techker Posted February 15, 2009 Share Posted February 15, 2009 hey guys i was wondering on how i can get info from my database from a select box. like i tryed this with no luck: <? require "config2.php"; $connection = mysql_connect($dbhost, $dbusername, $dbpass); $SelectedDB = mysql_select_db($dbname); $cat=$_GET['cat']; $QQQ1 = mysql_query(" SELECT id, pic FROM $cat "); if(!$QQQ1) die(mysql_error()); $err = mysql_num_rows($QQQ1); if($err == 0) die("No matches met your criteria."); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <form id="form1" name="form1" method="post" action=""> <label> <select onchange="reload(this.form)" name="cat" id="cat"> <option value="ABS">ABS</option> <option value="ARMS">ARMS</option> </select> </label> </form> <? while ($line = mysql_fetch_array($QQQ1)) { echo "<option value=\"$line[id]\">$line[pic] </option>"; }?> </body> </html> You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1 im shure it's because it is executing the query before i chose the categorie? Quote Link to comment Share on other sites More sharing options...
fenway Posted February 16, 2009 Share Posted February 16, 2009 That's so very exposed to sql injection... echo your query string. Quote Link to comment Share on other sites More sharing options...
techker Posted February 16, 2009 Author Share Posted February 16, 2009 What do you mean? Quote Link to comment Share on other sites More sharing options...
fenway Posted February 16, 2009 Share Posted February 16, 2009 The query that's causing the error -- I want to see it (not the php code). Quote Link to comment Share on other sites More sharing options...
techker Posted February 17, 2009 Author Share Posted February 17, 2009 ok so this is the box alone http://webworkout.info/sel.php Quote Link to comment Share on other sites More sharing options...
techker Posted February 17, 2009 Author Share Posted February 17, 2009 ok so this is the box alone http://webworkout.info/sel.php ah sorry wrong box..lol here is the code for the new probleme.i got the dselect part working but now i need to add more of the same on a same page ..but it refresh's the hole page so all the boxes reset.i have tryed changing all to select2,querry2..check it out to see.. <? require "config2.php"; $connection = mysql_connect($dbhost, $dbusername, $dbpass); $SelectedDB = mysql_select_db($dbname); $cat=$_GET['cat']; ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> <SCRIPT language=JavaScript> function reload(form) { var val=form.cat.options[form.cat.options.selectedIndex].value; self.location='sel2.php?cat=' + val ; } </script> </head> <body> <form id="cat" name="cat" method="post" action=""> <label> <select onchange="reload(this.form)" name="cat" id="cat"> <option value="selected"><? print"$cat" ?></option> <option value="ABS">ABS</option> <option value="ARMS">ARMS</option> <option value="LEGS">LEGS</option> <option value="SHOULDERS">SHOULDERS</option> <option value="CHEST">CHEST</option> </select> </label> </form> <p> <? if(isset($cat) and strlen($cat) > 0){ $quer=mysql_query("SELECT DISTINCT pic,id FROM $cat" ); }else{$quer=mysql_query("SELECT DISTINCT pic,id FROM $cat "); } echo "<select name='subcat'onChange='swapImage()' ><option value=''>Select one</option>"; while($noticia = mysql_fetch_array($quer)) { if($noticia['pic']==@$cat3){echo "<option selected value='$noticia[pic]'>$noticia[pic]</option>"."<BR>";} else{echo "<option value=/Account/PICS/$cat/$noticia[pic]>$noticia[pic]</option>";} } echo "</select>";?> </p> <p> </p> <table width="442" height="138" border="0" cellpadding="0" cellspacing="0"> <tr> <th scope="row"><form id="cat2" name="cat2" method="post" action=""> <label> <select onchange="reload(this.form)" name="cat2" id="cat2"> <option value="selected"><? print"$cat2" ?></option> <option value="ABS">ABS</option> <option value="ARMS">ARMS</option> <option value="LEGS">LEGS</option> <option value="SHOULDERS">SHOULDERS</option> <option value="CHEST">CHEST</option> </select> </label> </form> <p> <? if(isset($cat2) and strlen($cat2) > 0){ $quer2=mysql_query("SELECT DISTINCT pic,id FROM $cat2" ); }else{$quer2=mysql_query("SELECT DISTINCT pic,id FROM $cat2 "); } echo "<select name='subcat1'onChange='swapImage()' ><option value=''>Select one</option>"; while($noticia2 = mysql_fetch_array($quer2)) { if($noticia2['pic']==@$cat3){echo "<option selected value='$noticia2[pic]'>$noticia2[pic]</option>"."<BR>";} else{echo "<option value=/Account/PICS/$cat2/$noticia2[pic]>$noticia2[pic]</option>";} } echo "</select>";?></th> </tr> </table> <p> </p> </body> </html> http://webworkout.info/sel2.php Quote Link to comment Share on other sites More sharing options...
fenway Posted February 17, 2009 Share Posted February 17, 2009 All I want in the actual text of the mysql query...nothing more. Quote Link to comment Share on other sites More sharing options...
techker Posted February 17, 2009 Author Share Posted February 17, 2009 SELECT DISTINCT pic,id FROM $cat Quote Link to comment Share on other sites More sharing options...
fenway Posted February 17, 2009 Share Posted February 17, 2009 SELECT DISTINCT pic,id FROM $cat Not exactly... that's _still_ php code... echo that string. That's the one that's producing the error message, right? Quote Link to comment Share on other sites More sharing options...
techker Posted February 17, 2009 Author Share Posted February 17, 2009 my error is solved with the second script i posted.now the question was can i put it multiple times on one page? Quote Link to comment Share on other sites More sharing options...
fenway Posted February 17, 2009 Share Posted February 17, 2009 Sure, why not? Quote Link to comment Share on other sites More sharing options...
techker Posted February 17, 2009 Author Share Posted February 17, 2009 it refreshs tonly the fisrt box not the second one .. check http://webworkout.info/sel3.php Quote Link to comment Share on other sites More sharing options...
fenway Posted February 18, 2009 Share Posted February 18, 2009 Don't post links to code... no one's going to look at them. "Refreshing" is not a mysql problem... if that's really the issue, but the queries work just fine, then I'll move this thread to the PHP forum. Quote Link to comment Share on other sites More sharing options...
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