php - jquery php mysql - 显示所有项目时如何隐藏加载更多按钮

Question

我有一个用于显示数据库列表的代码和一个loadmore按钮。显示所有记录列表后，单击loadmore按钮显示空白位置。

所有记录显示后如何隐藏loadmore按钮？

index.php 文件:

<?php $chinp=$_GET['schinp'];?>
<div  id="schts"></div>
<button id="btnsch">load more</button>

<script>
$(document).ready(function() {
    var chinp="<?php echo $chinp;?>";
  var srchco = 1;
  var offsrch = 0;
  $("#btnsch").click(function() {
    $.ajax({
        method: "POST",
       url: "search.php",
        data: { srchcoun: srchco, offsrch: offsrch ,chinp:chinp}
      })
      .done(function(msg) {
        $("#schts").append(msg);
      });
    offsrch = offsrch + srchco;
  });

  $("button").trigger("click");
});
</script>

搜索.php:

$srchcoun=$_POST['srchcoun'];
$offsrch=$_POST['offsrch'];
$chinp=$_POST['chinp'];
$schql="SELECT id, name, lastname FROM t_users WHERE name LIKE '$chinp' ORDER BY name ASC limit $offsrch, $srchcoun";
$rsch=mysqli_query($conn,$schql);

while ($rch=mysqli_fetch_assoc($rsch)){
     $scid=$rch['id'];$snm=$rch['name'];$slnm=$rch['lastname'];?>

<div class="alsu">
<img class="sask" src="pic/<?php echo $scid;?>.png" alt="">
<span class="snm">Name : <?php echo $snm." ".$slnm;?></span>
</div>
<?php }?>
</div>

谢谢。

Answer 1

这是根据您的代码隐藏“加载更多”按钮的代码。

将此代码替换到您的第一个文件中。

<script type="text/javascript">
$(document).ready(function() {
  var chinp="<?php echo $chinp;?>";
  var srchco = 2;
  var offsrch = 0;
  var page_num = 0;

  $("#btnsch").click(function() {
    $.ajax({
      method: "POST",
      url: "search.php",
      data: { srchcoun: srchco, offsrch: offsrch ,chinp:chinp, page_num : page_num}
      })
      .done(function(msg) {
        if(msg == 'noMoreData'){
          $("#btnsch").hide();
        } else {
          $("#schts").append(msg);
        }
      });
      page_num = page_num + 1;
    offsrch = offsrch + srchco;
  });

  $("#btnsch").trigger("click");
});
</script>

现在将此代码添加到您的搜索文件中。

<?php
$srchcoun=$_POST['srchcoun'];
$offsrch=$_POST['offsrch'];
$chinp=$_POST['chinp'];
$page_num=$_POST['page_num'];

$total_records = 0;
$totalschql="SELECT id FROM t_users WHERE name LIKE 'krishna' ORDER BY name ASC";
if ($result=mysqli_query($conn,$totalschql))
    {
        $total_records=mysqli_num_rows($result);
    }

$last_records_count = ($page_num) * $srchcoun;
if($last_records_count >= $total_records) {
    echo 'noMoreData';exit();
} else {

    $schql="SELECT id, name, lastname FROM t_users WHERE name LIKE '$chinp' ORDER BY name ASC limit $offsrch, $srchcoun";;
    $rsch=mysqli_query($conn,$schql);

    while ($rch=mysqli_fetch_assoc($rsch)){
         $scid=$rch['id'];$snm=$rch['name'];$slnm=$rch['lastname'];
?>
    <div class="alsu">
    <img class="sask" src="pic/<?php echo $scid;?>.png" alt="">
    <span class="snm">Name : <?php echo $snm." ".$slnm;?></span>
    </div>
    <?php }?>
    </div>
<?php
}
?>

试试这个，如果您有任何问题请告诉我。