[RESOLVED] variable from php to mysql

[iiaustinll]

Hey all how ya doin, im fairly an amateur and having some problems, any help would be aweomse
I'm using the

on (release) {
getURL("phptest.php", "_blank", "GET");
}
function in flash to submit my variale to php, lets say i put in 'pizza' for the variable wand send it, it gives me the url
"www.mysite.com/phptest.php?value=pizza

now I have a database that with a PHP script i can run a query for 'pizza' and it returns all my items that have pizza in them. what i want to do is when i type 'pizza' or 'shrimp' for example into the flash input text box, and send it to phptest.php, it will enter "pizza" into the MySQL query, my problem is I don't know the correct syntax to do that. Can someone please help me? This is my php code as of now
/////////////////
<?php
$con = mysql_connect("**********.secureserver.net","","***");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db("***********", $con);

$value=$_GET['value'];
$result = mysql_query("SELECT M.MName, M.MPrice, R.Name, R.City, R.Website
FROM MenuItem M, Restaurants R
WHERE M.RID = R.RID
AND M.MName LIKE '%$value%'");////this is where i essentially want "pizza" to go, 😕 but im sure that's obvious for most of you. i doubt '%$value%'" is right, because it's not working, can someone show me the correct code i would put here, and anywhere else i need it? thanks!!

echo "<table border='1'>
<tr>
<th>Food Item</th>
<th>What's it cost?</th>
<th>Where's it at?</th>
<th>Where's the place?</th>
</tr>";

while($row = mysql_fetch_array($result))

{
$site = $row['Website'];
echo "<tr>";
echo "<td>" . $row['MName'] . "</td>";
echo "<td>" . $row['MPrice'] . "</td>";
echo "<td>" . $row['City'] . "</td>";
echo "<td>" . "<a href = '$site' target='_blank'>" . $row['Name'] . "</a>" . "</td>";
echo "</tr>";
}
echo "</table>";
mysql_close($con);
?>

Thanks again for your help!

[Scriptmaker]

I guess I'm getting a little confused with your database structure. Would you mind clarifying this part:

FROM MenuItem M, Restaurants R

Also, it helps if when you post code, you use the little PHP button on the top right corner of the forum editor so that the forum highlights it.

Also, when you say it doesn't work, what DOES it do? (try adding the same OR DIE statement you had above to the end of your mysql query)

[iiaustinll]

hey thanks for the reply, if you want to take a look at the database, go here
www.ueatout.com/letssee.php

i want to take a variable from flash lets say pizza from a user input text box, and send it to php which sends it to mysql and returns all the items that have the word pizza in it, let me know if u need anymore info

[rulian]

try '%".$value."%';

echo out your query or it's error to see what is actually being produced

[iiaustinll]

hey i tried that, nothing happened

i can get the variable to print out so i know its fetching it, i just need to get the MySQL query to recognize the variable, thats where im getting stuck, here is some of the code

$name = $_POST['name'];

print "Your food is ". $name . ".<br>";
echo mysql_select_db("jonpom", $con);
$result = mysql_query("SELECT M.MName, M.MPrice, R.Name, R.City, R.Website
FROM MenuItem M, Restaurants R
WHERE M.RID = R.RID
AND M.MName LIKE '%".$name."%'");////// this is where i think im losing it\\\

also i tried echoing it as you can see, and it returns this
.
1

also if you want to start at the oringinal input, the url is ueatout.com/passpost.swf

thanks again again again

[nemonoman]

here's the code to test with:

print "Your food is ". $name . ".<br>";
mysql_select_db("jonpom", $con);
$query="SELECT M.MName, M.MPrice, R.Name, R.City, R.Website
FROM MenuItem M, Restaurants R
WHERE M.RID = R.RID
AND M.MName LIKE '%".$name."%'";
echo $query;
$result = mysql_query($query);
if(!$result){
 die(mysql_error());
}
$row=mysql_fetch_array($result);
print_r($row);

[iiaustinll]

