我们应该在PHP中使用像"--->"这样的对象范围？

Question

我正在完成本教程

http://www.1stwebdesigner.com/tutorials/infinite-scrolling-tutorial/

它包含以下代码段:

<!--?php 
$con = mysql_connect("localhost", "username", "password"); 
mysql_select_db("database_name"); 
$result = mysql_query("select SQL_CALC_FOUND_ROWS * from scroll_images order by id asc   limit 12"); 
$row_object = mysql_query("Select Found_Rows() as rowcount"); 
$row_object = mysql_fetch_object($row_object); 
$actual_row_count = $row_object--->rowcount;
?>

这条线$actual_row_count = $row_object--->rowcount;让我困惑,该怎么--->rowcount办？当我在PHP中使用它时,我收到错误.

<?php
$sql_fetch = "SELECT * FROM articles ORDER BY time DESC limit 4;";
$dbresult = mysqli_query( $db, $sql_fetch );

$row_object = mysqli_query( $db, "Select Found_Rows() as rowcount" );
$row_object = mysqli_fetch_object( $row_object );
$actual_row_count = $row_object--->rowcount;
?>

Parse error: syntax error, unexpected T_OBJECT_OPERATOR in /Applications/MAMP/htdocs/index.php on line 14

Answer 1

<!-- Comment -->是你在HTML中注释掉东西的方式.这条线确实不正确(停止阅读这篇文章).

我想他的意思是

<?php 
$con = mysql_connect("localhost", "username", "password"); 
mysql_select_db("database_name"); 
$result = mysql_query("select SQL_CALC_FOUND_ROWS * from scroll_images order by id asc   limit 12"); 
$row_object = mysql_query("Select Found_Rows() as rowcount"); 
$row_object = mysql_fetch_object($row_object); 
$actual_row_count = $row_object->rowcount;
?>

另外一点:

请不要mysql_*在新代码中使用函数.它们不再维护,并且已被正式弃用.看到红色的盒子？了解准备好的语句,并使用 PDO或 MySQLi -本文将帮助您确定哪些.如果您选择PDO,这是一个很好的教程.