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contra10

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Posts posted by contra10

    improving my search

    in PHP Coding Help

    Posted

    Hi, I'm trying to improve my search so that it searches all words that i write in the search box and matches keywords

     

    my code right now is

    
<?php
$sub = $_GET['article'];

echo "<h2>search results for: $sub</h2><br><br>";


$sub = str_replace(" ", "%", $sub);

$alb = 	mysql_connect("localhost", "", "") or die(mysql_error());  
mysql_select_db("articles") or die(mysql_error()); 

mysql_set_charset('utf8', $alb); 

$query2 = "SELECT * FROM `blog` WHERE `title` like '%$sub%'";
$result2 = mysql_query($query2) or die(mysql_error());


while ($postede2 = mysql_fetch_assoc($result2))
{
$title = "{$postede2['title']}";
$author = "{$postede2['author']}";
$section = "{$postede2['section']}";
$description = "{$postede2['description']}";
$url = "{$postede2['url']}";

echo "<a href='$url'><h3><b>$title</b></h3></a><br>";
echo "By $author, Category: $section<br>";
echo "$description ...<br><br>";

}
?>

     

    How do i break down the search even further?

    cannot modify header error

    in PHP Coding Help

    Posted

    i'm having trouble working with my search bar, when i used wamp it worked but now online i keep getting this message

     

    Cannot modify header information - headers already sent by (output started at...)

     

    I placed that code for the search bar above the html tags but i keep getting the error here is my coding

     

    			<?php

				mysql_connect("localhost", "user", "pass") or die(mysql_error()); 

				if (isset($_POST['submit'])) {

				$search = mysql_real_escape_string($_POST['search']);
				$type = mysql_real_escape_string($_POST['type']);

				if ($type == true) {

				header("Location:http://www.link.com/searchresults.php?subdirectory=$search&type=$type"); 


				}
				}



				?>

    Get information from a website (i.e. search engine)

    in PHP Coding Help

    Posted

    I don't really understand preg_match when using it, although i understand the concept

     

    i did this as a trial

     

    <?php 

$html = file_get_contents('http://www.google.com/');

preg_match('google', "$html", $matches);

    preg_match_all("<a href", $html, $match);

foreach($match[1] as $val);  

{        

echo $val."<br>";   
    } 

?>

     

    i don't really get how to get the actual a href links from all the links on the page

    Get information from a website (i.e. search engine)

    in PHP Coding Help

    Posted

    Hello, this is more of a question since I don't know where to begin. I want to use a spider to search a website and gather information and then organize it into a table the specific information from the site that I need. For example lets say I wanted to search a sports website and only get the details of the teams and their scores and then place it into my database without getting other information of that website?

    syntax error message because of an apostrophe

    in PHP Coding Help

    Posted

    hello, i keep getting a syntax error message when I try to display my data from my database, heres the code that i use fo veiwing the data

     

    <?php
$alb = mysql_connect("localhost", "root", "") or die(mysql_error()); 
mysql_select_db("programs") or die(mysql_error()); 

mysql_set_charset('utf8', $alb); 

$query = "SELECT * FROM `canpro` WHERE `directory` = '$sub' ORDER BY area, school ASC";
$result = mysql_query($query) or die(mysql_error());

while ($postede = mysql_fetch_assoc($result))
{
$area = "{$postede['area']}";
$dn = "{$postede['dn']}";
$dl = "{$postede['dl']}";

echo "<font size='3'><b>$school</b></font><br>";
echo "<font size='2'><b>$dn</b></font><br><font size='2' color = '#3B5998'><b>$dl</b></font><br>";

mysql_select_db("schools") or die(mysql_error()); 
$query2 = "SELECT * FROM `canschools` WHERE `name` = '$area'";
$result2 = mysql_query($query2) or die(mysql_error());

while ($postede2 = mysql_fetch_assoc($result2))
{
$city = "{$postede2['city']}";
$area2 = "{$postede2['area']}";
echo "<font size='2'><b>$city, $area2</b></font><br><br>";
}
}
?>

     

    The display message i get 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 's Calender'' at line 1

     

    it shows all my information except when the word contains an apostrohpe then it doesn't shows the error and all the information afterwords won't get displayed

    creating a better search

    in PHP Coding Help

    Posted

    sry by better I meant allows more flexibility with the search

     

    I used implode seems to be working well

     

    but for example because my site is a music site when someone searches for

     

    black eyed peas I gotta feeling

     

    the song comes up in search results as

     

    black eyed peas - i gotta feeling

     

    which is what is in the db as the title

    but when someone searches

     

    i gotta feeling black eyed peas

     

    no results comes up

    creating a better search

    in PHP Coding Help

    Posted

    hey I'm trying to create a better search bar because I find LIKE to be very simple

     

    <?php
$max = 'limit ' .($pagenum - 1) * $page_rows .',' .$page_rows;
//This is your query again, the same one... the only difference is we add $max into it
$data_p = mysql_query("SELECT * FROM `musiclinks` WHERE `title` LIKE '%$sr%' $max") or die(mysql_error()); 
if ($data_p){
//This is where you display your query results
while ($postede = mysql_fetch_assoc($data_p))
{
$name = "{$postede['title']}";
$genre = "{$postede['genre']}";
$filename = "{$postede['filename']}";
$embed = "{$postede['embed']}";
$false = 'False';
$true = 'True';
}

