  14. Ok...i changed that and it filled the combo box, but it create a combobox for everyname. It didn't put them all in one and that is what i wanted? i appreciate your help...
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.