Question

所以基本上，这是我现在的情况。我正在使用xampp（apache friends）为我的localhost，我目前正在使用AJAX和PHP构建一个简单的聊天室窗口。目前，我有两个文件，分别是C:/xampp/htdocs/AJAX CHAT/index.php和C:/xampp/htdocs/AJAX CHAT/chat.php。

的index.php

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http:www.//w3.org/1999/xhtml">

<head>
  <meta http-equiv="Content-Type" content="text/html" />
  <title>Deontray's Chat Room!</title>

  <script src="jquery-1.7.2.js" type="text/javascript"></script>
  <script type="text/javascript">
    function chat_initial() {
      var user = document.getElementById("chat_user").value;

      $.post('./chat.php', {
        stage: "initial",
        user: user
      }, function(data) {
        alert(data);
      });

      /*
				get user
				check if taken
				hide the initial div
				display the primary div
				*/
    }
  </script>

  <style type="text/css">
    <!-- #chatbox {
      background-color: #DDD;
      border: 1px solid #000;
      width: 700px;
      height: 500px;
    }
    #chatbox #initial {
      text-align: center;
      margin: auto;
      width: 250px;
      padding-top: 100px;
    }
    -->
  </style>
</head>

<body>

  <div id="chatbox">

    <div id="initial">

      <table>
        <tr align="center">
          <td>Enter a username to start chatting:</td>
        </tr>
        <tr align="center">
          <td>
            <input type="text" name="chat_user" id="chat_user" style="width: 200px;" />
          </td>
        </tr>
        <tr align="center">
          <td>
            <br />
            <input type="button" value="Enter chat!" onClick="chat_initial();" />
          </td>
        </tr>
      </table>

    </div>

    <div id="primary"></div>

  </div>


</body>

</html>

chat.php

<?php
//connect to MySQL database
	mysqli_connect("localhost", "root", "deontray");
	mysqli_select_db("tutorials");

//read the stage
	$stage = $_POST['stage'];
//primary code
	if($stage == "initial"){
		//check the username
		$user = $_POST['user'];
		
		$query = mysqli_query("SELECT * FROM `chat_active` WHERE user = '$user'");
		if (mysqli_num_rows($query) == 0){
			$time = time();
			//
			mysqli_query("INSERT INTO `chat_active` VALUES ('$user', '$time')");
			//set the session
			$_SESSION['user'] = $user;
			
			echo "good";
		}
		else
			echo "Username Taken";
	}
	else
		echo "Error.";
?>

这是我的错误： 警告： mysql_num_rows期望参数1为资源，给定布尔值。

Answer 1

使用mysqli

连接数据库

$connent_mysqli = mysqli_connect($db_host,$db_username,$db_password, $db_name);

并获取类似的数据

$query = mysqli_query($connent_mysqli, "SELECT * FROM `chat_active` WHERE user = '$user'");

无缘无故地获取mysqli_num_rows（）错误？

1 个答案: