passing variables?

[Anon]

I'm stuck in a script that seems (of course) as though it should work. In HTML, I have created two drop down lists, and I select two items. I then pass these variables to the database, which returns a third column of data for me to edit. I'm stuck in that the two variables are assigned the correct data, but the query is not returning a row that matches them. If I enter a SQL command straight into the DB, it finds the row fine. So, the problem seems to be in passing the variables within the query to the DB. Anyone see where I'm going wrong? A condensed version of my script:

<form method=post action="<?php echo $PHP_SELF ?>">
<select name="animal">
<option selected value="dog">dog</option>
<option value="cat">cat</option>
</select>

<select name="category">
  <option selected value="feeding">feeding</option>
  <option value="habitat">habitat</option>
   </select>

<input type="submit" value="Load Comment" name="loaddata">
</form>

  <?php

if ($loaddata)
{
echo $animal;
echo "<br>";
echo $category;
$db = mysql_connect("localhost", "root");
mysql_select_db("bairey",$db);
$sql = "SELECT * FROM content WHERE animal = '$animal' AND category = '$category'";
$result = mysql_query($sql);
$myrow = mysql_fetch_array($result);
echo mysql_num_rows($result);
$editcomment = $myrow["comments"];
echo $editcomment

?>

When I click my "Load Comment" button, both variables (dog and feeding) display, so I know the variables are loaded correctly. When I enter the query "SELECT * FROM content WHERE animal = 'dog' AND category = 'feeding'" directly into MySQL, I see the correct row: 'dog', 'feeding', 'this is dummy text'. After much commenting and testing, it all seems to fall apart when no row is returned by the query in PHP.

Any ideas?

Thanks,

Steph

[Anon]

Did you try to hard-code the values 'dog' and 'feeding' in your query ? Does the query 'select count(*) from content' give the right value ?

[Anon]

Good testing tips. It does work when I hard-code the correct values, and the following code:

$testsql = "select count(*) from content";
$testresult = mysql_query($testsql) or die ("Didn't work.");
echo $testresult;

returns 2. There's only one row in the table.

Thanks!

[Anon]

One more thing you might try is echoing $sql, then pasting the output into MySQL. It looks to me like your script should work too, but maybe MySQL is seeing something we're not.

• Beth

[Anon]

Ugh! The troubles this thing is giving me.. my first issue has been resolved, though I don't know why. The code that works now is EXACTLY what I posted previously. Oh well.

Now I'm having trouble passing the variables to the next if() portion of the script.

<?php
if ($loaddata)
{
$db = mysql_connect("localhost", "user");
mysql_select_db("bairey",$db);
$sql = "SELECT * FROM content WHERE animal = '$animal' AND category = '$category'";
$result = mysql_query($sql);
$myrow = mysql_fetch_array($result);
$editanimal = $myrow["animal"];
$editcategory = $myrow["category"];
$editcomment = $myrow["comments"];
echo "edit-a: ";
echo $editanimal;
echo "<br>edit-c: ";
echo $editcategory;
?>

**These return correctly.

<center><form method=post action="<?php echo $PHP_SELF ?>">
<textarea name=newcomments><?php echo $editcomment ?></textarea>

<input type=submit value=Update name=submit></form>

<?php
} //end load data if
else
{
} // end load data else
?>

<?php
if ($submit)
{
$passedanimal = $HTTP_POST_VARS['animal'];
$passedcategory = $HTTP_POST_VARS['category'];
echo "<br>sub-a: ";
echo $passedanimal;
echo "<br>sub-c: ";
echo $passedcategory;
echo "<br>sub-ha: ";
echo $HTTP_POST_VARS['animal'];
echo "<br>sub-hc: ";
echo $HTTP_POST_VARS['category'];
echo "<br>sub-edit-a: ";
echo $editanimal;
echo "<br>sub-edit-c: ";
echo $editcategory;

** NONE of these have any value.

My host has track_vars disabled, and won't turn them on. I am putting <?php_track_vars?> at the top of this script, but it doesn't seem to be working - I put phpinfo() further down, and it shows track_vars disabled at both master and local.

ANYTHING that will pass these variables to the if($submit) function would be appreciated! I've tried writing my own functions, assigning them globally, etc. but I guess I'm missing something.

Thanks!

[Anon]

• To retrieve a count(*) info, proceed like this :

$sql = "select count(*) nb from content";
$result = mysql_query($sql);
$myrow = mysql_fetch_array($result);
echo $myrow["nb"];

• I don't understand why you try to use HTTP_POST_VARS at the end of the script. PHP automatically creates the variables, as well as it created $submit, $animal or $category.

• FYI, if you really want to enable track_vars, put this in your .htaccess file :

php3_track_vars on (PHP3)
php_flag track_vars on (PHP4)

Note : track_vars is automatically activated with PHP>=4.0.3

[Anon]

It doesn't look like you're passing the 'animal' or 'category' in the second form. Try adding a line like this for animal:

<input type="hidden" name="animal" value="<?php echo $animal;?>">

and a similar one for category.

• Beth