?>

     

    how can I make the search a lot better I know I have to work on variable $sr, should i break it down into strings. how should i go about doing that?

    everything echos on first page

    in PHP Coding Help

    Posted

    ok I'm using my pagination script and everything seems to be echoing on the first page and every other page

     

    <?php
//This checks to see if there is a page number. If not, it will set it to page 1 
$pagenum = $_GET['pagenum'];
$linkgenre = $_GET['l'];

if ($pagenum) 
{ 
}
else{
$pagenum = 1;  
} 

//Here we count the number of results 
$querye = "SELECT DISTINCT * FROM `musiclinks` WHERE `genre` ='$linkgenre' AND `embed` = 'True'";
$result = mysql_query($querye);
$rows= mysql_num_rows($result);

//This is the number of results displayed per page 
$page_rows = 5; 
$es = ($rows/$page_rows);
//This tells us the page number of our last page 
$last = ceil($es);

//this makes sure the page number isn't below one, or more than our maximum pages 
if ($pagenum < 1) 
{ 
$pagenum = 1; 
} 
elseif ($pagenum > $last) 
{ 
$pagenum = $last; 
} 
//This sets the range to display in our query 
$max = 'LIMIT ' .($pagenum - 1) * $page_rows .',' .$page_rows;
//This is your query again, the same one... the only difference is we add $max into it
$data_p = mysql_query("SELECT * FROM `musiclinks` WHERE `genre` = '$linkgenre' AND `embed` = 'True' $max"); 
if ($data_p){
//This is where you display your query results
while ($postede = mysql_fetch_assoc($result))
{
$name = "{$postede['title']}";
$genre = "{$postede['genre']}";
$filename = "{$postede['filename']}";

echo "<tr><td width='%50'><a href='http://www.site.com/music/plisten.php?f=$filename'>$name</a></td><td width='%50'>$genre</td>";

$queryear = "SELECT * FROM `ratings` WHERE `songtitle` = '$name'";
$resultear = mysql_query($queryear) or die(mysql_error());
while ($postedear = mysql_fetch_assoc($resultear))
{
$avsong = "{$postedear['songtitle']}";
$linksong = "{$postedear['song']}";
$ratedavg = "{$postedear['average']}";

if ($ratedavg > 1){
echo"<td>$ratedavg / 10</td></tr>";
}else{
echo "<td>No Rating</td></tr>";
}
}
}

// This shows the user what page they are on, and the total number of pages
echo " Page $pagenum of $last ($rows Results)<p>";?>

     

    I don't understand why its not working and the max variable changes when i click on a different page

    passing variables in url

    in PHP Coding Help

    Posted

    hey,

     

    I'm trying to pass a variable in my url and i think im suppose to use slashes for this one because the variable that I'm passing contains an & in it. In this case the variable l=R&B but when the page recieves the link it interprets l as just R because the & sign follows after making it seem as if there is another variable that is supposed to be passed along.

     

    <?php echo "<div class='date'><a href='http://www.site.com/music/plisting.php?l=R&B'>View All ($rnbcountp)</a></div>"; ?>

    php email problems

    in PHP Coding Help

    Posted

    I'm having trouble with emails and php when it sends to a hotmail account everything shows but in gmail or yahoo accout it shows a download link for a no name file, the actual writing doesn't show

     

    <?php
$to = ($email);
$subject = "Verification";
//create a boundary string. It must be unique 
//so we use the MD5 algorithm to generate a random hash
$random_hash = md5(date('r', time()));
$from = "admin@url.com";
$headers = "From: $from";
//add boundary string and mime type specification
$headers .= "\r\nContent-Type: multipart/alternative; boundary=\"PHP-alt-".$random_hash."\""; 
//define the body of the message.
ob_start(); //Turn on output buffering
?>

--PHP-alt-<?php echo $random_hash; ?>  
Content-Type: text/html; charset="iso-8859-1" 
Content-Transfer-Encoding: 7bit


<div style="width:800px; 
height:400px; 
border:4px dashed grey; 
">
<table bgcolor="black" width="800" >

<tr><td>
<img src="http://www.url.com/images/logo.png"/><br><br><font size="12" color="white">You have sucessfully become a member</font><br>

<?php
echo"<p align='center'>";
echo"<table background='#6D6D6D' width='700' height='250'>";
echo "<tr><td><b><font color='white'>Dear, $usere <br><br> 
To complete your activation please click on the \"Verify\" link below.";
echo "<b>To contine to verify your account please follow the link </b><a href='http://www.url.com/verify.php?v=$usere' target='_blank'><font color='red'>Verify Account</font></a>";
echo"</font></td></tr>";
echo "</table>";
echo"</p>";
?>
 </td></tr>
<tr><td align="center"><font size="1" color="white">Copyright 2009</font></td></tr>
</table>
</div>

--PHP-alt-<?php echo $random_hash; ?>--
<?php
//copy current buffer contents into $message variable and delete current output buffer
$message = ob_get_clean();
//send the email
$mail_sent = @mail( "$to", $subject, $message, $headers );

?>

    checking a website for a link - link directories

    in PHP Coding Help

    Posted

    I'm interested in creating a directory and as a result the users who post their link have to backlink back to the directory. I can tell users to backlink but is there a way where I can check the website to see if my link exists at any given time...I was thinking of using something like ob_get_contents i think thats the function to copy a page.

