PHP编码连接MYSQL问题

时间:2017-02-17 11:10:15

标签: javascript php

我正在创建一个数据库来制作我的'PHP'网站,但我无法做到这一点。我的网站cruzapp与rideshare公司相关,并将其更改为php以获取有关我们用户的详细信息。但是我无法使用以下MYSQL代码连接PHP

?php
$username = "name";
$password = "password";
$hostname = "host"; 

//connection to the database
$dbhandle = mysql_connect($hostname, $username, $password) 
 or die("Not connected to MySQL");
echo "Connected to MySQL<br>";

//select a database to work with
$selected = mysql_select_db("examples",$dbhandle) 
  or die("Could not select examples");

//execute the SQL query and return records
$result = mysql_query("SELECT id, model,year FROM cars");

//fetch tha data from the database 
while ($row = mysql_fetch_array($result)) {
   echo "ID:"$row{'id'}." Name:".$row{'model'}."Year: ". //display the results
   $row{'year'}.<br>";
}
//close the connection
mysql_close($dbhandle)
?>

有人可以帮我调试这段代码吗? 我将非常感谢你。

2 个答案:

答案 0 :(得分:1)

试一试。它使用MySQLi进行错误回显。

<?php

$username = "name";
$password = "password";
$hostname = "host";
$database = "examples";

$con = mysqli_connect($hostname, $username, $password, $database);

if (!$con) {
    exit("Connection failed: " . mysqli_connect_error());
}

$result = mysqli_query($con, "SELECT id, model,year FROM cars");

if (mysqli_error($con)) {
    exit("Error: " . mysqli_error($con));
}

while ($row = mysqli_fetch_array($result)) {
    echo "ID:" . $row['id'] . " Name:" . $row['model'] . "Year: " . $row['year'] . "<br>";
}

mysqli_close($con);

答案 1 :(得分:0)

首先你不应该使用mysql因为PHP 7 mysql扩展不再使用了。所以你必须考虑将其改为mysqli或PDO。建议使用PDO。任何如何快速修复$ selected = mysql_select_db($ dbhandle,“examples”)执行此操作,并检查所有值,如主机名数据库名称表名称,并确保没有错误。