php连接并显示mysql表 - LAMP服务器

时间:2016-06-08 10:42:28

标签: php mysql server lamp

我正在运行LAMP服务器并且我试图在网页上显示我的数据库表

这是我的php脚本

$servername = "127.0.0.1";
$username = "root";
$password = "none";
$dbname = "chegada";

$conn = mysqli_connect($servername, $username, $password, $dbname);
if ($conn->connect_error) {
     die("Connection failed: " . $conn->connect_error);
}else{echo "OK!";}

$sql = "SHOW TABLES FROM $dbname";
$result = mysql_query($sql);
if (!$result) {
    echo "DB Error, could not list tables\n";
    echo 'MySQL Error: ' . mysql_error();
    exit;
}

网页显示了这个

  

警告:mysqli_connect():( HY000 / 2002):拒绝连接   第16行/var/www/html/index.php

     

注意:尝试获取非对象的属性   第17行/var/www/html/index.php

     

OK!

     

致命错误:未捕获错误:调用未定义函数mysql_query()   在/var/www/html/index.php:22堆栈跟踪:#0 {main}被投入   第22行/var/www/html/index.php

有人可以帮我解决这个问题吗?

3 个答案:

答案 0 :(得分:0)

试试这个

$servername = "127.0.0.1";
$username = "root";
$password = ""; // maybe this should be an empty string if you dont have password setup
$dbname = "chegada";

$conn = mysqli_connect($servername, $username, $password, $dbname); // this should be mysqli_query
if ($conn->connect_error) {
     die("Connection failed: " . $conn->connect_error);
}else{echo "OK!";}

$sql = "SHOW TABLES FROM $dbname";
$result = mysqli_query($conn,$sql);
if (!$result) {
    echo "DB Error, could not list tables\n";
    echo 'MySQL Error: ' . mysqli_error();
    exit;
} else {
    $res = mysqli_fetch_all($result);
    var_dump($res);
}

答案 1 :(得分:0)

使用mysqli_query

$result = mysqli_query($conn,$sql);

答案 2 :(得分:0)

我怀疑你有一个默认的本地mysql,在这种情况下,密码“none”是一个空字符串。字面上没有密码,不是“无”字样。你需要:

$servername = "127.0.0.1";
$username = "root";
$password = "";
$dbname = "chegada";

你的第二个错误是由你在mysqli_中开始但后来恢复到mysql_的一半造成的。