[sarahmx]

First of all sorry for my English

Im a php coding newbie..i'm currently testing

doing a simple program for xoops cms...

i'm having trouble with this can anyone help me

ok i have this



two table

info_unit : log_id, occupation
myocc : id, occ_list


input

PHP Code:

<select name='occupation'>
 <option value="">Sellect Occupation</option>
   <?php
 global $xoopsDB; 
 $result = mysql_query("SELECT * FROM ".$xoopsDB->prefix("myocc")."");
 
 while($row = mysql_fetch_array($result))
{
$occ_list=$row['occ_list'];
echo "<option value='$occupation'>$occ_list</option>";
}
?>


</select>

the above code working properly...the drop down will display value from myocc..

and the value is submitted to info_unit table successfully

the problem is this code below when i want to edit a record in the dit form
the value that i selected in the input form is not display/choosen



update record

PHP Code:

<?php
$log_id=$_POST[log_id];

if($log_id=='')
{
echo "<center>Please Select a record</center><br /><br />";
}

else 
{
$result = mysql_query("SELECT * FROM info_unit WHERE log_id=$log_id");
while($row = mysql_fetch_array($result))
{
$log_id=$row['log_id'];
$occupation=$row['occupation'];
}
?>
<select name='occupation'>
<option value=''>Select occupation</option>
 
<?php
 
  global $xoopsDB; 
  $result = mysql_query("SELECT * FROM ".$xoopsDB->prefix("myocc")."");

 while($row = mysql_fetch_array($result))
{
$occ_list=$row['occ_list'];
 
        if($occupation == '$occ_list') {
    
    echo "<option value='$occupation' selected>$occ_list</option>";
}
 
else
{   

echo "<option value='$occupation'>$occ_list</option>";
}  

}
?>

  </select>

[Jcbones]

Try this, you was missing some syntax in the mysql_query. I added some validation to it as well.

PHP Code:

<?php
$log_id= (isset($_POST['log_id']) && is_numeric($_POST['log_id'])) ? intval($_POST['log_id']) : die('You did not specify a valid log id number');

if($log_id=='')
{
echo "<center>Please Select a record</center><br /><br />";
}

else 
{
$result = mysql_query("SELECT * FROM `info_unit` WHERE `log_id`='$log_id'");
if(mysql_num_rows($result) > 0) {
$row = mysql_fetch_array($result);
$log_id=$row['log_id'];
$occupation=$row['occupation'];
}
else {
    echo 'The log you selected does not exist!';
}
?>
<select name='occupation'>
<option value=''>Select occupation</option>
 
<?php
 
  global $xoopsDB; 
  $result = mysql_query("SELECT * FROM ".$xoopsDB->prefix("myocc")."");

 while($row = mysql_fetch_array($result))
{
$occ_list=$row['occ_list'];
 
        if($occupation == '$occ_list') {
    
    echo "<option value='$occupation' selected>$occ_list</option>";
}
 
else
{   

echo "<option value='$occupation'>$occ_list</option>";
}  

}
?>

  </select>

[sarahmx]

Hi jcbones

thanks a lot for the validation...i'm still learning....

but its not working

no value is selected ..i only see 'select occupation'

yeah.....in the input code

PHP Code:

<select name='occupation'>
 <option value="">Sellect Occupation</option>
   <?php
 global $xoopsDB; 
 $result = mysql_query("SELECT * FROM ".$xoopsDB->prefix("myocc")."");
 
 while($row = mysql_fetch_array($result))
{
$occ_list=$row['occ_list'];
$occupation=$occ_list;
echo "<option value='$occupation'>$occ_list</option>";
}
?>


</select>

i forgot to put
$occupation=$occ_list; other occupation is empty

[Jcbones]

I'm sorry, I mis-understood the situation.



All it does is:

PHP Code:

isset() //Check to make sure a variable is set.
is_numeric() //check to make sure this is a number, but it could be a positive number(+) or a float(.);
intval() // will change a positive, negative, or float to a integer(whole number).
die() //Stops execution of the page, and prints out the string. synonym for exit(); 

So, it's working now?
Why is $occupation empty from the first query? Is it not a valid column in the table?
Why couldn't you write your option line as : <option>$occ_list</option>? If you don't specify a value, the option returns the text, and since they are the same...

[sarahmx]

Thanks

ok nevermind i solved this

i just changed the if statement and mysql query...

now i have to validate using jcbones code....

PHP Code:

<?php
$log_id=$_POST[log_id];

if($log_id=='')
{
echo "<center>Please Select a record</center><br /><br />";
}

else 
{
$result = mysql_query("SELECT * FROM info_unit WHERE log_id=$log_id");
while($row = mysql_fetch_array($result))
{
$log_id=$row['log_id'];
$occupation=$row['occupation'];
}
?>
<select name='occupation'>
<option value=''>Select occupation</option>
 
<?php
 
  global $xoopsDB; 


 while($row = mysql_fetch_array($result))
{
$occ_list=$row['occ_list'];
$occupation=$row['occupation'];
 
        if($occ_list == $occupation) {
    
    echo "<option value='$occupation' selected>$occ_list</option>";
}
 
else
{   
$occupation=$occ_list;
echo "<option value='$occupation'>$occ_list</option>";
}  

}
?>

  </select>