getting my variables to work with each other

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November 30th, 2004, 08:26 AM

twistedtwig

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Join Date: Nov 2004

Location: Horsham - England

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passing variables from one card to another and using an sql query???

hey all,

not 100% sure if this should go in the wap or PHP section but thought it could be this one, sorry if i was wrong.. here goes anyway....

i have a problem with my code:

Quote:

<?php
header('Content-Type: text/vnd.wap.wml', true);
print "<?xml version=\"1.0\" encoding=\"ISO-8859-1\"?>\n";
?>
<!DOCTYPE wml PUBLIC "-//WAPFORUM//DTD WML 1.1//EN"
"http://www.wapforum.org/DTD/wml_1.1.xml">
<wml>
<card id="card1" title="Search For Film">
 <img src="logo2.wbmp" alt="Logo"/> 
stuff i have on video and dvd 

<do type="accept" label="Find">
<go href="#cardResult" />
</do>

FilmName: <input name="name" type="text"/> 

</card>

<card id="cardResult" title="Film results">
Film: 
<?php

$filmdb = mysql_connect("localhost", "YYY"QQQ");
$db = mysql_select_db("XXX",$filmdb);

$testvar = "$(name)";

$query = "select * from films where FilmName like '%$testvar%'";
$result = mysql_query($query);
if($result):
if(mysql_num_rows($result) > 0 ):
while($row = mysql_fetch_array($result)):
echo "$row[FilmName] ";
endwhile;
mysql_free_result($result);
endif;
else:
echo "No results found for $(mine)";
endif;

echo "$(name)";
echo "$testvar";

mysql_close($filmdb);
?>

</card>
</wml>

the problem is that i cant get the results to display. i am trying to get teh input from the user from teh input box called "name" and pass that into the sql query.. i wasnt having any luck at all so thought id try and put it into a normal variable and pass that into the query.

$testvar = "$(name)";

this didnt help at all.. the sql query works because i can hard code a result like %pay% and it will return a few results but with the variables it wont work.

i put some echos at the bottom to see if the variables held the input and it did.

i am completely confused.. any advice would be grand..
thx

Jon

November 30th, 2004, 01:51 PM

jabba_29

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I cheat when it comes top wml and use PHP variables. try this - (untested), not sure whether this would work posting to a card in the original deck but give it a go and see what happens. You were missing post field or method -may have caused the problem - if you just used get method, then still use php to capture the result: $_GET['name']

PHP Code:


		
			
<?php

// snip

<do type="accept" label="Find" method=post">

<go href="#cardResult" />

<postfield name="name" value="$name" /> 

</do>

<p>

FilmName: <input name="name" type="text"/><br/>

</p>

</card>

// snip

$testvar = $_POST['name'];

// snip

?>

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November 30th, 2004, 05:19 PM

twistedtwig

Contributing User

Join Date: Nov 2004

Location: Horsham - England

Posts: 302

Time spent in forums: 4 Days 4 h 34 m 19 sec
Reputation Power: 22

Confused

thx for the advice,

i am having a new problem now

its a bit werid... the code FULLY works on an online emulator, http://www.wapsilon.com/ , but wont run on my mobile.. wont even load now.. says "unsupported type" i know the database connection is good as it works on teh emulator and the html version works fine.. the emulator will run it perfectly.. but not on my phone, the phone is a samsung e700 btw. i tried wapalizer or something similar to that name and the first page loads, which is more than my mobile, but it wont action the query part of the script. here is my full and changed code:

Quote:

<?php
header('Content-Type: text/vnd.wap.wml', true);
print "<?xml version=\"1.0\" encoding=\"ISO-8859-1\"?>\n";
?>
<!DOCTYPE wml PUBLIC "-//WAPFORUM//DTD WML 1.1//EN"
"http://www.wapforum.org/DTD/wml_1.1.xml">
<wml>
<card id="card1" title="Search For Film">
 <img src="logo2.wbmp" alt="Logo"/> 
stuff i have on video and dvd 

<do type="accept" label="Find" >
<go href="#cardResult" method="post">
<postfield name="name" value="$name" />
</go>
</do>

FilmName: <input name="name" type="text"/> 

</card>

<card id="cardResult" title="Film results">
Film: 
<?php

$filmdb = mysql_connect("localhost", "YYY", "ZZZ");
$db = mysql_select_db("XXX",$filmdb);

$testvar = $_POST['name'];

$query = "select * from films where FilmName like '%$testvar%'";
$result = mysql_query($query);
if($result):
if(mysql_num_rows($result) > 0 ):
while($row = mysql_fetch_array($result)):
echo "$row[FilmName] ";
endwhile;
mysql_free_result($result);
endif;
else:
echo "No results found for $testvar";
endif;

mysql_close($filmdb);
?>

</card>
</wml>

i really dont ahve a clue whats going on to make it work on one place and not on another it doesnt make much sense to me, i know i am new to this stuff but its frying my brain.

any ideas plz?

Jon

December 1st, 2004, 01:44 AM