bigrossco Posted January 19, 2007 Share Posted January 19, 2007 I know how to make drop down box's but what I want to do is if a user selects something fromt he first drop down box i.e. Make of Car, the second drop down box will show models for that make of car only, I know that this is possible just dont know how to code itThanks for your help,Ross Quote Link to comment Share on other sites More sharing options...
ToonMariner Posted January 19, 2007 Share Posted January 19, 2007 you will be needing either;a, a MASSIVE javascript array populated when the age loads.or (better option)b, some ajax that will populate the select box based on the value of the other select boxes - this should be coded in such a way so that the page will work without javascript. Quote Link to comment Share on other sites More sharing options...
bigrossco Posted January 20, 2007 Author Share Posted January 20, 2007 do you know where i can get help on doing one of the two options?R Quote Link to comment Share on other sites More sharing options...
suzzane2020 Posted January 22, 2007 Share Posted January 22, 2007 here's a script tht 'll help:Just paste this script in the body tags and run it..before that a few changes where i have commented.This script is based on category table and its subcategory table:<?php//Your conection stringmysql_connect("localhost","root",'');mysql_select_db("test");$n=0;//here i have states in one table and destinations in another$res=mysql_query("SELECT states.stateid sID, states.name sname, destinations.did dID, destinations.name dnameFROM states, destinations WHERE states.stateid = destinations.sid "); while($row=mysql_fetch_array($res)) { extract ($row); echo $stateid; $TempArray1[n]=$sID.":".UCwords($sname).":".$dID.":".UCwords($dname)."~"; $arr1.=implode(",",$TempArray1); $n++;} ?>//Javascript starts here <script language="JavaScript" type="text/javascript">function populateCountry1(value,arr,len) { var va=value; var b=arr; var t=new Array(); t=b.split('~'); count=0; var j; var j=len; for(i=0;i<j;i++) { temp=t[i].split(':'); if(eval(temp[0])==eval(va)) { var newop=new Option(temp[3],temp[2]); document.world.destinations.options[count]=newop; count++; } else { // var newop=new Option('-- Select a Country First --',300); document.world.destinations.options[count]=null; var newop=new Option('-- Select a Level First --',300); document.world.destinations.options[count]=newop; } } }</script>//Ends here//then you you have the html tagsfor the two combo boxes<form name="world" action="samepage" method="POST"><select name="states" class="textbox" onChange="populateCountry1(this.options[selectedIndex].value,'<?php echo $arr1;?>','<?php echo $n;?>');"> <option value='' selected>Select level</option> <?php $res=mysql_query("select * from states"); while($row=mysql_fetch_array($res)) { extract ($row); ?> <option value="<?php echo $stateid;?>"><?php echo $name;?></option> <?php }?> </select><br> Education <select name="destinations" class="textbox"> <option value="300">-- Select a Level First --</option> </select> </form> Quote Link to comment Share on other sites More sharing options...
bigrossco Posted January 22, 2007 Author Share Posted January 22, 2007 When doing this i get the errorWarning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/ewencooper/domains/ewencooper.co.uk/public_html/index.php on line 63 Quote Link to comment Share on other sites More sharing options...
suzzane2020 Posted January 22, 2007 Share Posted January 22, 2007 cud u give the line where u get the error..so i cud check tht out Quote Link to comment Share on other sites More sharing options...
suzzane2020 Posted January 22, 2007 Share Posted January 22, 2007 its probabaly where u have te query codescud u paste the line where u get the error? Quote Link to comment Share on other sites More sharing options...
fenway Posted January 22, 2007 Share Posted January 22, 2007 [quote author=bigrossco link=topic=123137.msg510429#msg510429 date=1169469081]When doing this i get the errorWarning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/ewencooper/domains/ewencooper.co.uk/public_html/index.php on line 63[/quote]You probably just have a msql syntax error... check mysql_error().FYI, if this a relatively short list, there's no reason to hit the DB. Quote Link to comment Share on other sites More sharing options...
