Question

请指导我如何在每个页面中继续进行结果编号，直到最后一个数据。目前在第一页中发生的编号为1到10，但是一旦在下一页上移动，则该编号从1到10再次重复，应该是11到20。

我收到了这段代码并且只是对其进行了自定义Paginate result set having written the query with prepared statements,

<?php
// Script and tutorial written by Adam Khoury @ developphp.com
// Line by line explanation : youtube.com/watch?v=T2QFNu_mivw

$myConnection = new mysqli ($servername, $username, $password, $dbname);

// This first query is just to get the total count of rows
$stmt=$myConnection->prepare('SELECT COUNT(id) FROM questions');
// Don't use bind_result()...
// execute your statement
$stmt->execute();
// Get result set into a MySQLi result resource
$result = $stmt->bind_result($id);

// array to hold all rows
$rows = array();

// All results bound to output vars
while ($stmt->fetch()) {
  // Append an array containing your result vars onto the rowset array
  $rows[] = array(
    'id' => $id
  );
}
  $rows=$id;


// This is the number of results we want displayed per page
$page_rows = 10;
// This tells us the page number of our last page
$last = ceil($rows/$page_rows);
// This makes sure $last cannot be less than 1
if($last < 1){
    $last = 1;
}
// Establish the $pagenum variable
$pagenum = 1;
// Get pagenum from URL vars if it is present, else it is = 1
if(isset($_GET['pn'])){
    $pagenum = preg_replace('#[^0-9]#', '', $_GET['pn']);
}
// This makes sure the page number isn't below 1, or more than our $last page
if ($pagenum < 1) { 
    $pagenum = 1; 
} else if ($pagenum > $last) { 
    $pagenum = $last; 
}
$dynamicList = "";
$stmt = $myConnection->prepare('SELECT id, question, ans1, ans2, ans3, ans4 FROM questions ORDER BY id LIMIT ?,? ');
$begin= ($pagenum - 1) * $page_rows;
$end= $page_rows;
$stmt->bind_param('ii',$begin,$end);
$stmt->execute();
        /* store result */
        $stmt->store_result();
        /* get the row count */
        $count = $stmt->num_rows;
        if ($count >= 1) {
            $stmt->bind_result($id, $question, $ans1, $ans2, $ans3, $ans4);
// This shows the user what page they are on, and the total number of pages
$textline1 = "Questions (<b>$rows</b>)";
$textline2 = "Page <b>$pagenum</b> of <b>$last</b>";
// Establish the $paginationCtrls variable
$paginationCtrls = '';
// If there is more than 1 page worth of results
if($last != 1){
    /* First we check if we are on page one. If we are then we don't need a link to 
       the previous page or the first page so we do nothing. If we aren't then we
       generate links to the first page, and to the previous page. */
    if ($pagenum > 1) {
        $previous = $pagenum - 1;
        $paginationCtrls .= '<a href="'.$_SERVER['PHP_SELF'].'?pn='.$previous.'">Previous</a> &nbsp; &nbsp; ';
        // Render clickable number links that should appear on the left of the target page number
        for($i = $pagenum-4; $i < $pagenum; $i++){
            if($i > 0){
                $paginationCtrls .= '<a href="'.$_SERVER['PHP_SELF'].'?pn='.$i.'">'.$i.'</a> &nbsp; ';
            }
        }
    }
    // Render the target page number, but without it being a link
    $paginationCtrls .= ''.$pagenum.' &nbsp; ';
    // Render clickable number links that should appear on the right of the target page number
    for($i = $pagenum+1; $i <= $last; $i++){
        $paginationCtrls .= '<a href="'.$_SERVER['PHP_SELF'].'?pn='.$i.'">'.$i.'</a> &nbsp; ';
        if($i >= $pagenum+4){
            break;
        }
    }
    // This does the same as above, only checking if we are on the last page, and then generating the "Next"
    if ($pagenum != $last) {
        $next = $pagenum + 1;
        $paginationCtrls .= ' &nbsp; &nbsp; <a href="'.$_SERVER['PHP_SELF'].'?pn='.$next.'">Next</a> ';
    }
}

$number = 0;
while ($stmt->fetch()) {
                "$id, $question, $ans1, $ans2, $ans3, $ans4";
                $number = $number+1;
                $dynamicList .= "
                <div class='questions'>

                <span class='numbering'>$number</span>

                <span class='question' id='$id'>$question</span></br>

                 <input type='radio' name='choice1' value='$ans1' /></br>
                 <input type='radio' name='choice2' value='$ans2' /></br>
                 <input type='radio' name='choice3' value='$ans3' /></br>
                 <input type='radio' name='choice4' value='$ans4' />
                </div>


              ";


}
    }
// Close your database connection
mysqli_close($myConnection);
?>
<!DOCTYPE html>
<html>
<head>
<style type="text/css">
body{ font-family:"Trebuchet MS", Arial, Helvetica, sans-serif;}
div#pagination_controls{font-size:21px;}
div#pagination_controls > a{ color:#06F; }
div#pagination_controls > a:visited{ color:#06F; }
</style>
</head>
<body>
<div>
  <h2><?php echo $textline1; ?> Paged</h2>
  <p><?php echo $textline2; ?></p>
  <p><?php echo $dynamicList; ?></p>
  <div id="pagination_controls"><?php echo $paginationCtrls; ?></div>
</div>
</body>
</html>

Answer 1

对于你的编号，你可能想玩以下来看看它是如何运作的。

$pn=isset( $_GET['pn'] ) ? $_GET['pn'] : 0;
$startNumber=1;
$endNumber=$page_rows;
$i=0;

while ( $stmt->fetch() ) {

    if( $pn > 0 ) {
        if( $i % $page_rows == 0 ) {
            $startNumber=( $pn * $page_rows + 1 );
            $endNumber=( ( $startNumber + $page_rows ) - 1 );
        }
    }
    $displayedRecordNumber=abs( $startNumber + $i );
    $i++;

    /* The name of the radio group should be the same for each question */

    $dynamicList .= "
            <div class='questions'>
                <span class='numbering'>".$displayedRecordNumber."</span>
                <span class='question' id='".$id."'>$question</span><br />
                <input type='radio' name='choice".$i."' value='".$ans1."' /><br />
                <input type='radio' name='choice".$i."' value='".$ans2."' /><br />
                <input type='radio' name='choice".$i."' value='".$ans3."' /><br />
                <input type='radio' name='choice".$i."' value='".$ans4."' />
            </div>";

}

分页结果的编号

1 个答案: