php - 警告 : mysqli_connect() expects parameter 5 to be long, 中给出的字符串

Question

我知道已经有人问过这个问题，但似乎对我的问题没有任何帮助。我需要从数据库中获取一些数据并将其显示在 HTML 中。

代码是这样的:

<?php

$host="localhost"; // Host name 
$username="bluecode_power"; // Mysql username 
$password="bluecode123"; // Mysql password 
$db_name="bluecode_login"; // Database name 
$tbl_name="news"; // Table name 

$myConnection= mysqli_connect("$host", "$date", "$user", "$title", "$text")or die("cannot connect"); 
mysqli_select_db("$db_name")or die("cannot select database");

$query = "SELECT * FROM news"; //You don't need a ; like you do in SQL
$result = mysql_query($query);

echo "<table>"; // start a table tag in the HTML

while($row = mysql_fetch_array($result)){   //Creates a loop to loop through results
echo "<tr><td>" . $row['date'] . "</td><td>" . $row['news'] . "</td></tr>";  //$row['index'] the index here is a field name
}
echo "</table>"; //Close the table in HTML

mysql_close(); 

?>

感谢任何想法:)

Answer 1

你在哪里定义$text变量？ $text 的值是一个字符串，尽管 mysqli_connect() 中的第五个参数是端口 # 并期望传递一个长值或整数值。

$myConnection= mysqli_connect("$host", "$date", "$user", "$title", "$text")or die("无法连接");

应该是:

$myConnection = mysql_connect( $host, $username, $password, $db_name ) or die("无法连接");