Question

我正在使用下面的代码将数据库转换为json

<?php
header('Content-type:application/json');

mysql_connect('localhost','root','')or die('cannot connect');
mysql_select_db('sanpham');
$sql = "SELECT * FROM cpu";
$result = mysql_query($sql);
if(!$result)
{
    echo 'Lỗi SQL: '.mysql_error();
    echo "\r\n<br />";
    echo 'SQL: '.$sql;
    exit;
}
$rows = array();
while($row=mysql_fetch_array($result)){
    $rows[]=$row;
}
echo json_encode($rows);


?>

在我的php文件中没有错误，没有数据库，没有任何显示。请帮助我。非常感谢

Answer 1

不推荐使用：mysql_connect（）：不推荐使用mysql扩展程序，将来会删除它：在中使用mysqli或PDO

警告建议使用pdo或mysqli。

mysqli：

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

$sql    = "SELECT * FROM cpu";
$result = mysqli_query($conn, $sql);

if (mysqli_num_rows($result) > 0) {
    $rows = array();
    while ($row = mysqli_fetch_array($result)) {

        $rows[] = $row;
    }

    echo json_encode($rows);
} else {
    echo "no results found";
}

mysqli_close($conn);
?>

PDO：

<?php

$host    = 'localhost';
$db      = 'sanpham';
$user    = 'root';
$pass    = '';
$charset = 'utf8';

$dsn = "mysql:host=$host;dbname=$db;charset=$charset";
$opt = [
        PDO::ATTR_ERRMODE            => PDO::ERRMODE_EXCEPTION,
        PDO::ATTR_DEFAULT_FETCH_MODE => PDO::FETCH_ASSOC,
        PDO::ATTR_EMULATE_PREPARES   => false,
        ];


$dbh = new PDO($dsn, $user, $pass, $opt);

$sql  = $dbh->query("SELECT * FROM cpu");
$rows = array();
while ($row = $sql->fetchall()) {
    $rows[] = $row;
}
echo json_encode($rows);


?>

Answer 2

你可以简单地在存储mysql数据库数据的数组上使用json_encode方法：

<?php
$host = "localhost";
$db = "psychocodes";
$user = "your username";  //enter your database username
$pass = "your password";  //enter your database password
$conn = new mysqli($host,$user,$pass,$db); 
$rows = array();

$sql = "SELECT * FROM data";
$result = $conn->query($sql) or die("cannot write");
while($row = $result->fetch_assoc()){
    $rows[] = $row;
}

echo "<pre>";
print json_encode(array('serverres'=>$rows));
echo "</pre>";

?>

参考：How to Convert MySQL database table into JSON using PHP

Answer 3

使用此

<?php
    //open connection to mysql db
    $connection = mysqli_connect("hostname","username","password","dbname") or die("Error " . mysqli_error($connection));

    //fetch table rows from mysql db
    $sql = "select * from tablename";
    $result = mysqli_query($connection, $sql) or die("Error in Selecting " . mysqli_error($connection));

    //create an array
    $emparray = array();
    while($row =mysqli_fetch_assoc($result))
    {
        $emparray[] = $row;
    }
    echo json_encode($emparray);

    //close the db connection
    mysqli_close($connection);
?>

使用php将mysql数据库转换为json

3 个答案: