在數據庫中插入複選框和文本框中的數據

Question

我已經創建了兩個php文件，其中包含自助餐廳的訂單表單以及用於在數據庫中插入張貼值的代碼的相應代碼行。我在這裏介紹他們。

FOR addorder_form.php

**<?php 

     db_connect(); 

     $cats=array("Kafedes", "Rofhmata", "Pota", "Snack/Glyka"); 
     $arrlength=count($cats); 

    for($i=0;$i<$arrlength;$i++) { 
     $sql = mysql_query('SELECT title FROM products WHERE cname="'.$cats[$i].'"') or die(mysql_error()); 
     echo '<div id="main_content">'; 
     echo "<h4 style=color:#800000> ".$cats[$i]."</h4>"; 
     echo "<br />"; 


     while($row = mysql_fetch_array($sql, MYSQL_BOTH)){ 
      echo "<div id='center' style='align:center'>"; 
      echo "<input style='text-align:right;' type='checkbox' action='addorder.php' name='products[]' value='".$row["title"]."'>".$row["title"]; 
      echo '</div>'; 

      echo ' <div id="center_side" style="float:right"><form "method="post" action="addorder.php"><input type="text" size="4" padding-left="0.2em" name="quantity"/>'; 
      echo '</div>'; 
     echo '</div>'; 
     echo '<br />'; 
     } 
    } 
    echo '<form name="addorder" method="" action="addorder.php" onclick="addorder.php">'; 
       echo '<input type="submit" value="Add order" style="float: right;"><br/>'; 
       echo '</form>'; 
    ?>** 

CODE碼FOR addorder.php

**<?php 
     include_once("buzzcafe_fns.php"); 
     do_html_header(""); 




$quantity = ''; 
$title = ''; 

if (isset($_POST['quantity']) && isset($_POST['products'])) { 
    if(isset($_POST["Submit"])) { 
    $quantity = $_POST['quantity']; 
$title = $_POST['products']; 
if($_POST["Submit"] == "Submit") 
{ 
for ($i=0; $i<sizeof($title); $i++) { 
db_connect(); 
$insertOrder = mysql_query("INSERT INTO orders VALUES('".$title[i]."','".$quantity."')")or die(mysql_error()); 

    } 
echo "Record inserted"; 

} 
} 
} 
?>** 

當我運行他們，我有非任何語法錯誤，但它不工作。至於在我包含的buzzcafe_fns.php文件中設置了db_connect（），並檢查它是否正常工作。我怎樣才能讓我的「INSERT INTO」工作？

Answer 1

You can use another insert query 
First You will check if $title variable array or not. 
Next Using For loop 
<?php 
$con=mysql_connect("localhost","root",""); 
$db=mysql_select_db("yourdatabasename here" ,$con); 
$count=count($title); 
for($i=0;$i<$count;$i++){ 
mysql_query("INSERT INTO orders SET title='".$title[$i]."', quantity='".$quantity."'"); 
} 
?>