我有一个脚本，该脚本在其中使用了$_PHP_SELF常量错误的页面中显示SQL命令“ SELECT”的结果：
<html>
<head>
<title>Paging Using PHP</title>
</head>
<body>
<?php
$dbhost = '127.0.0.1';
$dbuser = 'guest';
$dbpass = 'guest123';
$rec_limit = 10;
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn ) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db('test_db');
/* Get total number of records */
$sql = "SELECT count(emp_id) FROM employee ";
$retval = mysql_query( $sql, $conn );
if(! $retval ) {
die('Could not get data: ' . mysql_error());
}
$row = mysql_fetch_array($retval, MYSQL_NUM );
$rec_count = $row[0];
if( isset($_GET{'page'} ) ) {
$page = $_GET{'page'} + 1;
$offset = $rec_limit * $page ;
}else {
$page = 0;
$offset = 0;
}
$left_rec = $rec_count -($page * $rec_limit);
$sql = "SELECT emp_id, emp_name, emp_salary ".
"FROM employee "."LIMIT $offset, $rec_limit";
$retval = mysql_query( $sql, $conn );
if(! $retval ) {
die('Could not get data: ' . mysql_error());
}
while($row = mysql_fetch_array($retval, MYSQL_ASSOC)) {
echo "EMP ID :{$row['emp_id']} <br> ".
"EMP NAME : {$row['emp_name']} <br> ".
"EMP SALARY : {$row['emp_salary']} <br> ".
"--------------------------------<br>";
}
if( $page > 0 ) {
$last = $page -2;
echo "<a href = \"$_PHP_SELF?page = $last\">Last 10 Records</a> |";
echo "<a href = \"$_PHP_SELF?page = $page\">Next 10 Records</a>";
}else if( $page == 0 ) {
echo "<a href = \"$_PHP_SELF?page = $page\">Next 10 Records</a>";
}else if( $left_rec < $rec_limit ) {
$last = $page -2;
echo "<a href = \"$_PHP_SELF?page = $last\">Last 10 Records</a>";
}
mysql_close($conn);
?>
</body>
我已经看到了很多答案，这些答案对我帮助不大，例如：
$PHP_SELF = &$_SERVER['PHP_SELF'];
echo $PHP_SELF;
如何使脚本正常工作？
每页应该只显示10个结果，并通过单击页面底部的超链接转到下一个10个结果的下一页。
最佳答案
if(isset($_GET["page"]))
{
$page=$_GET["page"];
$back=$page-5;
}
else
{
$page=0;
$back=0;
}
$forward=$page+5;
while($page<=$back and $page>=$forward)
{
//code to execute
}
这样的事情行吗？
echo '<a href="?=' . $page+10 . '">next</a>';