Mysql_query help

baris22 · 02-16-2007, 02:58 PM

Hello,
????: NamePros.com http://www.namepros.com/programming/295084-mysql_query-help.html

I have got a field in mysql. There is <br> after each entry.
The first entry is a link for an image all the time.

So in the row it looks like this

http://www.google.com/a.gif<br>http://www.yahoo.com<br>http://www.aa.com<br>

Ofcourse the links changes for every row. But the first entry is an image file with bmp, gif and jpg extantion at all the time.

Is it possible to disregard the first data in this row when i display the results.
I mean i do not want to show the first data which is the link of the image.

Instead of showing

http://www.google.com/a.gif
http://www.yahoo.com
http://www.aa.com

in the results, I want to show

http://www.yahoo.com
http://www.aa.com

Thank you all.

Dan · 02-16-2007, 03:01 PM

You couldn't stip out the first part through MySQL, but it's simple with PHP.

Query for whichever rows you want and then

PHP Code:

  // $var is the row/column data

$var = explode('<br>', $var);

????: NamePros.com http://www.namepros.com/showthread.php?t=295084
$var = array_shift($var);

$var = implode('<br>', $var);

Barrucadu · 02-16-2007, 03:02 PM

This is how I would do it, except there is probably a better way.

PHP Code:

  <?php

$row_data = 'blah<br>blah<br>etc<br>';

$i=0;

$row_data2 = explode('<br>', $row_data);

 
while($i < count($row_data2)){

     if($i != 0){

          echo $row_data2[$i] . '<br>';

     }

     $i++;

}

?>

<edit>
????: NamePros.com http://www.namepros.com/showthread.php?t=295084
Dan beat me to it - and taught me a new function, yay! lol

baris22 · 02-16-2007, 03:16 PM

How can i use that code for

$row['links']

Thanks

PHP Code:

  
<?php

ob_start();

session_start();

 
$hostname="x";

$username="x";

$password=" x";

$dbname="x";

 
$con = mysql_pconnect ($hostname,$username,$password) or print "No database connection exists or invalid query<br>".mysql_error();

mysql_select_db ($dbname, $con) or print "Can't select databse <br>".mysql_error();

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

 
 
  $find[] = '1';

  $find[] = '2'; 

  $find[] = '3';

  $find[] = '4';

  $find[] = '5';

  $replace[] = 'audio';

  $replace[] = 'video';

  $replace[] = 'document';

  $replace[] = 'program';

  $replace[] = 'others';

 
 
echo "<table width='100%' border='1' cellspacing='0' cellpadding='16' bordercolor='#000000'>

<tr  >

<th bgcolor='#FFCC00'>Num.</th>

<th bgcolor='#FFCC00'>Type</th>

<th bgcolor='#FFCC00'>Title</th>

<th bgcolor='#FFCC00'>Description</th>

<th bgcolor='#FFCC00'>Links</th>

</tr>";while($row = mysql_fetch_array($result))

  {

  echo "<tr>";

  echo "<td>" . $row['id'] . "</td>";

  echo "<td>" . str_replace($find, $replace, $row['type']) .  "</td>";

  echo "<td>" . $row['title'] . "</td>";

????: NamePros.com http://www.namepros.com/showthread.php?t=295084
????: NamePros.com http://www.namepros.com/showthread.php?t=295084
  echo "<td>" . $row['description'] . "</td>";

  echo "<td nowrap='nowrap'>"  . $row['links'] .  "</td>";

  echo "</tr>";

  }

  

echo "</table>";mysql_close($con);

?>

baris22 · 02-16-2007, 06:00 PM

Originally Posted by Mikor

This is how I would do it, except there is probably a better way.

PHP Code:

  <?php

$row_data = 'blah<br>blah<br>etc<br>';

$i=0;

$row_data2 = explode('<br>', $row_data);

????: NamePros.com http://www.namepros.com/showthread.php?t=295084
 
while($i < count($row_data2)){

     if($i != 0){

          echo $row_data2[$i] . '<br>';

     }

     $i++;

}

?>

<edit>
Dan beat me to it - and taught me a new function, yay! lol

Thanks

It worked.

