better.in.design Posted November 14, 2010 Share Posted November 14, 2010 This is just a snippet of my code. The php script itself is called test.php and so it calls itself once the form is submitted. I keep having problems retrieving the data back correctly; I am testing in retrieving the data on the same script from the same page before retrieving the POST data from another webpage. If $decimalSum is a variable that assigns a value that is hard-coded then I would be able to retrieve the same value. However, if the value is computed I can not retrieve it unless I click on the "Find" button twice. This is such a strange behavior and I don't know why. Does anyone have any suggested solutions to this? <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title></title> </head> <body> <?php echo '<form enctype="multipart/form-data" method="post" action="test.php">'; echo ' <label for="_firstName">First name : </label>'; echo ' <input type="text" id="_firstName" name ="_firstName">'; echo ' <input type="checkbox" value="1" name="_firstNameChecked"/>'; echo ' <label for="_middleName">Middle name : </label>'; echo ' <input type="text" id="_middleName" name ="_middleName">'; echo ' <input type="checkbox" value="1" name="_middleNameChecked" />'; echo ' <label for="_lastName">Last name : </label>'; echo ' <input type="text" id="_lastName" name ="_lastName">'; echo ' <input type="checkbox" value="1" name="_lastNameChecked" />'; echo '<br />'; echo '<input type="submit" name="Find" value="Find" />'; $firstNameChecked = intval(($_POST['_firstNameChecked'])); $middleNameChecked = intval(($_POST['_middleNameChecked'])); $lastNameChecked = intval(($_POST['_lastNameChecked'])); $decimalSum = (int)((2*2*2)*$firstNameChecked + (2*2)*$middleNameChecked + (2*1)*$lastNameChecked); //$decimalSum = 28; echo '<br />'; echo '$decimalSum = ' . $decimalSum . '<br />'; echo '<input type="hidden" name="_decimalSum" value = "' . $decimalSum . '" />'; $decimalSum2 = ($_POST['_decimalSum']); echo '$decimalSum2 = ' . $decimalSum2 . '<br />'; echo '</form>'; echo '</body>'; echo '</html>'; ?> Quote Link to comment Share on other sites More sharing options...
ohdang888 Posted November 14, 2010 Share Posted November 14, 2010 have you tried debugging by putting this code before the page starts to see what all its receiving.... echo "<pre>"; print_r($_POST); echo "</pre>"; Quote Link to comment Share on other sites More sharing options...
better.in.design Posted November 15, 2010 Author Share Posted November 15, 2010 I've tried that and I see the value for $decimalSum is 0 after submitting the form with the first name field filled with a name. It seems to imply that there is something wrong on the line where $decimalSum is computed. This doesn't make sense to me since the echo statement right after it's computation shows its value. For some reason, this value is not sent to POST. If I resend the same checked field again, the value goes through. Do you know something I don't? Quote Link to comment Share on other sites More sharing options...
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