Save multiple field in db from dropdown selection - php

Question

this is what i successfully done.
User key in the data into the text field and preview back the data.
this is what i failed to do:
capture back the data from db because the data that i select from drop down didn't store in the db. thats why i cannot print back the page after clicked submit.
it works well for other field but not for my drop down selection. is there somethings wrong with my code for this part?
this is my index.html file
<tr>
<td colspan="4" align="left" class="form-group form-inline"><b>Section
B</b></td>
</tr>
<script type="text/javascript">
function show() {
var dropdown = document.getElementById("employment");
var current_value =
dropdown.options[dropdown.selectedIndex].value;
if(current_value =="--Choose one--"){
didfv3.style.display = 'none';
didfv2.style.display = 'none';
}
if(current_value == "Unemployed"){
didfv3.style.display = 'none';
didfv2.style.display = '';
}
if(current_value == "Working"){
didfv3.style.display = '';
didfv2.style.display = 'none';
}
}
</script>
<td>employment</td>
<td colspan="4">
<select name="employment" id="employment" onChange="show();">
<option value="--Choose one--">--Choose one--</option>
<option value="Unemployed">Unemployed</option>
<option value="Working">Working</option>
</select>
<table id="didfv2" width="90%" border="0" align="center"
style="display:none">
<tr>
<td width="50%">Balance</td>
<td width="1%" align="center">:</td>
<td width="73%" ><input type="text" name="balance" readonly maxlength="10"
value="<?php echo $balance?>" size="5" > months</td>
</tr>
</table>
<table id="didfv3" width="90%" border="0" align="left"
style="display:none">
<tr>
<td class="form-group form-inline">Basic Salary</td>
<td width="0.5%" align="knk" class="form-group form-inline">:</td>
<td width="73%" class="form-group form-inline">
<input name="basic" type="text" id="basic" maxlength="20"/></td>
</tr>
<tr>
<td>Gross Salary</td>
<td width="0.5%" align="center" class="form-group form-inline">:</td>
<td width="73%" class="form-group form-inline">
<input name="gross" type="text" id="gross" maxlength="20"/></td>
</tr>
<tr>
<td class="form-group form-inline">Nett Salary</td>
<td width="0.5%" align="center" class="form-group form-inline">:</td>
<td width="73%" class="form-group form-inline">
<input name="nett" type="text" id="nett" maxlength="20"/></td>
</tr>
<tr>
</table>
here is my insert.php
<?
$employment = $_POST['xxx'];
if (isset($_POST['xxx'])) {
$employment =$_POST['xxx'];
}
if($employment =='Working'){
$basic = $_POST['basicsalary00'];
$gross = $_POST['grosssalary00'];
$nett = $_POST['nettsalary00'];
}
elseif($xxx =='Unemployed'){
$balance = $_POST['balance'];
}
$sqlinsert = "INSERT INTO application
(
xxx,basicsalary00,grosssalary00,nettsalary00)
VALUES(0,'$xxx','$basic','$gross','$nett')";
?>
// here is to print back the output
<form name="formName" action="form1" method="post" class="form-group form-
inline">
<tr>
<td rowspan="2">Employment</td>
<td colspan="3">
<option value="<?php echo $employment;?>"><?php echo $employment;?>
</option>
<table id="didfv2" width="90%" border="0" align="center"
style="display:none">
<tr>
<td colspan="3">
<input type="hidden" id="balance" value="<?php echo $balance;?>" />
</td>
</tr>
</table>
<table id="didfv3" width="90%" border="0" align="center"
style="display:none">
<tr></tr>
<tr>
<td>Basic Salary</td>
<td>
<input name="basic" type="hidden" id="basic" value="<?php echo $basic;?>" />RM <?php echo $basic;?></td>
</tr>
<tr>
<td>Gross Salary</td>
<td>
<input name="gross" id="gross" type="hidden" value="<?php echo $gross;?>"
/>RM <?php echo $gross;?></td>
</tr>
<tr>
<td>Nett Salary</td>
<td>
<input name="nett" type="hidden" id="nett" value="<?php echo $nett;?>"
/>RM <?php echo $nett;?></td>
</tr>
</table>
</td>
</tr>
</form>
So there will be 4 pages,
1) for user input (index.html)
2) Preview back the input (i didnt include the file in here - successful)
3) Insert the input to db (insert.php)
4) Print back the result (code at the bottom of the insert.php...fetch from db- which i failed. Working for other field but not for drop down section)
I really need your help.. i've been stuck here for a week.... thank you

Answer 1

A proper dropdown should have a <select> and the <option> elements should be inside it!
In your insert.php, I can see that you have only used this line:
<option value="<?php echo $employment;?>"><?php echo $employment;?></option>
How about your try it like this:
<select readonly>
<option value="--Choose one--">--Choose one--</option>
<option <?php echo ($employment == 'Unemployed') ? 'selected' : ''; ?> value="Unemployed">Unemployed</option>
<option <?php echo ($employment == 'Working') ? 'selected' : ''; ?> value="Working">Working</option>
</select>
Here, we are checking whether any of those <option> element is same as the one that the user has submitted. If so, make it as selected
Hope it helps.
EDIT
If you want to store the value of $employment to display it later, you definitely have to store it in your db.