從數據庫中獲取數據以創建表-PHP
[英]Fetch data from database to create a table - PHP
問題描述
我創建了一個從數據庫中檢索數據的函數，以及一個將數據放入表中的函數。
function get_order()
{
$order_query = "SELECT order_number From tbl_order_header";
$data = mysqli_query($con, $order_query);
$order = array();
while ($object = mysqli_fetch_object($data))
{
$order[] = $object;
}
mysqli_close($con);
return $order;
}
function get_table()
{
$table_str = '<table>';
$get_orders = get_order();
foreach ($get_orders as $get_order)
{
$table_str .= '<td>';
$table_str .= '<td>'.$get_order->order_number.'</td>';
$table_str .= '</td>';
}
$table_str .= '</table>';
return $table_str;
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
</head>
<body>
<?php echo get_table();?>
</body>
</html>
但是我收到如下錯誤消息。
line 8: $data = mysqli_query($con, $order_query);
Warning: mysqli_query() expects parameter 1 to be mysqli, null given in C:\xampp\htdocs\cmg-logistics\Testing.php on line 8
line 10: while ($object = mysqli_fetch_object($data))
Warning: mysqli_fetch_object() expects parameter 1 to be mysqli_result, null given in C:\xampp\htdocs\cmg-logistics\Testing.php on line 10
line 14: mysqli_close($con);
Warning: mysqli_close() expects parameter 1 to be mysqli, null given in C:\xampp\htdocs\cmg-logistics\Testing.php on line 14
我該如何解決錯誤？
3 個解決方案
解決方案1
0 2016-06-15 04:37:26
這是因為未定義$ con。 嘗試在函數內傳遞$ con。
$con = mysqli_connect(HOST, USER, PASS, DB_NAME);
解決方案2
0 2016-06-15 04:53:47
可能是您的連接變量問題。
這樣連接
//db connection
global $conn;
$servername = "localhost"; //host name
$username = "username"; //username
$password = "password"; //password
$mysql_database = "dbname"; //database name
//mysqli prepared statement
$conn = mysqli_connect($servername, $username, $password) or die("Connection failed: " . mysqli_connect_error());
mysqli_select_db($conn,$mysql_database) or die("Opps some thing went wrong");
函數內部應該是
function get_order()
{
global $conn;
}
解決方案3
0 2016-06-15 04:59:51
您得到的所有錯誤都是警告，而不是錯誤。 如果您的PHP代碼有錯誤，它將返回錯誤狀態500 INTERNAL SERVER ERROR 。 您可以使用以下命令禁用警告。
// Turn off all error reporting
error_reporting(0);
ini_set("display_errors", "0");
通過從MySQL數據庫中選擇數據，通過PHP + HTML創建表
[英]Create table via PHP + HTML by selecting data from MySQL database
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