php form validation returns mismatched error

Question

I am new to php and this is my first post here, so new to writing post also. I am using a script to vaidate a form input. Part of the script is as below:

$data=$_POST;
print_r($data); 

if(($data['dm']="No") and (!empty($data['dmsince']))):
    $error=$error." dm mismatch N";  
endif;  

if(($data['dm']="Yes") and (empty($data['dmsince']))):
    $error=$error." dm mismatch Y";  
endif;

if ($error):
    print $error;
    $error="";   
    print $form;
else:
    print "OK";

I am accepting $_POST['dm'] through select statement either as yes or no.

The problem that occurs is when the form is processed:

$_POST['dm']='No' and $_POST['dmsince'] is blank. I get error as 'dm mismatch Y'.

$_POST['dm']=Yes and $_POST['dmsince']='some valid date' I get error as 'dm mismatch N'.

Logically in both these cases I should not get any error. It appears that it is entering one of the loops forcibly. I have also tried with elseif with no result. What is wrong with this code?

$data['dm']=="No" and use {} brackets if(2 > 1) { ... }else{ ... } — Peter
– Peter, Commented Sep 7, 2012 at 16:34

Mike Brant · Accepted Answer · 2012-09-07 16:33:58Z

3

Update your conditionals to actually use conditional comparison operators (i.e. == or ===). Right now you are assigning $data['dm'] to equal "No" in the first conditional and "Yes" in the second.

You might also want to get in the habit of writing your comparisions like this:

if(true === $var)
if("Yes" === $var2)

With the item you are comparing the variable to on the left side. In this manner if you accidentally did something like:

if(true=$var)

PHP would throw an error rather than a case like this

if($var=true)

Which would set $var = true and always evaluate the conditional as true.

It makes debugging your code a lot easier.

answered Sep 7, 2012 at 16:33

Mike Brant

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3 Comments

Ratnam Mani Over a year ago

Thanx. I tried and my code now is: if(("No"===$data['dm']) and (!empty($data['dmsince']))): $error=$error." dm mismatch N"; endif; if(("Yes"===$data['dm']) and (empty($data['dmsince']))): $error=$error." dm mismatch Y"; endif; But now I have exactly opposite situation. Though an error condition is fulfilled, it does not report error at all. It is not entering any of the loops.

Mike Brant Over a year ago

@RatnamMani You may want to add a sample of var_dump($_POST) to your question to give more context. Without knowing what sort of values you are posting which are causing unexpected results, it is hard to troubleshoot any farther.

Ratnam Mani Over a year ago

I do not know how, but it got solved thus: Prior to checking consistency between $data['dm'] and $data['dmsince'], I was checking the empty string with if(!$data['dm']||$data['dm']=''): and capturing the error. I changed it to if(empty($data['dm']): and with that everything is fine. Thanks a lot.

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