如何修复php中未定义的变量？

Question

I have an index.php page.我有一个index.php页面。 The function of this page is infinite scrolling using AJAX, PHP and MySQL.该页面的功能是使用AJAX、PHP和MySQL无限滚动。 The top portion contains PHP MySQL codes and bottom contains JavaScript.顶部包含 PHP MySQL 代码，底部包含 JavaScript。
I want to print the total number of rows in center of the page, but every time I try it shows "undefined variable" error.我想在页面中央打印总行数，但每次尝试时都会显示“未定义变量”错误。
I think when loading the page, the total number of variable tries to print first and then the PHP query takes place, so it shows "undefined variable", but when I put the total number of variable inside the PHP codings, there is no problem.我认为在加载页面时，变量总数尝试先打印，然后进行PHP查询，因此显示“未定义变量”，但是当我将变量总数放入PHP编码中时，没有问题.
How can I prevent this?我怎样才能防止这种情况？

My index.php is我的index.php是

//my php part here
<?php

if(isset($_POST["anotherID"])){
require_once("config.php");

$limit = (intval($_POST['limit']) != 0 ) ? $_POST['limit'] : 10;
$offset = (intval($_POST['offset']) != 0 ) ? $_POST['offset'] : 0;
$id = $_POST["anotherID"];
$query = $id;
$sql = "SELECT SQL_CALC_FOUND_ROWS * FROM x where title like '%xx%'  ORDER BY rand() LIMIT $limit OFFSET $offset";

try {
$stmt = $DB->prepare($sql);
$stmt->execute();
$results = $stmt->fetchAll();

$row_object = $DB->prepare("Select Found_Rows() as rowcount");
$row_object->execute();
$roww_object =$row_object->fetchobject();
$actual_row_count = $roww_object->rowcount;


} catch (Exception $ex) {
 echo $ex->getMessage();
}
if (count($results) > 0) {
foreach ($results as $res) {
echo'something';
}
}
 $count = $actual_row_count;
 exit;
 }
 ?>

//my html part here

 <html>
  //some html codes

  <?php echo $count; ?>

 //some html codes here

 //my java scripts here

<script type="text/javascript">
  var busy = false;
  var limit = 6
  var offset = 0;
  var anotherID = 5


   function displayRecords(lim, off) {
    $.ajax({
      type: "POST",
      async: false,
      data: "limit=" + lim + "&offset="+ off+"&anotherID="+anotherID,
      cache: false,
      beforeSend: function() {
        $("#loader_message").html("").hide();
        $('#loader_image').show();
      },
      success: function(html) {
        $("#results").append(html);
        $('#loader_image').hide();
        if (html == "") {
          $("#loader_message").html('<button class="btn btn-default btn-block" type="button">No more records.</button>').show()
        } else {
          $("#loader_message").html('<button class="btn btn-default btn-block"  type="button"><div id="loader_image"><img src="loader.gif" alt="" width="24" height="24">Loading please wait...</button>').show();
        }
        window.busy = false;


      }
    });
  }

  $(document).ready(function() {
    // start to load the first set of data
    if (busy == false) {
      busy = true;
      // start to load the first set of data
      displayRecords(limit, offset);
    }


    $(window).scroll(function() {
      // make sure u give the container id of the data to be loaded in.
      if ($(window).scrollTop() + $(window).height() > $("#results").height() && !busy) {
        busy = true;
        offset = limit + offset;

        // this is optional just to delay the loading of data
        setTimeout(function() { displayRecords(limit, offset); }, 500);

        // you can remove the above code and can use directly this function
        // displayRecords(limit, offset);

      }
    });

  });

</script>
//some html codes her

</html>

I know when a page is loading, the HTML parts are first loaded and then my jQuery stimulates the PHP part and then my results appear.我知道当页面加载时，首先加载 HTML 部分，然后我的 jQuery 刺激 PHP 部分，然后我的结果出现。
How can I fix this?我怎样才能解决这个问题？
Why does $count always show "undefined variable"?为什么$count总是显示“未定义的变量”？

Thanks in advance.提前致谢。

Answer 1

You get an error of undefined $count because $count is defined only inside the if statement.您会收到 undefined $count错误，因为$count仅在if语句中定义。
When the if clause doesn't apply $count is not defined.当if子句不适用时， $count未定义。

Add an else clause to the if and initialize $count=0;在if添加else子句并初始化$count=0; and it will solve your problem.它会解决你的问题。

Example:例子：

....
$count = $actual_row_count;
exit;
}
else $count = 0;
?>

如何修复php中未定义的变量？

问题描述

1 个解决方案

解决方案1
0 2015-08-28 07:15:34

如何修复php中未定义的变量？

问题描述

1 个解决方案

解决方案1 0 2015-08-28 07:15:34

解决方案1
0 2015-08-28 07:15:34