PHP - Colums And Links

hello,

normally when i query mysql, it returns my results line by line.

im trying to make a print view now that will put 3 results in a row before going to the next row.

any help appreciated.

I have been working on a way to split my data into 2 columns as a way to make the page shorter and easier to read. I have been able get it to split like this

1 2
3 4
5 6

But I would rather it go like this, because it would be easier on the eyes.

1 4
2 5
3 6

This is what I have now, I have no idea if I'm even on the right track.
Code: [Select]
  <?php
include"scripts/connect.php" ;

mysql_connect('localhost',$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM movies WHERE type LIKE 'tv' ORDER BY title";
$result=mysql_query($query);

$num=mysql_numrows($result);

mysql_close();
?>
<table width="70%" align="left">
 
<?php
$i=0;
while ($i < $num/2)
 {

$title=mysql_result($result,$i,"title");

?>
<tr>
<td><a class="nav" href=/show.php?title=<?php echo urlencode($title); ?>><?php echo $title; ?></td>



<?php
while ($i >= $num/2)
 {
?>

<td><a class="nav" href=/show.php?title=<?php echo urlencode($title); ?>><?php echo $title; ?></td>
</tr>


<?php
}
$i++;

}

?>
</table>

This will show only the first half of the titles.

Thanks in advance.

I am pulling some info from a mysql table, checking some logic, and displaying in on a page. The way this is written now it shows all of the items in a one column table, I simply want to break it up into multiple columns (ideally by specifying say 20 items per column). I know I need a counter and a loop but I'm just not sure how it all fits in with my existing program. Any help would be greatly appreciated.

Here is my code
Code: [Select]
$result2 = mysql_query("SELECT * FROM propertyids ORDER BY propertyname ASC");
while($row2 = mysql_fetch_array($result2)){
${'resp'.$row2['id']} = curl_multi_getcontent(${'ch'.$row2['id']});
${'status'.$row2['id']} = strpos(${'resp'.$row2['id']},$httpvar);
if(${'status'.$row2['id']} === false) {
echo "<tr><td>";
echo "<img src='images/down.png'> <a href='" . $row2['primaryurl'] . "' target='_blank'>" . $row2['propertyname'] . "</a>";
echo "</td></tr>";
}
else {
echo "<tr><td>";
echo "<img src='images/up.png'> <a href='" . $row2['primaryurl'] . "' target='_blank'>" . $row2['propertyname'] . "</a>";
echo "</td></tr>";
}
}
echo "</td></table>";

I am using the following code to generate a dynamic drop down menu from a column in table1(name) - and then inserting the selected option into table2.
What I am trying to do is, select data from 2 columns from table1(name,age) and then store and insert both of those values into table2

I am hoping someone can help me out or at least point me in the rigth direction. ..I would like to insert into table 2 the age that matches the person selected by the drop down menu from table1.

For example - if from the drop down menu I choose John - when the form is sumbitted i want to insert into table2 John's name and age (based on what is stored in table1, in this case John,22)

...Any ideas???

table1
id|name|age|score
01-john-22-1547
02-jane-22-1245

table2
id|county|name|age


the dynamic drop down is generated from table1 (name) with following code

Code: [Select]
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">

<table width="100%" border="0" cellpadding="5" cellspacing="0">

 

<?php

require_once('sql.php');

$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);



$query="SELECT name FROM table1  ORDER BY name ASC";

$result = mysqli_query ($dbc,$query) or die(mysqli_error());



$dropdown = "<select name='name'>";

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

$dropdown .= "\r\n<option value='{$row['name']}'>{$row['name']}</option>";

}

$dropdown .= "\r\n</select>";

echo $dropdown;

?>



      <tr>

          <td height="50" align="right">

                <label for="scheduled_time">Country</label><br/>

</td>

          <td align="left">

<input type="text" id="country" name="country" class="input" maxlength="30"" />

                </td>

      </table>

      <p></p>

    <input type="submit" value="Add Info" name="submit" />

</form>

I am inserting into table2 with the following code

Code: [Select]
<?php
$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);

if (isset($_POST['submit'])) {

$county = mysqli_real_escape_string($dbc, trim($_POST['country']));

$name = mysqli_real_escape_string($dbc, trim($_POST['name']));

$query = "INSERT INTO table2 (country, name) VALUES ('$country', '$name'')";
$result = mysqli_query($dbc, $query);

if (!$result) {

printf("query error : <br/> %s\n", mysqli_error($dbc));

}



if ($result) {

echo 'Success';

}



      // close dbc

      mysqli_close($dbc);

      exit();

      }

?>

hirealimo.com.au/code1.php

this works as i want it:
Quote

SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1



but apparelty selecting * is not a good thing????

but if I do this:
Quote

SELECT priceID, price FROM price INNER JOIN vehicle....etc


it works but i lose the info from the vehicle table.

but how do i make this work:
Quote

SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc


so that i am specifiying which colums from which tables to query??

thanks

<html>
<?php




$id = $_GET['id'];

$dbusername="web148-matt";
   $dbpassword="matt";
   $dbdatabase="web148-matt";
   mysql_connect(localhost,$dbusername,$dbpassword);
   @mysql_select_db($dbdatabase) or die( "Unable to select database");      
   
mysql_query("UPDATE count SET clicks=clicks+1 WHERE id='$id'");