     

    What approach should i take?

    can't echo out else statement

    in PHP Coding Help

    Posted

    i see what your saying

    in this case one of my $ratedavg = 5

     

    5 > 1 so it would show the result but if there is no rating and the $ratedavg is equal to nothing shouldn't it show. Or would I have to put something like

     

    <?php
if ($ratedavg == true){
echo"<td>$ratedavg / 10</td></tr>";
}else{
echo "<td>No Rating</td></tr>";
}
?>

     

    can't echo out else statement

    in PHP Coding Help

    Posted

    I'm trying to echo out a statement if somethin gis untrue but its not working

    <?php
$queryear = "SELECT * FROM `ratings` WHERE `songtitle` = '$name'";
$resultear = mysql_query($queryear) or die(mysql_error());
while ($postedear = mysql_fetch_assoc($resultear))
{
$avsong = "{$postedear['songtitle']}";
$linksong = "{$postedear['song']}";
$ratedavg = "{$postedear['average']}";

if ($ratedavg > 1){
echo"<td>$ratedavg / 10</td></tr>";
}else{
echo "<td>No Rating</td></tr>";
}
}
?>

    THe average shows for those that have been rated but the other statement is suppose to show for those that havn't but it doesn't[/code]

    mix php with java

    in PHP Coding Help

    Posted

    hi I'm trying to put a javascript in my php tags but it doesn't seem to echo anything out

     

    <?php
$querye = "SELECT * FROM `musiclinks`";
$resulte = mysql_query($querye) or die(mysql_error());

while ($postede = mysql_fetch_assoc($resulte))
{
$filename = "{$postede['filename']}";
$date = "{$postede['date']}";
$link = "{$postede['link']}";
$type = "{$postede['type']}";

	echo"<p id='audioplayer_1'>$date</p>";
        echo"<script type='text/javascript'>"; 
        echo"AudioPlayer.embed('audioplayer_1', {soundFile: 'http://www.website.com/songs/$filename'});";   
        echo"</script>";

echo"<tr><td width='%50'><a href=''></a></td><td width='%50'></td></tr>";

}
?>

     

     

     

    I also tried to change it to this thinking it was the quotations:

     

     

    <?php
$querye = "SELECT * FROM `musiclinks`";
$resulte = mysql_query($querye) or die(mysql_error());

while ($postede = mysql_fetch_assoc($resulte))
{
$filename = "{$postede['filename']}";
$date = "{$postede['date']}";
$link = "{$postede['link']}";
$type = "{$postede['type']}";

	echo"<p id='audioplayer_1'>$type</p>";
        echo"<script type=\"text/javascript\">\n"; 
        echo"AudioPlayer.embed(\"audioplayer_1\", {soundFile: \"http://www.website.com/songs/$filename\"});\n";   
        echo"</script>";

echo"<tr><td width='%50'><a href=''></a></td><td width='%50'></td></tr>";

}
?>

    [SOLVED] mix php with java

    in Applications

    Posted

    hi I'm trying to put a javascript in my php tags but it doesn't seem to echo anything out

     

    <?php
$querye = "SELECT * FROM `musiclinks`";
$resulte = mysql_query($querye) or die(mysql_error());

while ($postede = mysql_fetch_assoc($resulte))
{
$filename = "{$postede['filename']}";
$date = "{$postede['date']}";
$link = "{$postede['link']}";
$type = "{$postede['type']}";

	echo"<p id='audioplayer_1'>$date</p>";
        echo"<script type='text/javascript'>"; 
        echo"AudioPlayer.embed('audioplayer_1', {soundFile: 'http://www.website.com/songs/$filename'});";   
        echo"</script>";

echo"<tr><td width='%50'><a href=''></a></td><td width='%50'></td></tr>";

}
?>

     

    I also tried to change it to this thinking it was the quotations:

     

    <?php
$querye = "SELECT * FROM `musiclinks`";
$resulte = mysql_query($querye) or die(mysql_error());

while ($postede = mysql_fetch_assoc($resulte))
{
$filename = "{$postede['filename']}";
$date = "{$postede['date']}";
$link = "{$postede['link']}";
$type = "{$postede['type']}";

	echo"<p id='audioplayer_1'>$type</p>";
        echo"<script type=\"text/javascript\">\n"; 
        echo"AudioPlayer.embed(\"audioplayer_1\", {soundFile: \"http://www.website.com/songs/$filename\"});\n";   
        echo"</script>";

echo"<tr><td width='%50'><a href=''></a></td><td width='%50'></td></tr>";

}
?>

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