Using a Form to Search Site. Why Doesn't My Form Work?

OK, guys. Try not to laugh. This is my first attempt at a form using PHP. Why doesn't it work?

I want people to be able to search for the online stores on my site. The stores are in a MySQL table called Store_Table.

Form:

<form action="search-results.php" method="post">
<input type="text" name="search" />
<input type="submit" value="Search Stores" />
</form>

Code (markup):

Script on search-results.php:

<?php

$search = $_POST["search"];

$con = mysql_connect("localhost","root");

if (!$con)

{
die('Could not connect: ' . mysql_error());
}

mysql_select_db("ctyi", $con);

$result = mysql_query("SELECT * from Store_Table WHERE store_name LIKE '%$search%'");

echo $result;

else echo "Sorry! Couldn't find store.";

?>

PHP:

 

does it print any errors?

 

Sorry! I should have mentioned that. Here's the error:

Parse error: syntax error, unexpected T_ELSE in C:\xampp\htdocs\xampp\search-results.php on line 19

When I remove the ELSE statement, it still doesn't work. I don't get an error; it just says "Resource id #3", no matter what I search for.

 

this line must be causing the error

else echo "Sorry! Couldn't find store.";

 

I realize that, dude. Do you know how I can fix it?

 

first this line

if you are using xampp and installed everything as default try this

$con = mysql_connect("localhost","root","") or die('cant connect to server'.mysql_error());

the tell me if it contains errors

 

Yes, I am using XAMPP, and I did install everything as default. However, when I tried what you said, I got the same error:

Parse error: syntax error, unexpected T_ELSE in C:\xampp\htdocs\xampp\search-results.php on line 15

 

alright, can you try this code

<?php

$search = $_POST["search"];

$con = mysql_connect("localhost","root","") or die('cant connect: '.mysql_error());
mysql_select_db("ctyi", $con);


$sql = "SELECT * from Store_Table WHERE store_name LIKE '%$search%'";
$result = mysql_query($sql) or die('cant select: '.mysql_error());

while($row = mysql_fetch_assoc($result) )
{
print_r($row);
}


?>

 

Didn't work, dude. This is what it returned:

Array ( [advertiser_id] => 1 [advertiser_name] => AbeBooks.com [advertiser_logo] => abebooks.png [advertiser_url] => abebooks-coupons.php )

Note that, in fact, the table's name is Advertiser_Table, not Store_Table, which is why all the columns' names begin with "Advertiser", too.

Basically, the code you provided listed the contents of an entire row, depending on which store/advertiser I searched for.

Thanks for trying, though. I appreciate it. I have added reputation points to your account.

 

hmm, but that query is what you have in the code, i basically just modified it, the other parts like formatting, you do on your own i guess

 

Does my problem have something to do with the fact that there is no IF statement at the beginning?

But then, it wouldn't make sense for me to have an IF statement.

 

The above one would work for you

but why you are putting a if else loop for a database connection?

 

Thanks for the help, Freewebspace! However, it still didn't work! Whatever I search for, it just keeps echoing: "Sorry! Couldn't find store."

I really don't know what I'm doing. All I want is for people to be able to search for stores on my site, all of which are in a MySQL table. My form seems to work, but not my action script.

 

It should work. Besides, if it display message "Sorry! Couldn't find store." thats doesn't mean that it doesn't work, just there are no such as store with that name in mysql database table.

<?php
$search = $_POST["search"];
if($search<>''){
$con = mysql_connect("localhost","root");
if (!$con){die('Could not connect: ' . mysql_error());}
mysql_select_db("ctyi", $con);

$result = mysql_query("SELECT * FROM Store_Table WHERE store_name LIKE '%$search%'");
while ( $row = mysql_fetch_array(  $result ) )
	{
	$store_name = $row['store_name'];

	echo $store_name;
	}else{echo"Sorry! Couldn't find store.";}
}else{echo "Search field can't be empy.";}
?>

PHP:

 

<?php

$search = $_POST["search"];

$con = mysql_connect("localhost","root");

if (!$con)

{
die('Could not connect: ' . mysql_error());
}

mysql_select_db("ctyi", $con);

$result = mysql_query("SELECT * from Store_Table WHERE store_name LIKE '%$search%'");
$num=mysql_num_rows($result);
if($num>0){
echo $result;
}
else{ echo "Sorry! Couldn't find store.";}

?>

PHP:

 

Didn't work:

Parse error: syntax error, unexpected T_ELSE in C:\xampp\htdocs\xampp\search-results.php on line 20

Thank you for the effort.

 

This one keeps echoing "Resource id #3" no matter what I search for.

Nonetheless, thank you for the effort.

Reputation points added to all of your accounts.

 

it keeps echoing "Resource id #3" because that's what the current code does

$num=mysql_num_rows($result);
if($num>0){
echo $result;
}
else{ echo "Sorry! Couldn't find store.";}

if you need professional help, i'll gladly lend a hand

 

OK. I've found a code which retrieves the data that I want. All I need is for someone to please tell me how to incorporate an else statement. If a search query doesn't match any advertiser_names, I want it to echo, "No matches found!".

<?php

$search = $_POST["search"];

$con = mysql_connect("localhost","root","") or die('cant connect: '.mysql_error());

mysql_select_db("ctyi", $con);

$sql = "SELECT advertiser_name from Advertiser_Table WHERE advertiser_name LIKE '%$search%'";

$result = mysql_query($sql) or die('cant select: '.mysql_error());

while($row = mysql_fetch_assoc($result) )

{
echo $row['advertiser_name'];
}

?> 

PHP:

 

<?php

$search = $_POST["search"];

$con = mysql_connect("localhost","root","") or die('cant connect: '.mysql_error());

mysql_select_db("ctyi", $con);

$sql = "SELECT advertiser_name from Advertiser_Table WHERE advertiser_name LIKE '%$search%'";

$result = mysql_query($sql) or die('cant select: '.mysql_error());
$numRows = mysql_num_rows($result)

if( $numRows > 0 )
{
while($row = mysql_fetch_assoc($result) )
{
echo $row['advertiser_name'];
}
}
else
{
echo "No matches found!";
}

?>