Write PHP Paging Code mysql_fetch_arry () just can't execute

PHP Paging MySQL

/*
* Created on 2013-7-30
*
* To change the template for this generated file go to
* Window-preferences-phpeclipse-php-code Templates
*/

$page = $_get[page];
if (!function_exists (Pageft)) {
function Pageft ($totle, $displaypg =20, $url = ") {
Define several global variables:
$page: current page number;
$firstcount: The starting item for the (database tutorial) query;
$pagenav: the page navigation bar code, inside the function does not output it;
$_server: Read this page url "$_server[" Request_uri "]" must.
Global $page, $firstcount, $pagenav, $_server;
$globals ["Displaypg"]= $displaypg;
if (! $page) $page = 1;
If you use the default, or null value, for the value of the page, the URL is assigned:
if (! $url) {$url =$_server["Request_uri"];}

URL Analysis:

$parse _url=parse_url ($url);
$url _query= $parse _url["Query"]; Query string to take out the URL separately
if ($url _query) {
$url _query=preg_replace ("/(^|&) page= $page/", "", $url _query);
Replace the query string for the processed URL with the query string for the original URL:
$url =str_replace ($parse _url["Query"], $url _query, $url);
Add page query information after URL, but to assign value:
if ($url _query) $url. = "&page"; else $url. = "page";
}else {
$url. = "? page";
}

Page Count:

$LASTPG =ceil ($totle/$DISPLAYPG); Last page, also the total number of pages
$page =min ($LASTPG, $page);
$PREPG = $page-1; Previous page
$NEXTPG = ($page = = $lastpg? 0: $page + 1); Next page
$firstcount = ($page-1) * $DISPLAYPG;

Start Page navigation bar code:

$pagenav = "Show ". ($totle? ($firstcount + 1): 0). "- ". Min ($firstcount + $DISPLAYPG, $totle)."Records, a total of $totle records ";
If only one page jumps out of the function:
if ($lastpg <=1) return false;
$pagenav. = "Home";
if ($PREPG)
$pagenav. = "front page";
else $pagenav. = "front page";
if ($NEXTPG)
$pagenav. = "Back Page";
else $pagenav. = "Back Page";
$pagenav. = "Last";

Drop-down Jump List, looping through all page numbers:
$pagenav. = "To \ n "; for ($i =1; $i <= $lastpg; $i + +) {if ($i = = $page) $pagenav. = "$i \ n "; else $pagenav. = "$i \ n "; } $pagenav. = "Page, total $lastpg page ";
}
}


Include ("conn.php");

$result =mysql_query ("SELECT * from ' Test ');
$total =mysql_num_rows ($result);
Call Pageft (), displaying 10 messages per page (this parameter can be omitted using the default 20 o'clock), using the page URL (default, so omitted).
Pageft ($total, 3);
Echo $pagenav;

$result = "SELECT * from ' Test ' limit $firstcount, $displaypg";
echo $result;

$query =mysql_query ($result);
while ($row =mysql_fetch_array ($query)) {

echo " ". $row [name]." | ". $row [Sex];

}


?>

Reply to discussion (solution)

May be a problem with SQL, it is recommended that you post an error message for analysis.
Echo Mysql_error (); can see what

It's so complicated ... It's OK to replace the $firstcount with the parameters after get the link.

The output SQL statement. Then copy the phpMyAdmin inside, and if there is an error, it will be reported directly.

echo $result;
There's no problem with this.
Description SQL Statement no problem.

This error means you didn't find anything.

If I find out, I won't tell you: The parameter you gave is not a resource but a Boolean value

Check SQL statements

SELECT * from ' test ' limit 0,

There is a problem here, the paging length value is not loaded in

$DISPLAYPG =20

It should be set as a global variable and declared before your function definition.

$DISPLAYPG =20
It should be set as a global variable and declared before your function definition.


For this DISPLAYPG variable i define it as a global variable before it is placed in function and it seems useless. I think there are two problems with this code, one is that the paging length value is not loaded, and the second is that the SQL statement did not execute successfully, I was just in touch with PHP soon, ask the great God detailed guidance.
______3q!

You have $globals in your function ["DISPLAYPG"]= $displaypg;
The intention is to save the number of rows per page for subsequent use
But you didn't notice that the variable names in PHP are case-sensitive.
$GLOBALS ["Displaypg"]= $displaypg;
is what you need.

There's another place.
$page = $_get[page];
should be for
$page = $_get[page];
Otherwise it will not jump page

You have $globals in your function ["DISPLAYPG"]= $displaypg;
The intention is to save the number of rows per page for subsequent use
But you didn't notice that the variable names in PHP are case-sensitive.
$GLOBALS ["Displaypg"]= $displaypg;
is what you need.

There's another place.
$page = $_get[page];
should be for
$page = $_get[page];
Otherwise it will not jump page


Thank you, that's all.