$sql = mysql_query("SELECT link FROM count WHERE id='$id'");
$fetch = mysql_fetch_row($sql);


$result = mysql_query("SELECT * FROM count");

while($row = mysql_fetch_array($result))
  {
 
  echo "<a href=" .$row['link'].   ">Link</a>";

  }


?>


<a href='http://www.google.com'>Google</a>
<a href='/index.php?id=2'>link2</a>
</html>

So I have been working on my website for a while which all is php&mysql based, now working on the social networking part building in similar functions like Facebook has. I encountered a difficulty with getting information back from a link. I've checked several sources how it is possible, with title 'Facebook Like URL data Extract Using jQuery PHP and Ajax' was the most popular answer, I get the scripts but all of these scripts work with html links only. My site all with php extensions and copy&paste my site links into these demos do not return anything . I checked the code and all of them using file_get_contents(), parsing through the html file so if i pass 'filename.php' it returns nothing supposing that php has not processed yet and the function gets the content of the php script with no data of course. So my question is that how it is possible to extract data from a link with php extension (on Facebook it works) or how to get php file executed for file_get_contents() to get back the html? here is the link with code&demo iamusing: http://www.sanwebe.c...-php-and-jquery thanks in advance.

Hi... I am really new to this. Please help me with this...  as part of a gallery, I have categories and a thumbnail. As of now the title/link to the gallery shows up on the side of the thumbnail. How can I place the link below the thumbnail?

Here is the code    Code: [Select]
$result_array[] = "<a href='viewgallery.php?cid=".$row[0]."'>".$row[1]."<img src='/photos/categoryimages/category".$row[0].".jpg' border=0 height='133px' width='200px' </a>";
Thank you for your help... 

I am trying to create a link with an ID from one of my MySQL records.

$data = mysql_query("SELECT * FROM USERS");
$info = mysql_fetch_array($data);
echo "<a href=a.php?'$info['id']'>hello</a>";

is what I am attempting, but this does not work. Any  ideas?

ok so im trying to make a news website and i want to be able to include links inside me headline like most news website they talk about the stuff and include some links along the way how would i manage to do that. I would put all my story in the Release textarea

Code: [Select]
<form class="createheadline" action="createrelease.php" method="post" enctype="multipart/form-data">
<table>
<tr>
<td></td>
<td><?php echo $msg;?></td>
</tr>
<tr>
<td>Author:</td>
<td><input type="text" name="author" size="30"/></td>
</tr>
<tr>
<td>Date:</td>
<td><input type="text" name="date" /></td>
</tr>
<tr>
<td>Title:</td>
<td><input type="text" name="title" size="30"/></td>
</tr>
<tr>
<td>Main Picture</td>
<td><input type="file" name="picture" /></td>
</tr>
<tr>
<td>Release:</td>
<td><textarea name="release" cols="75" rows="27"></textarea></td>
</tr>
<tr>
<td></td>
<td><input type="submit" name="submitbtn" /></td>
</tr>
</table>
</form>

With my custom CMS I'm developing I have a menu structures type module where the admin can create a menu structures for their website. Inside the db is the links table and I'm trying to figure out what I can do if the link has an actual image. I created a field called link_image however I just thought of what if the link is apart of a sprite which would have most of the time 3 parts of the image (active, hover, normal) or whatever or has its own sprite. How should I handle this if I want to store this information in the database.

Hi all,

Im sorry if this is in the wrong section.


But nonetheless, I use php to display my news:
I.e - news.php?id=1 that displays all the news on that id.



What I want to know is using php how can I get the URL from news.php?id=1 too something like /news/titleofthestory ?

Hope I've given enough detail.

Thanks!!

In the MySQL help forum someone asked how to list out all the files in a directory. Another person gave the following code: Code: [Select]
<?php

$handle = opendir('PUT NAME OF YOUR DIR HERE');
while (false !== ($file = readdir($handle)))
{
    if ($file != "." && $file != "..")
{
        echo $file,'<br />';
    }
}
closedir($handle);
?>
I found this to be very helpful! However, is there a way to modify it so that if there are subdirectories, it will list those as hyperlinks so the code would run off of the subdirectory if you were to click that link?

When I asked that question in the MySQL forum, I was told that it had nothing to do with PHP which I know. So that's why I am posting it here. Can anyone help?

I have some links created dynamicaly with php. But i have the following problem. When i am in the root folder ieg. localhost/myPage/index.php the links work fine because they link in a file named entry.php. but when i am in a folder ieg. localhost/myPage/users/authors.php and i click the links they try to go to  localhost/myPage/users/entry.php and i want them to link not inside the folder but to the root folder localhost/myPage/entry.php.

Hi! I have an odd problem... I inserted a flash file in a php doc, and basically what happens is that on some computers, the link does not work without the www extension ... by that i mean if people go to http://mysite.com they will not be able to click the banner but if they put http://www.mysite.com it works...
Very odd as the link is part of the flash movie and it only happens on some computers not all ... anyone would have an idea as to what could cause this ?

Alternativly  what can I do with this piece of code that does not seem to do the trick either

Code: [Select]
<div class="subtitulo"><a href="http://hostelsuites.com/hostel-banner.php?Buenos+Aires+Hostel+Suites+Florida&id_hostel=30&idioma=3&id_banner=2"><?php echo ucfirst($lang["expanish"]) ?></a><br />
<br/></div>
Many thanks for any help !