Search engine code

[Johnnyboy123]

Can someone help me with my code. Used a tutorial as aid but still struggling. Basically I have a table called student. I created a search engine so that if a user searches for either the sname,fname or sno of a student then that student would appear. Apart from some notices I got it working and displaying students when searching for only their sname. Now, I want to be able to make it possible to find a student by searching either sname,fname or sno and I'm getting errors. Can anyone have a look at my code:

 

<?php

//get data
$button = $_GET ['submit'];
$search = $_GET ['search'];

if (!$button)
echo "You didn't submit a term.";
else
{
if (strlen($search)<=0)
	echo "search term too short.";
	else
	{
		echo "You searched for <b>$search</b><hr size='1'>";
		include 'includes/config.php';
		include 'includes/functions.php';

		connect();




			//explode search term
			$search_exploded = explode(" ",$search);

			foreach($search_exploded as $search_each)
			{
			//construct query
			$x++;                                                                                         //Line 30
			if ($x==1)
			$construct .= "sname,fname,sno LIKE '%$search_each%'";      //Line 32
			else
			$construct .= "OR sname,fname,sno LIKE '%$search_each%'";



			}



	$construct = "SELECT * FROM student WHERE $construct";
	$run = mysql_query($construct);
	$foundnum = mysql_num_rows($run);                                      //Line 44

	if ($foundnum==0)
		echo "No results found.";
		else
		{
		echo"$foundnum results found!<p>";

		while ($runrows = mysql_fetch_assoc($run))
		{
		//get data
		$sname = $runrows['sname'];
		$fname = $runrows['title'];
		$sno = $runrows['sno'];

		echo "
		<b>$fname</b>
		$sname
		$sno";

		}
	}
	}
}
?>

 

Errors:Notice: Undefined variable: x in C:\Program Files\EasyPHP-5.3.3\www\Project\search.php on line 30

 

Notice: Undefined variable: construct in C:\Program Files\EasyPHP-5.3.3\www\Project\search.php on line 32

 

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\Program Files\EasyPHP-5.3.3\www\Project\search.php on line 44

No results found.

 

First time attempting this, so any tips on what I did wrong would be great. Thank you in advance.

[PaulRyan]

Try the following code...

<?php

//get data
$button = $_GET ['submit'];
$search = $_GET ['search'];

if (!$button)
echo "You didn't submit a term.";
else
{
if (strlen($search)<=0)
	echo "search term too short.";
	else
	{
		echo "You searched for <b>$search</b><hr size='1'>";
		include 'includes/config.php';
		include 'includes/functions.php';

		connect();




			//explode search term
			$search_exploded = explode(" ",$search);
			$construct = '';
			$x = 0;

			foreach($search_exploded as $search_each)
			{
			//construct query
			$x++;
			if ($x==1)
			$construct .= "sname,fname,sno LIKE '%{$search_each}%'";
			else
			$construct .= " OR sname,fname,sno LIKE '%{$search_each}%'";

			}



	$construct = "SELECT * FROM student WHERE $construct";
	$run = mysql_query($construct);
	$foundnum = mysql_num_rows($run);

	if ($foundnum==0)
		echo "No results found.";
		else
		{
		echo"$foundnum results found!<p>";

		while ($runrows = mysql_fetch_assoc($run))
		{
		//get data
		$sname = $runrows['sname'];
		$fname = $runrows['title'];
		$sno = $runrows['sno'];

		echo "
		<b>$fname</b>
		$sname
		$sno";

		}
	}
	}
}
?>

 

Tell me how it goes buddy

 

Regards, PaulRyan.

[xyph]

Before your foreach on line 27, add a $x = 0; $construct = ''; That will define the variable before your script attempts to increment, check or add to it.

 

Echo out $construct before you execute the query. Also, try this out

 

$run = mysql_query($construct) or die(mysql_error());

 

The error will tell you where your query is going wrong.

[matthew9090]

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\Program Files\EasyPHP-5.3.3\www\Project\search.php on line 44

No results found.

 

that means it is not connecting to the database propibly and like xyph said to use the error thing to check.

 

try this

//explode search term
			$search_exploded = explode(" ",$search);
			$x = 0;
			foreach($search_exploded as $search_each)
			{
			//construct query
			$x++;
			if ($x==1)
			$construct = "sname,fname,sno LIKE '%{$search_each}%'";
			else
			$construct .= " OR sname,fname,sno LIKE '%{$search_each}%'";

			}

 

[Johnnyboy123]

Thanks guys, I adjusted it with the code that matthew provided, which cleared up all the notices. I used die(mysql_error()); to find the error. Seems the error is:

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 'fname,sno LIKE '%sdfgf%'' at line 1

 

which refers to:

<?php
if ($x==1)
$construct = "sname,fname,sno LIKE '%{$search_each}%'";
else
$construct .= " OR sname,fname,sno LIKE '%{$search_each}%'";
}
?>

 

When I only use 1 field however e.g $construct = "sname LIKE '%{$search_each}%'";

It seems to work fine. The original tutorial from where I worked off also only used 1 field. I believe the tut was based on using entered info from only 1 field to find a student. However, I want to be able to use either any 1 of 3 fields: either sname,fname or sno. Any ideas?

 

[xyph]

I think this has to do with your usage of LIKE. This is a MySQL problem, not PHP

 

Check out this tutorial. It might help

 

http://www.tutorialspoint.com/mysql/mysql-like-clause.htm