[RESOLVED] Inserting 2 values to MySql from a single selection tag of drop down menu

polarexpress

I'm finding a way about how to insert 2 values to MySql from a single selection option of drop down menu under following circumstance.

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));
}

and I'm trying to insert possible values (Pizza, pizza.jpg) of 2 columns (food & image_extension) from each selected tag of the following dropdown list to MySql but it can't insert data into image_extension column. It only update food column. Here the values are stored in the selection tag only from which these should be inserted into Mysql database.

<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>

Is it how I should do this?

Thanks for any help.

dagon

the easy option i see is to explode() on the comma. Depending where the values comes from for the select, this may not be needed, if they are from an array for instance, i would just reuse that.

polarexpress

The option values aren't stored anywhere. This is the drop down list from which 2 values under each option shall be stored into 2 columns of MySql database after the users submit the form.

I'm very much confused about how to do it. I've learned the basic of explode function by this time but I can't figure out how to implement it properly under the circumstances I explained in my previous post.

I shall remain grateful if you guide me to a step by step solution of this problem.

Kind regards,

dagon

$insertSQL = sprintf("INSERT INTO menu (food, image_extension) VALUES (%s, %s)",
GetSQLValueString($_POST['food'], "text"),
GetSQLValueString($_POST['image_extension'], "text"),

turn in to this

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

$insertSQL = sprintf("INSERT INTO menu (food, image_extension) VALUES (%s, %s)",
GetSQLValueString(trim($food[0]), "text"),
GetSQLValueString(trim($food[1]), "text"),

polarexpress

Dagon, many many thanks for your short but precious guideline that works perfectly to solve the problem.

For the convenience of the other viewers I'm presenting the whole solved scenario below:

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

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

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 shall 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>

Kind regards,

polarexpress

polarexpress;11002518 wrote:

Dagon, many many thanks for your short but precious guideline that works perfectly to solve the problem.

For the convenience of the other viewers I'm presenting the whole solved scenario below:

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

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

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 shall 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>

Kind regards,

In case ',' comma within single quote doesn't work it should be replaced by double quote like ","

Weedpacket

polarexpress wrote:
In case ',' comma within single quote doesn't work it should be replaced by double quote like ","

In this case it wouldn't make any difference.

polarexpress

Weedpacket;11002528 wrote:
In this case it wouldn't make any difference.

Thanks for the tip.