从下拉菜单的单个选择标签向MySql插入2个值

Question

Is there a way to insert 2 values to Mysql from a single selection of drop down menu?

The example of mysql syntax for inserting the form values is as following:

if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form1")) {
$insertSQL = sprintf("INSERT INTO menu (food, image_extension) VALUES (%s, %s)",
GetSQLValueString($_POST['food'], "text"),
GetSQLValueString($_POST['image_extension'], "text"),

mysql_select_db($database_menu, $menu);
$Result1 = mysql_query($insertSQL, $menu) or die(mysql_error());

$insertGoTo = "menu.php?status=choosen";
if (isset($_SERVER['QUERY_STRING'])) {
$insertGoTo .= (strpos($insertGoTo, '?')) ? "" : "?";
$insertGoTo .= $_SERVER['QUERY_STRING'];
}
header(sprintf("Location: %s", $insertGoTo));
}

I'm trying to insert values of 2 columns (food & image_extension) from each selected tag of the following dropdown list to MySql but it can't insert data to image_extension column. It only update food column.

<select name="food, image extension" class="dropdownmenu" input id="food" value="<?php echo $_POST['food'].$_POST['image_extension']; ?>"> 
<option value="selected="selected">Select Food</option>
<option value="Pizza, pizza.jpg">Pizza</option>
<option value="French Fry' frenchfry.jpg">French Fry</option>
</select>

I'm confused about how to put values in the following three attributes of the above dropdown list properly in this case?

1. <select name="food, image extension" 
2. <select value="<?php echo $_POST['food'].$_POST['image_extension']; ?>" 
3. <option value="Pizza, pizza.jpg">Pizza</option>

Any guideline shall be gighly appreciated.

Answer 1

Instead of using two different names for the select just grab the value from the select menu and turn it into an array

<?php

$food = explode (",", $_POST['food']);

//$food[0] will equal Pizza, $food[1] will equal pizza.jpg
$insertSQL = "INSERT INTO menu (food, image_extension) VALUES ({$food[0]}, {$food[1]})";

?>

<select name="food" class="dropdownmenu" id="food">
<option value="Pizza, pizza.jpg">Pizza</option>
<option value="French Fry, frenchfry.jpg">Pizza</option>
</select>

Answer 2

The problem has been solved anyway with the precise guideline of an expert fellow in www.phpbuilder.com I'm presenting the whole solved scenario below:

if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form1")) { 

//("," Comma within the double quote shall be used as delimiter here) 
$food=explode(",",$_POST['food']); 
$insertSQL = sprintf("INSERT INTO menu (food, image_extension) VALUES (%s, %s)", 

// The column names in the following string should be replaced by the newly created 
// array elements along with trim function to remove any unexpected white space.       

GetSQLValueString (trim($food[0]), "text"), 
GetSQLValueString (trim($food[1]), "text");

mysql_select_db($database_menu, $menu); 
$Result1 = mysql_query($insertSQL, $menu) or die(mysql_error()); 

$insertGoTo = "menu.php?status=choosen"; 
if (isset($_SERVER['QUERY_STRING'])) { 
$insertGoTo .= (strpos($insertGoTo, '?')) ? "" : "?"; 
$insertGoTo .= $_SERVER['QUERY_STRING']; 
} 
header(sprintf("Location: %s", $insertGoTo)); 
} 

The html part should be changed to the following:

 <select name="food" class="dropdownmenu" input id="food" 
 value="<?php echo  $_POST['food']; ?>">
 <option value="selected="selected">Select Food</option>
 <option value="Pizza, pizza.jpg">Pizza</option>
 <option value="French Fry' frenchfry.jpg">French Fry</option>
 </select>