如何从动态表单获取值和设置名称

Question

Maybe the title made confused. But I have a case like this,

I have a form that contains some input text (textbox). But this input is depend on my database.

<label class="strong"><?php echo $rowexp['mexpense_name']; ?></label>
<input type="text" class="span4" placeholder="Expense <?php echo $rowexp['mexpense_name']; ?>" name="" />

Here's my database for that input text:

tbl_mexpensive -> mexpensive_id, mexpensive_name
tbl_expensive -> expensive id, expensive_year
tbl_texpensive -> mexpensive_id, expensive_id

So the number of textfield will be same as much as the data at tbl_mexpensive and the POST page will set to tbl_texpensive.

For example, if mexpensive got 3 value then: [textbox1] [textbox2] [textbox3] If i click submit It return to POST page and get the value each of the textbox BUT the query is like this: insert into texpensive (exp_id, mexp_id) VALUE (exp_id, XXX)

From above, my question is, how to get the value at the POST page (XXX)? I mean, what name should I give to that input text? I have try an array but it's not work properly. I try give it name as the table value, still no result because it's a dynamic textbox.

Sorry for my bad English.

--EDIT

My POST PAGE

 $asal = $_POST['asal']; $unit = $_POST['unit']; $tahun = $_POST['tahun']; $querydata = mysql_query("SELECT * FROM m_mainexpense WHERE mexpense_status=1"); $last = mysql_query("SELECT * FROM t_expense"); $getlast = mysql_fetch_array($last); $nom = (int)$getlast['expense_id'] + 1; while($rowdata = mysql_fetch_array($querydata)) { if(isset($_POST[mexpense_name])) { //actually don't know what to do in here $arr = $_POST[mexpense_name]; $query = "INSERT INTO t_expense (texp_mext, texp_exp) VALUES ('$arr','$nom')"; } } $query = "INSERT INTO m_expense(expense_unit, expense_school, expense_year, expense_type, expense_status) VALUES ('$unit', '$asal', '$tahun', 1)"; if(mysql_query($query)) { echo "<script> alert('Penambahan Data Sukses'); window.location.href='return_index.php'; </script>"; echo '<noscript> Please Enable Your JavaScript. <a href="welcome.php">Main Menu</a> </noscript>'; die(); } 

Answer 1

Using key($array) and get the value by foreach function. And return it to database. Finally it solved....