Using PHP to Search MYSQL DB and Display Result

[mattyd]

I am seeking to learn more about the noted subject, how to use PHP to allow a user to enter search terms and search a database.  I have experimented with this with little results save for errors.  Please see code listed below:

 

search.php

<?
//// filename = search.php
<form method="post" action="result.php3">
<select name="metode" size="1">
<option value="row_name1">metode1</option>
<option value="row_name2">metode2</option>
</select>
<input type="text" name="search" size="25">
<input type="submit" value="Begin Searching!!">
</form>
?> 

 

results.php

//// filename = result.php3
<?
$hostname = "mysql7.000webhost.com"; // Usually localhost.
$username = "a4542527_root"; // If you have no username, leave this space empty.
$password = "*******"; // The same applies here.
$usertable = "people"; // This is the table you made.
$dbName = "a4542527_test1"; // This is the main database you connect to.
MYSQL_CONNECT($hostname, $username, $password) OR DIE("Unable to connect to database");
@mysql_select_db( "$dbName") or die( "Unable to select database");
?>
<?
//error message (not found message)
$XX = "No Record Found";
$query = mysql_query("SELECT * FROM $usertable WHERE $metode LIKE '%$search%' LIMIT 0, 30 ");
while ($row = mysql_fetch_array($query))
{
$variable1=$row["row_name1"];
$variable2=$row["row_name2"];
$variable3=$row["row_name3"];
print ("this is for $variable1, and this print the variable2 end so on...");
}

//below this is the function for no record!!
if (!$variable1)
{
print ("$XX");
}
//end
?>

 

Upon viewing search.php I receive the error message:

Parse error: syntax error, unexpected '<' in /home/a4542527/public_html/search.php on line 3

 

I believe I may be missing something and am a bit lost.

Thank-you in in advance for any help or suggestions.

~Matty

[Pikachu2000]

You're using short <? open tags. Change the tags from <? To <?php tags and see if that helps.

 

EDIT: that isn't what the problem is, but it won't hurt to do it.

The problem is you're trying to echo within php tags, without actually issuing an echo . . . 

[mattyd]

You're using short <? open tags. Change the tags from <? To <?php tags and see if that help.

 

Thank-you for your reply.  I made the change as you suggested but I am still getting the same error message as before.

 

Matty

[Pikachu2000]

I jumped the gun. See my edit above . . .

[mattyd]

I jumped the gun. See my edit above . . .

 

In the file "result.php"?  I am sort of confused (and not sure how to fix this).

 

Thanks.

~Matty

[Pikachu2000]

In search.php, you open the php tags, then have an html block in there without using 'echo'. If you simply remove the <?php tags around that block of html, it should fix that error. I didn't look over all of the code, so fixing that one may reveal other errors that need attention.

[mattyd]

In search.php, you open the php tags, then have an html block in there without using 'echo'. If you simply remove the <?php tags around that block of html, it should fix that error. I didn't look over all of the code, so fixing that one may reveal other errors that need attention.

 

Thank-you for all of your help. 

 

Matty

[mattyd]

That error has been resolved. Now I am able to enter search data and perform a search.  When attempting to do this, though, I receive the error message:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a4542527/public_html/result.php on line 15

 

Note: I am using this code from a tutorial and did not write it myself so I am a bit stuck with some of it.  I am attempting to connect to a pre-existing DB and search for data that already resides there.

 

Here is the actual page: http://bluelinedown.netau.net/search.php

 

Thanks,

~Matty

 

[Pikachu2000]

Frankly, that tutorial is already teaching you some bad habits. Look this over and see my comments within the code. Let me know if you have any questions.

