maxves1 Posted April 5, 2010 Share Posted April 5, 2010 Hello , Can anyone help me with displaying results of my php onto another page? Heres the code I wrote for search category and it works fine : <?php include ('db_connect.php'); // DATABASE CONNECTION ?> <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"><html> <head> <META http-equiv="Content-Type" content="text/html; charset=utf-8"> <meta http-equiv="refresh" content="180; url=http://faculty.poly.edu/~yjeanpie/searchcat.php"> <SCRIPT language=JavaScript> function reload(form) { var val=form.cat.options[form.cat.options.selectedIndex].value; self.location='searchcat.php?cat=' + val ; } </script> <script Language="JavaScript"> <!-- function confirm_dropdown() { if (f1.cat.options.selectedIndex == 0) { alert("Please select the Category."); f1.cat.focus(); return (false); } if (f1.subcat.options.selectedIndex == 0) { alert("Please select the Products."); f1.subcat.focus(); return (false); } return (true); } //--></script> <script language=" JavaScript" ><!-- function MyReload() { window.location.reload(); } //--></script> </head> <body onLoad=" MyReload()"> <center> <p> </p> <p> </p> <p> </p> <div><center> <hr align="center" size="1" width="550" color="#99CCFF"> <hr align="center" size="1" width="550" color="#99CCFF"> <p><img src="fitl reslogo.jpg" width="223" height="67"></p> <p><font face="Arial, Helvetica, sans-serif" size="2" align="Left">Search by Category</font></p> <div align="center"> <? @$cat=$_GET['cat']; if(strlen($cat) > 0 and !is_numeric($cat)){ echo "Data Error"; exit; } $quer2=mysql_query("SELECT DISTINCT category,cat_id FROM category order by category"); if(isset($cat) and strlen($cat) > 0){ $quer=mysql_query("SELECT DISTINCT prod_name FROM PRODUCTS where cat_id=$cat order by prod_name"); } echo '<form method=post name=f1 action="dd-check1.php" onsubmit="return confirm_dropdown()">'; echo "<select name='cat' onchange=\"reload(this.form)\"><option value='' style= width:13em;>Select category</option>"; while($noticia2 = mysql_fetch_array($quer2)) { if($noticia2['cat_id']==@$cat){echo "<option selected value='$noticia2[cat_id]'>$noticia2[category]</option>"."<BR>";} else{echo "<option value='$noticia2[cat_id]'>$noticia2[category]</option>";} } echo "</select>"; echo "<br>"; echo "<br>"; echo "<select name='subcat' style= width:15em;><option value=''>Select products</option>"; while($noticia = mysql_fetch_array($quer)) { echo "<option value='$noticia[prod_name]'>$noticia[prod_name]</option>"; } echo "</select>"; echo "<br>"; echo "<br>"; echo '<input type=image src="back.jpg" srcname=btnback id=btnback onclick="history.go(-1);return false;"/>'; echo '<span style="padding-left:10px">'; echo '<input type=image src="button.JPG" >'; echo '</span>'; echo "</form>"; ?> </div></center> <br> <hr align="center" size="1" width="550" color="#99CCFF"> <hr align="center" size="1" width="550" color="#99CCFF"> </div></center> </div> </center> </body></html> Now I saved this file onto my database as searchcat.php you can check the link at: http://faculty.poly.edu/~yjeanpie/searchcat.php only problem is upon submission ,I want it to display results onto a new page "dd-check1" in this case, in such a way, that it displays the category and product selected and along with it, it displays relevant information related with the specific product for the category selected, like say for example "if i select software and photoshop " I want it to show that on the other page along with the tutorial links i have in database and other relevant info " My databse PRODUCTS looks like the file attached ( from where i want to retrieve other relevant results" PLEASE HELP ME ASAP , i need to get this done within 2 weeks any help will be honestly appreciated [attachment deleted by admin] Link to comment https://forums.phpfreaks.com/topic/197653-help-with-displaying-selected-dropdown-menu-on-another-page-with-related-queries/ Share on other sites More sharing options...
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