bigrossco Posted January 26, 2007 Author Share Posted January 26, 2007 The line is the querey line. I am wanting to use a Database as it is going to be a large list (its going to be a list for makes and models of cars!)$res=mysql_query("SELECT states.stateid sID, states.name sname, destinations.did dID, destinations.name dnameFROM states, destinations ");This is what I am using just now (the same as the one put in, yes I do have the database for it) while($row=mysql_fetch_array($res)) { extract ($row); echo $stateid; $TempArray1[n]=$sID.":".UCwords($sname).":".$dID.":".UCwords($dname)."~"; $arr1.=implode(",",$TempArray1); $n++;} the first part (while) is line 63ThanksR Quote Link to comment Share on other sites More sharing options...
bigrossco Posted January 26, 2007 Author Share Posted January 26, 2007 I changed the querey to this;$res=mysql_query("SELECT states.stateid sID, states.name sname, destinations.did dID, destinations.name dnameFROM states, destinations WHERE states.stateid = destinations.dID");and dont get the message now, in the drop down for the first part it dose show the state name but in the second box nothing is shown, but I know this is be cause of it having only 1 thing to show Education <select name="destinations" class="textbox"> <option value="300">-- Select a Level First --</option> </select>what I want it to do is show all destionations releated to the state selectedR Quote Link to comment Share on other sites More sharing options...
fenway Posted January 26, 2007 Share Posted January 26, 2007 Well, you'll have to parse the result set to generate a hash by state, and then use that array inside for the 2nd dropdown. Quote Link to comment Share on other sites More sharing options...
bigrossco Posted January 26, 2007 Author Share Posted January 26, 2007 this seems to be more difficult that i thought it would be lol is their any other way I can do this without a database for example?As I thought it would be easyer than this lol unless its just something I have done wrong with the code?The code I have in it is:[code]<?php//Your conection stringmysql_connect("localhost","ewencooper_cars",'password');mysql_select_db("ewencooper_cars");$n=0;//here i have states in one table and destinations in another$res=mysql_query("SELECT states.stateid sID, states.name sname, destinations.did dID, destinations.name dnameFROM states, destinations WHERE states.stateid = destinations.dID"); while($row=mysql_fetch_array($res)) { extract ($row); echo $stateid; $TempArray1[n]=$sID.":".UCwords($sname).":".$dID.":".UCwords($dname)."~"; $arr1.=implode(",",$TempArray1); $n++;} ?><script language="JavaScript" type="text/javascript">function populateCountry1(value,arr,len) { var va=value; var b=arr; var t=new Array(); t=b.split('~'); count=0; var j; var j=len; for(i=0;i<j;i++) { temp=t.split(':'); if(eval(temp[0])==eval(va)) { var newop=new Option(temp[3],temp[2]); document.world.destinations.options[count]=newop; count++; } else { // var newop=new Option('-- Select a Make --',300); document.world.destinations.options[count]=null; var newop=new Option('-- Select a Make First --',300); document.world.destinations.options[count]=newop; } }}</script><form name="world" action="samepage" method="POST">Make<select name="states" class="textbox" onChange="populateCountry1(this.options[selectedIndex].value,'<?php echo $arr1;?>','<?php echo $n;?>');"><option value='' selected>Select Make</option> <?php $res=mysql_query("select * from states"); while($row=mysql_fetch_array($res)) { extract ($row); ?> <option value="<?php echo $stateid;?>"><?php echo $name;?></option> <?php }?> </select><br><Br> Model <select name="destinations" class="textbox"> <option value="300">-- Select a Make First --</option> </select> </form>[/code]Could it be something to do with the java script coding?R Quote Link to comment Share on other sites More sharing options...
fenway Posted January 26, 2007 Share Posted January 26, 2007 I've confused about your PHP/JS code... you need to hash the results by state name, and in each key's value, store the array of destinations. You have implodes, joins, and splits for no reason. Quote Link to comment Share on other sites More sharing options...
bigrossco Posted January 26, 2007 Author Share Posted January 26, 2007 [quote author=fenway link=topic=123137.msg513874#msg513874 date=1169814216]you need to hash the results by state name, and in each key's value, store the array of destinations. [/quote]How would I go about doing this? Quote Link to comment Share on other sites More sharing options...
fenway Posted January 26, 2007 Share Posted January 26, 2007 I don't know the PHP syntax... but in psuedo-code:1) Make a states hash2) Iterate though the result set3) Push the destination onto an array for the matching state hash keyIn Perl, this is easyl Quote Link to comment Share on other sites More sharing options...
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