 

<?php
$hostname = "mysql7.000webhost.com"; // Usually localhost.
$username = "a4542527_root"; // If you have no username, leave this space empty.
$password = "*******"; // The same applies here.
$usertable = "people"; // This is the table you made.
$dbName = "a4542527_test1"; // This is the main database you connect to.
MYSQL_CONNECT($hostname, $username, $password) OR DIE("Unable to connect to database");
mysql_select_db( "$dbName") or die( "Unable to select database");
//error message (not found message)
$XX = "No Record Found"; // assign a value for the 'empty result set' message
$query = "SELECT * FROM $usertable WHERE $metode LIKE '%$search%' LIMIT 0, 30"; // Query string, separated from the query execution
if( !$result = mysql_query($query) ) { // check query execution success/failure
echo '<br>Query string: ' . $query . '<br>Resulted in an error: ' . mysql_error() . '<br>';  // if execution fails, echo debugging info incl. error and query string
} else {  // if query succeeds
if( mysql_num_rows($result) > 0 ) {  // if result set is not empty, echo the result set
	while ($row = mysql_fetch_array($query)) {
		$variable1=$row["row_name1"];
		$variable2=$row["row_name2"];
		$variable3=$row["row_name3"];
		print ("this is for $variable1, and this print the variable2 end so on...");
	}
} else { // else, if the result set is empty, echo the message for an empty result set.
	echo $XX;
}
}
//end
?>

[mattyd]

Frankly, that tutorial is already teaching you some bad habits. Look this over and see my comments within the code. Let me know if you have any questions.

 

<?php
$hostname = "mysql7.000webhost.com"; // Usually localhost.
$username = "a4542527_root"; // If you have no username, leave this space empty.
$password = "*******"; // The same applies here.
$usertable = "people"; // This is the table you made.
$dbName = "a4542527_test1"; // This is the main database you connect to.
MYSQL_CONNECT($hostname, $username, $password) OR DIE("Unable to connect to database");
mysql_select_db( "$dbName") or die( "Unable to select database");
//error message (not found message)
$XX = "No Record Found"; // assign a value for the 'empty result set' message
$query = "SELECT * FROM $usertable WHERE $metode LIKE '%$search%' LIMIT 0, 30"; // Query string, separated from the query execution
if( !$result = mysql_query($query) ) { // check query execution success/failure
echo '<br>Query string: ' . $query . '<br>Resulted in an error: ' . mysql_error() . '<br>';  // if execution fails, echo debugging info incl. error and query string
} else {  // if query succeeds
if( mysql_num_rows($result) > 0 ) {  // if result set is not empty, echo the result set
	while ($row = mysql_fetch_array($query)) {
		$variable1=$row["row_name1"];
		$variable2=$row["row_name2"];
		$variable3=$row["row_name3"];
		print ("this is for $variable1, and this print the variable2 end so on...");
	}
} else { // else, if the result set is empty, echo the message for an empty result set.
	echo $XX;
}
}
//end
?>

 

Hi.  I looked this over and replaced my file with this code.  I received the following result when running a search query:

 

Query string: SELECT * FROM people WHERE LIKE '%%' LIMIT 0, 30

Resulted in an error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'LIKE '%%' LIMIT 0, 30' at line 1

 

I am not really sure what that means or how to address the issue.

 

Page: http://bluelinedown.netau.net/search.php

 

Thanks.

~Matty

 

[Pikachu2000]

That means the $search and $metode variables are empty or undefined. Is there anything in the tutorial that looks like this? If not, add this to the results.php file, right after //error message (not found message)

 

if( isset($_POST['search']) ) {
     $search = mysql_real_escape_string($_POST['search']);
}

if( isset($_POST['metode']) ) {
     $metode = mysql_real_escape_string($_POST['metode']);
}

[mattyd]

That means the $search and $metode variables are empty or undefined. Is there anything in the tutorial that looks like this? If not, add this to the results.php file, right after //error message (not found message)

 

if( isset($_POST['search']) ) {
     $search = mysql_real_escape_string($_POST['search']);
}

if( isset($_POST['metode']) ) {
     $metode = mysql_real_escape_string($_POST['metode']);
}

 

I did as you suggested and am getting a different error now:

 

Query string: SELECT * FROM people WHERE row_name2 LIKE '%kate%' LIMIT 0, 30

Resulted in an error: Unknown column 'row_name2' in 'where clause'

 

 

[Pikachu2000]

You're supposed to substitute your actual database column names into those . . .