Essentially I'm attempting to quasi mimic the effect of phpMyAdmin. When you activate the checkbox after you click submit it will return to the previous page to re-enter another item
Input Page
PHP Code:
<center>
<form action="secured.php">
<input type="hidden" name="loc" value="cars">
<input type="hidden" name="sloc" value="choose_model">
<input type="hidden" name="s2loc" value="create_model">
<input type="hidden" name="manu" value="<?echo $manu?>">
<input type="text" name="new_model" class="form"><br>
<div class="main">Insert New Model After Insert
<input type="checkbox" name="insertnew" <?echo $insert_again?> value="setnew" class="form">
</div>
<input type="submit" value="Create New Model" class="button_nowidth">
</form>
</center>
And.. The result page. It enters the information into the DB then determines if the checkbox was checked then includes a "meta" tag to refresh back to the previous page... But, php refuses to accept the information passed to it in the check box... I didn't want to clean up my debugging code b/c it shows what I have already tried to get the variable to show up in the second page
PHP Code:
$sql = "INSERT INTO `car_model` VALUES ('', '$manu', '$new_model')";
//$res = mysql_query($sql);
echo "<br>"."--".$insertnew."--";
if (isset($insertnew))
{
?><META HTTP-EQUIV=Refresh CONTENT="0; URL=secured.php?loc=cars&sloc=choose_model&manu=$manu&insert_again=checked"><?
}
if ($insertnew == true)
echo "testing";
if ($insertnew)
echo "boolean";
if ($insertnew == "on")
echo "on";
if ($insertnew == "setnew")
echo "setnew";
When run... Nothing is echoed by the if statements and $insertnew echo's blank...
The page output at the echo "--".$insertnew."--" prints out as "----"....
Anyone have any clues about to why this isn't working?
I can provide for enviroment details about whats going on if need be...