PHP Paging Display Database related issues, there is code, analysis and analysis

PHP Paging Display database problems, there is code, analysis and analysis
A warning prompt appears the first time you open it. The reason is undefined index:page
How to solve the problem of the transfer of value AH?

The code is as follows


Include ("conn.php");
$perNumber = 10; The number of records displayed per page
$page =$_get[' page '; Get the current page value
$count =mysqli_query ($conn, "SELECT COUNT (*) from user"); Total Records obtained
$rs =mysqli_fetch_array ($count);
$totalNumber = $rs [0];
$totalPage =ceil ($totalNumber/$perNumber); Calculate the total number of pages
if (!isset ($page)) {
$page = 1;
}//If there is no value, assign a value of 1
$startCount = ($page-1) * $PERNUMBER; Start with pagination and calculate the starting record based on this method
$result =mysqli_query ($conn, "select * from User limit $startCount, $perNumber"); Based on the previous calculation, the number of records and records started
while ($row =mysqli_fetch_array ($result)) {
echo "user_id:". $row [' user_id ']. " ";
echo "user_name:". $row [' user_name ']. "
"; Displaying the contents of a database
}
if ($page! = 1) {//pages not equal to 1
?>
"> Previous page
}
for ($i =1; $i <= $totalPage; $i + +) {//Cycle Show page
?>
">
}
if ($page < $totalPage) {//If page is less than total pages, show next link
?>
"> Next page
}
?>






------to solve the idea----------------------
$page =$_get[' page '; Get the current page value
Switch
$page =isset ($_get[' page ')? $_get[' page ': 1; Get the current page value
------to solve the idea----------------------
$page = @$_get[' page ']; Get the current page value

Change into
$page =isset ($_get[' page ')? $_get[' page ': 1; Get the current page value
Yes, but
if (!isset ($page)) {
$page = 1;
}//If there is no value, assign a value of 1
It's superfluous.

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