[pkcidstudio]

What I am trying to do is, call a row from my table and allow there to be numbers to select different rows, or the user can hit next. but for some reason I am not able to get the information to appear, and am reciving this error.

Code:

Parse error: syntax error, unexpected ';' in /misc/20/103/291/210/0/user/web/uorf.org/new/example.php on line 18

and here is my code. any thoughts? please assist.

PHP Code:

<?php
include("_php/config.php");
// database connection stuff here
$rows_per_page = 1;
$sql = "SELECT * FROM sponsor WHERE checkSponsorChild='no'";
$result = mysql_query($sql);
$total_records = mysql_num_rows($result);
$pages = ceil($total_records / $rows_per_page);
//mysql_free_result($result);
if (!isset($screen))
  $screen = 0;
$start = $screen * $rows_per_page;
//$sql = "SELECT * FROM sponsor ";
$sql .= "LIMIT $start, $rows_per_page";
$result = mysql_query($sql);
$rows = mysql_num_rows($result);
while ($i = 0; $i < $rows; $i++) {
  $childrenInfo = mysql_fetch_array($result, $i, 0);
  //$description = mysql_result($result, $i, 0);
  echo "\t\t". $childrenInfo['childsName'] ."\r\n";
  //echo "\$description = $description<br>\n";
mysql_free_result($result);
}
echo "<p><hr></p>\n";
// let's create the dynamic links now
if ($screen > 0) {
  $url = "example.php?screen=" . $screen - 1;
  echo "<a href=\"$url\">Previous</a>\n";
}
// page numbering links now
for ($i = 0; $i < $pages; $i++) {
  $url = "example.php?screen=" . $i;
  echo " | <a href=\"$url\">$i</a> | ";
}
if ($screen < $pages) {
  $url = "example.php?screen=" . $screen + 1;
  echo "<a href=\"$url\">Next</a>\n";
}
?>

I think I am probolly just misusing a function, and or missing a ";" thankx agian for your help.

[YourPHPPro]

You use 'while' on line 18 when I think you mean 'for'

http://us2.php.net/while

http://us2.php.net/manual/en/control-structures.for.php

[pkcidstudio]

Originally Posted by YourPHPPro

You use 'while' on line 18 when I think you mean 'for'

YourPHPPro, when using "for" I only get 1 letter or 1 number for instance; say the word "frenchfry" is in my databse, when using "for" I only recive "f", instead of "frenchfry". any thoughts? maybe I am using the wrong call to with the "for". thank you for your help, I look to here what you or anyone else has to say.

[YourPHPPro]

Well, the original problem - parse error - was solved, but for people to assist with your new question, it would be good for you to - at least - list the schema for the databases/tables and explain in detail what you want the program to do...

[Nico]

EDIT:

A bit too late...

Try,

PHP Code:

while ($childrenInfo = mysql_fetch_array($result))

{

  echo "\t\t". $childrenInfo['childsName'] ."\r\n";

  //echo "\$description = $description<br>\n";

}



mysql_free_result($result); 

[pkcidstudio]

What the script is suppose to do is display the data, similar to the way Nico used the while script, (that part now works Thank you nico) but also with the "for" section should allow the user to go forward and backword in the database.

Details:

This site is for an orphange, and it will have probolly 100 kids in it at a time.
Table = sponsor
row in table = childsName
row in table = childsAge
...

I need the ability for a user to cycle through the orphans on the database, which is where the

PHP Code:

for($i = 0; $i < $rows; $i++) { 

I was trying to use this to allow the user to go to next or previus in the databse.

This is how i was trying to do the next and previus buttons.

PHP Code:

 
if ($screen > 0) {
 
$url = "example.php?screen=" . $screen - 1;
 
echo "<a href=\"$url\">Previous</a>\n";
 
}
 
// page numbering links now
 
for ($i = 0; $i < $pages; $i++) {
 
$url = "example.php?screen=" . $i;
 
echo " | <a href=\"$url\">$i</a> | ";
 
}
 
if ($screen < $pages) {
 
$url = "example.php?screen=" . $screen + 1;
 
echo "<a href=\"$url\">Next</a>\n";
 
} 

Hope this helps to explain what I am trying to do., thanks for the help agian.

[YourPHPPro]

Here are a few sites that might help:

http://www.xlinesoft.com/phprunner/
http://www.scriptcase.net/siteV2/index/index.php
http://www.hkvstore.com/phpmaker/
http://www.phpmyedit.org/
http://www.yessoftware.com/products/...p?product_id=1

[pkcidstudio]

Thanks for the sites, but would rather not get into a third party. Thank you though.

[pkcidstudio]

Has anyone ever done this with out a third party? If so, If you could help it would be much appreciated. Thank you in advance.

[pkcidstudio]

So I got it to display the kids names, thank you to all really am thankful.
Now I am having trouble with the next button working, it doesnt seem to be grabing the next child in line. any thoughts? also the previous is not showing up? Thank you in advance.

PHP Code:

<?php
include("_php/config.php");
// database connection stuff here
$rows_per_page = 1;
$sql = "SELECT * FROM sponsor WHERE checkSponsorChild='no'";
$result = mysql_query($sql);
$total_records = mysql_num_rows($result);
$pages = ceil($total_records / $rows_per_page);
//mysql_free_result($result);
if (!isset($screen))
  $screen = 0;
$start = $screen * $rows_per_page;
//$sql = "SELECT * FROM sponsor ";
$sql .= "LIMIT $start, $rows_per_page";
$result = mysql_query($sql);
$rows = mysql_num_rows($result);
//this might be gone
//for ($i = 0; $i < $rows; $i++) {
//this might be gone

//test
$i= 0;
while ($i < $rows){
//test
//reset ($childrenInfo);
 while ($childrenInfo = mysql_fetch_array($result)) 
 //$description = mysql_result($result, $i, 0);
{ 
  echo "\t\t". $childrenInfo['childsName'] ."\r\n"; 
  //echo "\$description = $description<br>\n"; 
} 
$i++;
//return $children;

//$i++; try this if above doesnt work
}
//test

mysql_free_result($result); 

echo "<p><hr></p>\n";
// let's create the dynamic links now
if ($screen > 0) {
  $url = "example.php?screen=" . $screen - 1;
  echo "<a href=\"$url\">Previous</a>\n";
}
// page numbering links now
/*for ($i = 0; $i < $pages; $i++) {
  $url = "example.php?screen=" . $i;
  echo " | <a href=\"$url\">$i</a> | ";
}*/
if ($screen < $pages) {
  $url = "example.php?screen=" . $screen + 1;
  echo "<a href=\"$url\">Next</a>\n";
}
?>

What should be happening: when user clicks next it selects the next child in line. same with previous, selects the past child. Thank you for your help agian.