Question

的search.php

<form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>" method="post">
      <input type="text" name="search" id="search" autocomplete="off">
      <button type="submit" class="btn btn-primary">Search</button>
</form>
    <?php $search = $_POST['search']; ?> // This is what I've tried.
    <script type="text/javascript" src="js/search.js"></script>

我尝试添加$search = $_POST['search'];并将其放入search.js这样var searchTerm = "<?php echo $search; ?>";

search.js

 var ajax_arry=[];
 var ajax_index =0;
 var sctp = 100;
 $(function(){
   $('#loading').show();
   var searchTerm = "<?php echo $search; ?>"; // This is what I've tried.
 $.ajax({
     url:"scroll.php",
              type:"POST",
              data:"actionfunction=showData&page=1&search="+searchTerm, // This is what I've tried.
    cache: false,
    success: function(response){
       $('#loading').hide();
      $('#demoajax').html(response);

    }

   });
$(window).scroll(function(){

   var height = $('#demoajax').height();
   var scroll_top = $(this).scrollTop();
   if(ajax_arry.length>0){
   $('#loading').hide();
   for(var i=0;i<ajax_arry.length;i++){
     ajax_arry[i].abort();
   }
}
   var page = $('#demoajax').find('.nextpage').val();
   var isload = $('#demoajax').find('.isload').val();

     if ((($(window).scrollTop()+document.body.clientHeight)==$(window).height()) && isload=='true'){
       $('#loading').show();
   var ajaxreq = $.ajax({
     url:"scroll.php",
              type:"POST",
              data:"actionfunction=showData&page="+page,
    cache: false,
    success: function(response){
       $('#demoajax').find('.nextpage').remove();
       $('#demoajax').find('.isload').remove();
       $('#loading').hide();

      $('#demoajax').append(response);

    }

   });
   ajax_arry[ajax_index++]= ajaxreq;

   }
return false;

if($(window).scrollTop() == $(window).height()) {
   alert("bottom!");
}
});

});

但是当我从search data:"actionfunction=showData&page=1&search="+searchTerm获取scroll.php进行查询时，它无效。

function showData($data,$con,$limit){
$page = $data['page'];
if($page==1){
    $start = 0;
}
else{
    $start = ($page-1)*$limit;
}
$name = $data['search'];
$sql = "SELECT * FROM product WHERE p_name LIKE '%$name'";

Answer 1

你应该在你的php文件中包含你的Js代码。因为你不能在js文件中编写php代码。而js代码可以用php编写。

所以你的代码应该是。

<form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method="post">
    <input type="text" name="search" id="search" autocomplete="off">
    <button type="submit" class="btn btn-primary">Search</button>
</form>

<?php if ($_POST) {
    $search = $_POST['search'];
    ?> 
    <script>
        var ajax_arry = [];
        var ajax_index = 0;
        var sctp = 100;
        $(function () {
            $('#loading').show();
            var searchTerm = "<?php echo $search; ?>"; // This is what I've tried.
            $.ajax({
                url: "scroll.php",
                type: "POST",
                data: "actionfunction=showData&page=1&search=" + searchTerm, // This is what I've tried.
                cache: false,
                success: function (response) {
                    $('#loading').hide();
                    $('#demoajax').html(response);

                }

            });
            $(window).scroll(function () {

                var height = $('#demoajax').height();
                var scroll_top = $(this).scrollTop();
                if (ajax_arry.length > 0) {
                    $('#loading').hide();
                    for (var i = 0; i < ajax_arry.length; i++) {
                        ajax_arry[i].abort();
                    }
                }
                var page = $('#demoajax').find('.nextpage').val();
                var isload = $('#demoajax').find('.isload').val();

                if ((($(window).scrollTop() + document.body.clientHeight) == $(window).height()) && isload == 'true') {
                    $('#loading').show();
                    var ajaxreq = $.ajax({
                        url: "scroll.php",
                        type: "POST",
                        data: "actionfunction=showData&page=" + page,
                        cache: false,
                        success: function (response) {
                            $('#demoajax').find('.nextpage').remove();
                            $('#demoajax').find('.isload').remove();
                            $('#loading').hide();

                            $('#demoajax').append(response);

                        }

                    });
                    ajax_arry[ajax_index++] = ajaxreq;

                }
                return false;

                if ($(window).scrollTop() == $(window).height()) {
                    alert("bottom!");
                }
            });

        });
    </script>

<?php } ?>

Answer 2

解释

你没有在PHP中“存储”PHP变量。 PHP是服务器端，javascript在客户端工作。他们必须沟通。

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2 个答案: