MySQLi將表導出爲CSV

Question

我一直在嘗試多個日期來查詢.csv與MySQLi中的某些值，但導出時出現問題，所有內容都導出到單個列，而我需要導出應該像SQL代碼（我有3個選定的列，我需要3個導出的列在.cvs）。 婁你可以找到兩個代碼：

1）該代碼創建數據的基礎上，表和插入值：

<?php 
$servername = "localhost"; 
$username = "root"; 
$password = ""; 
$dbname = "myDB"; 
// Create connection 
$conn = new mysqli($servername, $username, $password); 
// Check connection 
if ($conn->connect_error) { 
    die("Connection failed: " . $conn->connect_error); 
} 
// Create database 
$sql = "CREATE DATABASE myDB"; 
if ($conn->query($sql) === TRUE) { 
    echo "Database created successfully"; 
} else { 
    echo "Error creating database: " . $conn->error; 
} 
// Create connection pt a crea tabela 
$conn = new mysqli($servername, $username, $password, $dbname); 
// Check connection 
if ($conn->connect_error) { 
    die("Connection failed: " . $conn->connect_error); 
} 
// sql to create table 
$sql = "CREATE TABLE MyGuests (
    id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, 
    firstname VARCHAR(30) NOT NULL, 
    lastname VARCHAR(30) NOT NULL, 
    email VARCHAR(50), 
    reg_date TIMESTAMP 
)"; 
if ($conn->query($sql) === TRUE) { 
    echo "Table MyGuests created successfully"; 
} else { 
    echo "Error creating table: " . $conn->error; 
} 
// Create connection pt a insera valori 
$conn = new mysqli($servername, $username, $password, $dbname); 
// Check connection 
if ($conn->connect_error) { 
    die("Connection failed: " . $conn->connect_error); 
} 
// Insert value in db 
$sql = "INSERT INTO MyGuests (firstname, lastname, email) 
    VALUES ('John', 'Doe, Mark', 'john@example.com'), 
    ('Mary', 'Moe-Johny', 'mary@example.com'), 
    ('Julie', 'Dooley', 'julie@example.com')"; 
if ($conn->query($sql) === TRUE) { 
    echo "New record created successfully"; 
} else { 
    echo "Error: " . $sql . "<br>" . $conn->error; 
} 
$conn->close(); 
?> 

2）該代碼可用於在.csv出口：

<?php 
//header to give the order to the browser 
header('Content-Type: text/csv'); 
header('Content-Disposition: attachment;filename=export.csv'); 
$servername = "localhost"; 
$username = "root"; 
$password = ""; 
$dbname = "myDB"; 
// Create connection 
$conn = new mysqli($servername, $username, $password, $dbname); 
// Check connection 
if ($conn->connect_error) { 
    die("Connection failed: " . $conn->connect_error); 
} 
$sql = "SELECT id, firstname, lastname FROM MyGuests"; 
$result = $conn->query($sql); 
if ($result->num_rows > 0) { 
    // output data of each row 
    while($row = $result->fetch_assoc()) { 
     echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>"; 
    } 
} else { 
    echo "0 results"; 
} 
$conn->close(); 
?> 

我使用這個版本XAMP：

Apache/2.4.18 (Win32) OpenSSL/1.0.2e PHP/7.0.1

Database client version: libmysql - mysqlnd 5.0.12-dev - 20150407 - $Id: 7e72f9690b1498a1bead7a637c33a831c0d2f655 $

PHP extension: mysqli Documentation

PHP version: 7.0.1

這個版本爲MySQL代碼創建了一些問題，這就是爲什麼我更喜歡MySQLi中的代碼。 非常感謝！

Answer 1

基於新信息，我將此作爲解決方案。

<?php 

// call export function 
exportMysqlToCsv('export_csv.csv'); 


// export csv 
function exportMysqlToCsv($filename = 'export_csv.csv') 
{ 

    $conn = dbConnection(); 
// Check connection 
    if ($conn->connect_error) { 
     die("Connection failed: " . $conn->connect_error); 
    } 
    $sql_query = "SELECT id, firstname, lastname FROM MyGuests"; 

    // Gets the data from the database 
    $result = $conn->query($sql_query); 

    $f = fopen('php://temp', 'wt'); 
    $first = true; 
    while ($row = $result->fetch_assoc()) { 
     if ($first) { 
      fputcsv($f, array_keys($row)); 
      $first = false; 
     } 
     fputcsv($f, $row); 
    } // end while 

    $conn->close(); 

    $size = ftell($f); 
    rewind($f); 

    header("Cache-Control: must-revalidate, post-check=0, pre-check=0"); 
    header("Content-Length: $size"); 
    // Output to browser with appropriate mime type, you choose ;) 
    header("Content-type: text/x-csv"); 
    header("Content-type: text/csv"); 
    header("Content-type: application/csv"); 
    header("Content-Disposition: attachment; filename=$filename"); 
    fpassthru($f); 
    exit; 

} 

// db connection function 
function dbConnection(){ 
    $servername = "localhost"; 
    $username = "root"; 
    $password = ""; 
    $dbname = "myDB"; 
    // Create connection 
    $conn = new mysqli($servername, $username, $password, $dbname); 
    return $conn; 
} 


?>