I know php is recieving the variable becuase it shows it when i use the print function. The problem i'm having if passing that variable into my query, i want my data base to return any name that has pizza in it, but it just gives me the table and nothing more.
Here's the full code im using:

<?php
$con = mysql_connect("xxxxxxxxx","xxxxxxx","xxxxxxxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

$name = $_POST['name'];

print "Your food is ". $name . ".<br>";
mysql_select_db("jonpom", $con);
$query="SELECT M.MName, M.MPrice, R.Name, R.City, R.Website
FROM MenuItem M, Restaurants R
WHERE M.RID = R.RID
AND M.MName LIKE '%".$name."%'";


echo "<table border='1'>
<tr>
<th>Food Item</th>
<th>What's it cost?</th>
<th>Where's it at?</th>
<th>Where's the place?</th>
</tr>";

while($row = mysql_fetch_array($result))

  {
  $site = $row['Website'];
  echo "<tr>";
  echo "<td>" . $row['MName'] . "</td>";
  echo "<td>" . $row['MPrice'] . "</td>";
  echo "<td>" . $row['City'] .  "</td>";
  echo "<td>" . "<a href = '$site' target='_blank'>" . $row['Name'] .  "</a>" . "</td>";
  echo "</tr>";
  }
echo "</table>";
mysql_close($con);
?>

thanks for all your help already, this is starting to baffle me, ive spent like 6 hours researching this, can't figure it out.

[nemonoman]

if you use the code I provided for testing, you'll get a better sense of your problem from the debugging information that will print out.

[iiaustinll]

Ok I tried that and it threw back that "the query was empty" so I'm assuming that the input from flash is in the wrong format? does it matter if the size is wrong or the font? seems strange to me. Because when i make $name = pizza, and then put $name in the query, it comes back fine, anyone familiar with this?

[nemonoman]

Please print the full output returned by the debugging code I posted. Not what you posted. In your last posted code, you don't even query the database.

[rulian]

if your query return 1 then it was successful
see what this prints out after you do your $results

while($row=mysql_fetch_array($result))
{
echo "<pre>";
print_r($row); 
echo "</pre>";

}

[xxbplxx]

Hey. I'm working with iiaustinll on this site that we're having issues with. I did run that most recent code snippet that was posted, and I got this output.

Your food is

pizza

That's all. Now I'm not sure exactly how big the "pizza" is, but it certainly displays it much bigger than the "Your food is". There is no database result. The query runs simply fine... I'll post the code here and explain it.

<?php
$con = mysql_connect("xxxxxxx.secureserver.net","xxxxx","xxxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

//$name = 'pizza';***************************
$name = $_POST['name'];*********************

print "Your food is ". $name ."<br>";
mysql_select_db("jonpom", $con);
$result = mysql_query("SELECT M.MName, M.MPrice, R.Name, R.City, R.Website
FROM MenuItem M, Restaurants R
WHERE M.RID = R.RID
AND M.MName LIKE '%$name%'");

while($row=mysql_fetch_array($result))
{
echo "<pre>";
print_r($row);
echo "</pre>";

} 

mysql_close($con);

?>

Now, where the asteriks are, I run the code with "$name = $POST['name'];" ... in theory the PHP should get the input variable from Flash, then run it through the SQL query. It does get the input variable, but I don't think it's in the right "format" because there is never a result from the DB. So, I believe it runs the query, but instead of having "pizza" as the variable, it's running "messed up format of pizza" as the variable. Definitely don't have that in any database fields. Now I know the query runs fine, because if you look by the asteriks, I also have "$name = 'pizza';" ... I commented out the flash $POST and used "$name = 'pizza', and it returned

Array
(
    [0] => 12" Margherita Pizza
    [MName] => 12" Margherita Pizza
    [1] => 7.50
    [MPrice] => 7.50
    [2] => Leo's Italian Restaurant & Pizzeria
    [Name] => Leo's Italian Restaurant & Pizzeria
    [3] => Wappingers Falls
    [City] => Wappingers Falls
    [4] => http://www.uEatOut.com/leos.swf
    [Website] => http://www.uEatOut.com/leos.swf
)

Array
(
    [0] => 12" Tre Colori Pizza
    [MName] => 12" Tre Colori Pizza
    [1] => 9.00
    [MPrice] => 9.00
    [2] => Leo's Italian Restaurant & Pizzeria
    [Name] => Leo's Italian Restaurant & Pizzeria
    [3] => Wappingers Falls
    [City] => Wappingers Falls
    [4] => http://www.uEatOut.com/leos.swf
    [Website] => http://www.uEatOut.com/leos.swf
)

Array
(
    [0] => 12" Di Melonzane Pizza
    [MName] => 12" Di Melonzane Pizza
    [1] => 9.00
    [MPrice] => 9.00
    [2] => Leo's Italian Restaurant & Pizzeria
    [Name] => Leo's Italian Restaurant & Pizzeria
    [3] => Wappingers Falls
    [City] => Wappingers Falls
    [4] => http://www.uEatOut.com/leos.swf
    [Website] => http://www.uEatOut.com/leos.swf

There's more it returns, but it's just every "pizza" result in the database ... which is what it SHOULD return. So I believe that shows that the problem is with the flash input variable and something with the format/size of it. Any thoughts?

[nemonoman]

I don't understand why you won't use the code I provided which would print out the query that is actually being sent to the database, and the actual rows returned by the database. Accordingly, I think I've about had it with offering opinions. Please have fun debugging your problem your own way.

[xxbplxx]

nemonoman wrote:

I don't understand why you won't use the code I provided which would print out the query that is actually being sent to the database, and the actual rows returned by the database. Accordingly, I think I've about had it with offering opinions. Please have fun debugging your problem your own way.

Wow, I don't see the need of getting angry over this. I did try your code, and it produced no rows, just like every other code. Here's what I tried:

<?php
$con = mysql_connect("xxxxxxxxxxx.secureserver.net","xxxxxx","xxxxxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

//$name = 'pizza';
$name = $_POST['name'];

print "Your food is ". $name . ".<br>";
mysql_select_db("jonpom", $con);
$query="SELECT M.MName, M.MPrice, R.Name, R.City, R.Website
FROM MenuItem M, Restaurants R
WHERE M.RID = R.RID
AND M.MName LIKE '%".$name."%'";
echo $query;
$result = mysql_query($query);
if(!$result){
die(mysql_error());
}
$row=mysql_fetch_array($result);
print_r($row); 
mysql_close($con);
?>

If entering 'pizza' on my page, it returns this...

Your food is

pizza
.
SELECT M.MName, M.MPrice, R.Name, R.City, R.Website FROM MenuItem M, Restaurants R WHERE M.RID = R.RID AND M.MName LIKE '%

pizza
%'

Once again, it returns no database rows. No code whatsoever reproduces database rows when it accepts the variable from Flash like we've been doing. Now if I do what I did last time and switch out the $name from flash to simply assigning it a text field ...

[CODE]
<?php
$con = mysql_connect("xxxxxxxxxxx.secureserver.net","xxxxxx","xxxxxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

$name = 'pizza';
//$name = $_POST['name'];

print "Your food is ". $name . ".<br>";
mysql_select_db("jonpom", $con);
$query="SELECT M.MName, M.MPrice, R.Name, R.City, R.Website
FROM MenuItem M, Restaurants R
WHERE M.RID = R.RID
AND M.MName LIKE '%".$name."%'";
echo $query;
$result = mysql_query($query);
if(!$result){
die(mysql_error());
}
$row=mysql_fetch_array($result);
print_r($row); 
mysql_close($con);
?>

