我有一个名为displayMenu()的函数,它从表中提取数据..这个函数位于一个名为functions.php的php文件中
if (isset($_POST['Starters'])) {
$sql = "SELECT * FROM product WHERE category_id=1";
$sql1= "SELECT cat_name,cat_description FROM category WHERE cat_id=1";
displayMenu();
}
function displayMenu(){
global $db, $sql,$sql1,$nb;
if($result = mysqli_query($db, $sql1)){
if(mysqli_num_rows($result) > 0){
while($row = mysqli_fetch_array($result)){
echo '<h1 class="header">' . $row['cat_name'] . '</h2>';
echo '<h3 class="content">' . $row['cat_description'] . '</h3>';
}
}
}
if($result = mysqli_query($db, $sql)){
if(mysqli_num_rows($result) > 0){
echo "<table>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['p_code'] . "</td>";
echo "<td>" . $row['p_name'] . "</td>";
echo "<td>" . $row['p_description'] . "</td>";
echo "<td>" . $row['p_price'] . "</td>";
echo "<td>" ;
echo '<form class="nbItem" action="order.php" method="post">Nb of items:';
echo '<input type="number" name="item"> <button type="submit" class="btn" name="addtocart">Add To Cart</button>';
echo '</form>';
echo "</td>";
echo "</tr>";
}
echo "</table>";
// Free result set
mysqli_free_result($result);
} else{
echo "No records matching your query were found.";
}
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
这实际上有效..但我想在另一个名为order.php的文件中创建表,并从此文件中调用此函数 我应该怎么做?
答案 0 :(得分:0)
如果我理解正确,您只需要在orders.php文件中包含您的functions.php文件,然后从包含您的订单文件调用displayMenu?
include 'functions.php';
echo displayMenu();