search problem returning blank page

[jgsketch]

Hello,
I have a simple search on my site. It searches through by database of products by keyword. I'm sure my coding is wrong. When I submit, I recieve a blank page. I'm using MySQL, Apache server and PHP4. Any help would be great. I've tried changing some code, but still get the same results.

Here is the code on my main page.

<form action="search.php" method="post"> 
<input type="hidden" name="content" value="search.php"> 
<input type="text" name="searchstring" cols="5"> 
<input type="submit" value="GO" /> 
</form>

Here is search.php

<?php

/This is just my DBconnect and select.
include ("startmarket.php");


$search_result = @mysql_query("SELECT * FROM products WHERE product_name LIKE '%$searchstring%' ");

while ($row =@ mysql_fetch_array($result)) 
{ 
echo "<a href=\"" . $row["url"] . "\"><img src=\"" . $row["imagefile"] . "\" border=0 alt=\"" . $row["name"] . 
"\"></a><BR>";

echo "<b>" . $row["name"] . "</b><br>";
echo $row["description"] . "<br>";
echo "<b>Price: $</b>";
echo "<b>" . $row["price"] . "</b><br><hr /><br>";
}

?>

Thanks again for any help.
Jeremy

[torrent]

Maybe register globals are off and so you need to reference the $searchstring var with either:

$_POST['searchstring'] // if php v4.1 or later
$HTTP_POST_VARS['searchsrting'] // if earlier version than 4.1

[MrAlaska]

You are assigning the results set to $search_result, but asigning $row to the result set in $result

I would also take those @ signs off, at least until you get the code working. Error messages are your friends when developing.h

[sarahk]

Hi Jeremy

I use a database class to manage most of this so had to look up the tutorial at http://www.phpbuilder.com/columns/joe20010104.php3?page=3

Now, assuming that your query can be echo'd onto the page and copied into phpMyAdmin (or equivalent) and create a result then the following should work...

while($row = mysql_fetch_array($result))
{
    echo "<a href='" . $row["url"] . "'><img src='" . $row["imagefile"] . "' border=0 alt='" . $row["name"] . 
"'></a><BR>"; 
  echo "<b>" . $row["name"] . "</b><br>"; 
  echo $row["description"] . "<br>"; 
  echo "<b>Price: $" . $row["price"] . "</b><br><hr /><br>"; 
}

[jgsketch]

Hello, Thanks for the quick response.

MrAlaska, I changed the $result to $search_result. No change.
I also took out the @. It came up with a invalid MySQL result resource. On
while ($row =@ mysql_fetch_array($result))
but there was no error message in the line above with the @ taken out.

Sarahk, I tried your code, But got the same results. I've used this echo code in other parts of my site, and it works fine.

I have not tried the HTTP_POST_VARS yet. I have the most recent version of PHP4.

My code must have the problem in the main site or the query part. I can't seem to figure this one out. Thanks again for the help, any other suggestions?

Here is my updated code

/Here is my main page

<form action="search.php" method="post"> 
<input type="text" name="searchstring" cols="10"> 
<input type="submit" value="GO" /> 
</form> 

/Here is my search.php

<?php 
   include ("startmarket.php");

$search_result = mysql_query ("SELECT * FROM products WHERE product_name LIKE '%$searchstring%' ");

while ($row = mysql_fetch_array ($search_result)) 
{ 
echo "<a href=\"" . $row["url"] . "\"><img src=\"" . $row["imagefile"] . "\" border=0 alt=\"" . $row["name"] . 
"\"></a><BR>";

echo "<b>" . $row["name"] . "</b><br>";
echo $row["description"] . "<br>";
echo "<b>Price: $</b>";
echo "<b>" . $row["price"] . "</b><br><hr /><br>";
}

?>

Jeremy

[jgsketch]

Hello again,
Does anyone have any suggestions on this error that I'm recieving? Thanks for the help.

Jeremy

[torrent]

I still think this is related to register globals being Off. You can test this by instead of executing the query, echoing it out to the screen instead. Make sure you see the posted search string in the WHERE clause.

My guess is that it won't be there, in which case you need to use $_POST['searchstring']. E.g.

$mysearch = &$_POST['searchstring']
echo "SELECT ...... WHERE .... LIKE '%{$mysearch}%'";

[jgsketch]

Got IT!! I had the wrong column name in there. I do not know what I was thinking. But the other post did help. Thanks again.

Jeremy

[torrent]

😃
Cool. Glad you got it sorted.