I receive this

Your food is pizza.
SELECT M.MName, M.MPrice, R.Name, R.City, R.Website FROM MenuItem M, Restaurants R WHERE M.RID = R.RID AND M.MName LIKE '%pizza%'Array ( [0] => 12" Margherita Pizza [MName] => 12" Margherita Pizza [1] => 7.50 [MPrice] => 7.50 [2] => Leo's Italian Restaurant & Pizzeria [Name] => Leo's Italian Restaurant & Pizzeria [3] => Wappingers Falls [City] => Wappingers Falls [4] => http://www.uEatOut.com/leos.swf [Website] => http://www.uEatOut.com/leos.swf )

Funny because it doesn't return all the rows that have pizza in it like the other query suggested in this post, or my original HTML table code. Nevertheless, notice how there is no size difference in the "pizza" variable, and how this time it actually returns rows. The only plausible solution is that the variable from Flash does not properly get thrown into the query as regular text. Once again, any thoughts?

[iiaustinll]

nemonoman wrote:

I don't understand why you won't use the code I provided which would print out the query that is actually being sent to the database, and the actual rows returned by the database. Accordingly, I think I've about had it with offering opinions. Please have fun debugging your problem your own way.

:glare: Sarcasm not needed, unless you feel you are too cool to help; then don't. We used your code, and it returned "1" so we know that isn't the problem, we actually have decided that we probably need to use the loadVars() function in Flash, rather than just a post method, if you even know what I'm talking about. So I'm going to try that, and when I mark it as resolved, i'll put 'with all the help from "nemonoman"' to make you feel better = because you sound upset, don't let me ruin your day. :o

[nemonoman]

I assure you I am not being sarcastic. You are not working with me on solving your problem. I work with PHP/MySQL and Flash, and I assure you, I have learned how to debug these problems.

The code I asked you to run:

$query="SELECT M.MName, M.MPrice, R.Name, R.City, R.Website 
FROM MenuItem M, Restaurants R 
WHERE M.RID = R.RID 
AND M.MName LIKE '%".$name."%'"; 
echo $query; 

Would have returned something. Something useful: your query as the database would get it.

Since you didn't use that code, you never found out what was being sent to the database. Or if you did, you didn't post it.

In fact the later comment "query was empty" suggests that NOTHING was being sent to the database. Didn't that strike you as odd?? Not a 1064 error (syntax), but 1065 -- Query was Empty??

You post code that fetches arrays without sending queries -- You post information being returned without posting the exact query that returns it. In short, you are approaching this debugging problem with little or no effort to support or work with persons attempting to assist you.

Consider my comments to be addressed to the helpful experts on this board, to wave them off from attempting to help you. Your answers suggest that you are not interested in following actual advice.

The above is not sarcasm, by the way. When I use sarcasm, you'll know it. I sincerely wish you fun in debugging your project. You certainly will have very little company in the endeavor.

[nemonoman]

You have finally posted the actual query being used. Thank you.

The ACTUAL value you are inserting into your ery is:
<br />
<br />
<font size="7">pizza</font><br />

This is not what you want. You need to apply [man]strip_tags[/man] to $name before creating the query.

$name=strip_tags($name);

You may also need to [man]trim[/man] the value:

$name=trim(strip_tags($name));

[iiaustinll]

nemonoman wrote:

I assure you I am not being sarcastic. You are not working with me on solving your problem. I work with PHP/MySQL and Flash, and I assure you, I have learned how to debug these problems.

You post code that fetches arrays without sending queries -- You post information being returned without posting the exact query that returns it. In short, you are approaching this debugging problem with little or no effort to support or work with persons attempting to assist you.

Consider my comments to be addressed to the helpful experts on this board, to wave them off from attempting to help you. Your answers suggest that you are not interested in following actual advice.

The above is not sarcasm, by the way. When I use sarcasm, you'll know it. I sincerely wish you fun in debugging your project. You certainly will have very little company in the endeavor.

OOOPPPPFFFHHH, put me in my place huh? :bemused: 😉
anyhow, the trim/strip tags did the trick, i appreciate your time over the past 3 days, and all your help as well, im sure ill be back later tonite lol. THANKS AGAIN!!!

[nemonoman]

I'll just point out that when you actually used the code I posted, which displayed the query being passed to the database -- the answer was obvious.