2015-10-22 26 views
0

比方說,我已經產生了像這樣5個陣列:PHP刀片上產生陣列MySQL數據庫

$MyArr = array(); 
    for($i = 0; $i < 15; $i++){ 
     $data[] = array(
     'title'  => 'Title', 
     'description'  => 'Description', 
     'created_on' => 'Time' 
     ); 
    } 

我怎樣才能將所生成的陣列的每個迭代到我MySQL數據庫

UPD。在W3Schools的網站(程序),我使用mysqli_connect();功能連接到我的數據庫(我已經按照教程

+2

取決於您的數據庫連接上 - PDO,mysqli,mysql,一些框架? – Philipp

+0

我已更新問題。我正在使用[this](http://www.w3schools.com/php/php_mysql_select.asp)教程來設置連接。 –

回答

0

你可以通過你的數據數組這樣的循環:

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

$yourData = array(); 
for($i = 0; $i < 5; $i++){ 
    $yourData[] = array(
    'title'  => md5(rand()), 
    'description'  => rand(0, 1)?'active':'banned', 
    'created_on' => rand(0, 1)?'now':'2014/12/31 14:51:52' 
    ); 
} 

for($j = 0; $j < count($yourData); $j++) { 
    $title = $yourData[$j]["title"]; 
    $description = $yourData[$j]["description"]; 
    $created_on = $yourData[$j]["created_on"]; 
    $sql = "INSERT INTO yourTable (title,description,created_on) VALUES ($title, $description, $created_on)"; 
    if (mysqli_query($conn, $sql)) { 
     echo "New record created successfully"; 
    } else { 
     echo "Error: " . $sql . "<br>" . mysqli_error($conn); 
    } 
    mysqli_close($conn); 
} 
+0

非常感謝!這已經幫了我很多了! –

+0

像一個魅力工作 –

+0

不客氣:) – zachu