all entries from db passed to output file in error

[Bazzaah]

Hi

 

Hope someone can help me please.

 

I have constructed an audio dictionary and have discovered an error now that I have added a few entries to the database.

 

A user can search the database and can click on a result to be taken to the content associated with the entry he chooses.

 

This is the search function;

 

$search=$_POST["search"];


//get the mysql and store them in $result

$result = mysql_query("SELECT word FROM pro_words WHERE word LIKE '%$search%'");
		        
//get the db content that is specified above

if (mysql_num_rows($result) < 1) {

echo "<br><br><h2> We didn't find anything. Sorry - we did look though.</h2>";

}else {
         
echo '<table align="center" cellspacing="8" cellpadding="8" width="85%"><tr><td align="left"><b>Word</b></td></tr>';

echo "<br><br><h2>Success! Here's what we found:</h2><br>";

while ($r=mysql_fetch_array($result, MYSQLI_ASSOC))

echo '<h2><tr><td align="left"><a href="word.php?w=' . $r['word'] . '">' . $r['word'] . '</a></td></tr></h2>';

}

 

You will see that content is displayed in a new file called word.php.

 

This is relevant code from word.php

 

$query = "SELECT word,word_type1,sentence1,word_type2,sentence2,word_type3,sentence3 FROM pro_words"; 

$result = mysql_query($query) or die(mysql_error());

while($row = mysql_fetch_array($result)){


echo '<div class="colmask rightmenu">';
echo '<div class="colleft">';
echo '<div class="col1">';
echo '<p>Here are some example sentences that show how we use the word; </p>';
echo '<div id="small"><i> ' . $row['word_type1'] . '</i></div>';	
echo '<p><div id="small">' . $row['sentence1'] . '</p></div>';
echo '<div id="small"><i> ' . $row['word_type2'] . '</i></div>';
echo '<p><div id="small">' . $row['sentence2'] . '</p></div>';
echo '<div id="small"><i> ' . $row['word_type3'] . '</i></div>';
echo '<p><div id="small">' . $row['sentence3'] . '</p></div>';

}

 

The problem is that every entry on the database is echoed in word.php whereas I would like only the entries for the word selected to appear.

 

Thanks in advance for any help; do say if you need more info.

 

 

[chintansshah]

You can save your search criteria in session. 

$_SESSION['where_caluse'] = " WHERE word LIKE '%".$search."%'";
$result = mysql_query("SELECT word FROM pro_words ".$_SESSION['where_clause']);

 

 

Get session value in word.php file your query looks like

$query = "SELECT word,word_type1,sentence1,word_type2,sentence2,word_type3,sentence3 FROM pro_words".$_SESSION['where_clause']; 

 

[Bazzaah]

Thanks for that.

 

I should have said that the facility can only be used by registered users - there's already a session variable in existence for the user; where does that leave the session variable you create?

 

I popped that code you kindly provide into the relevant files but doesn't seem to have an effect..I wonder if I'm missing something - is there any significance behind your calling 'where_clause'?

 

Thanks

[Bazzaah]

Thanks again - it does work but only if there is one search result. If there are two results, it displays two results  on word.php.

 

How do I get it so that, if there are multiple results displayed, then only the chosen entry is displayed on word.php?

 

 

 

[chintansshah]

have you written session_start(); at first line of the php file?

 

If you write on both the files then debug $_SESSION variable.

[Bazzaah]

Yes the files have session start().

 

How do i debug the session variable? Sorry, I'm a bit of a noob with this...

[chintansshah]

write below code after session_start();

 

<?php

session_start();

 

echo "<pre>";

print_r($_SESSION);

echo "</pre>";

?>

 

After that, you can find out session values on screen.

[Bazzaah]

I ran that and no errors, it just returned this;

 

Array

(

    [user_id] => 1

    [first_name] => Myname

    [user_level] => 0

    [where_clause] =>  WHERE word LIKE '%e%'

)

 

which is what I'd expect; no errors though.