02-16-2007, 02:58 PM	THREAD STARTER #1 (permalink)
baris22 NamePros Member Join Date: Jan 2007 Posts: 98	Mysql_query help Hello, ????: NamePros.com http://www.namepros.com/programming/295084-mysql_query-help.html I have got a field in mysql. There is <br> after each entry. The first entry is a link for an image all the time. So in the row it looks like this http://www.google.com/a.gif<br>http://www.yahoo.com<br>http://www.aa.com<br> Ofcourse the links changes for every row. But the first entry is an image file with bmp, gif and jpg extantion at all the time. Is it possible to disregard the first data in this row when i display the results. I mean i do not want to show the first data which is the link of the image. Instead of showing http://www.google.com/a.gif http://www.yahoo.com http://www.aa.com in the results, I want to show http://www.yahoo.com http://www.aa.com Thank you all.

02-16-2007, 03:01 PM	#2 (permalink)
Dan Buy my domains. Join Date: Feb 2006 Posts: 2,796	You couldn't stip out the first part through MySQL, but it's simple with PHP. Query for whichever rows you want and then PHP Code: `// $var is the row/column data $var = explode('<br>', $var); ????: NamePros.com http://www.namepros.com/showthread.php?t=295084 $var = array_shift($var); $var = implode('<br>', $var);`

02-16-2007, 03:02 PM	#3 (permalink)
Barrucadu Senior Member Join Date: Aug 2005 Location: East Yorkshire, England Posts: 2,689 Follow @barrucadu	This is how I would do it, except there is probably a better way. PHP Code: `<?php $row_data = 'blah<br>blah<br>etc<br>'; $i=0; $row_data2 = explode('<br>', $row_data); while($i < count($row_data2)){ if($i != 0){ echo $row_data2[$i] . '<br>'; } $i++; } ?>` <edit> ????: NamePros.com http://www.namepros.com/showthread.php?t=295084 Dan beat me to it - and taught me a new function, yay! lol __________________ Website \| Blog Arch Hurd developer; GNU Webmaster; FSF member #8385

02-16-2007, 03:16 PM	THREAD STARTER #4 (permalink)
baris22 NamePros Member Join Date: Jan 2007 Posts: 98	How can i use that code for $row['links'] Thanks PHP Code: <?php ob_start(); session_start(); $hostname="x"; $username="x"; $password=" x"; $dbname="x"; $con = mysql_pconnect ($hostname,$username,$password) or print "No database connection exists or invalid query<br>".mysql_error(); mysql_select_db ($dbname, $con) or print "Can't select databse <br>".mysql_error(); $result = mysql_query("SELECT * FROM articles"); $find[] = '1'; $find[] = '2'; $find[] = '3'; $find[] = '4'; $find[] = '5'; $replace[] = 'audio'; $replace[] = 'video'; $replace[] = 'document'; $replace[] = 'program'; $replace[] = 'others'; echo "<table width='100%' border='1' cellspacing='0' cellpadding='16' bordercolor='#000000'> <tr > <th bgcolor='#FFCC00'>Num.</th> <th bgcolor='#FFCC00'>Type</th> <th bgcolor='#FFCC00'>Title</th> <th bgcolor='#FFCC00'>Description</th> <th bgcolor='#FFCC00'>Links</th> </tr>";while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['id'] . "</td>"; echo "<td>" . str_replace($find, $replace, $row['type']) . "</td>"; echo "<td>" . $row['title'] . "</td>"; ????: NamePros.com http://www.namepros.com/showthread.php?t=295084 ????: NamePros.com http://www.namepros.com/showthread.php?t=295084 echo "<td>" . $row['description'] . "</td>"; echo "<td nowrap='nowrap'>" . $row['links'] . "</td>"; echo "</tr>"; } echo "</table>";mysql_close($con); ?>

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