php same code not returning value

Question

my php code not executing correctly , i am echoing a category and it is echoing perfectly but when i am freeing my variable i am getting error saying mysqli_free_result() expects parameter 1 to be mysqli_result, null given ....

<li><a href="#" > <?php while($result3 = mysqli_fetch_array($row2)){ 
echo $result3['category'] ; } ?> </a></li>
<?php mysqli_free_result($result3) ; ?>

and aditionally none of my php code below this is exectuing , its just returning no data (page source is giving no data )

what mistake am i doing ?

you should free the $row2

talsibony
– talsibony

2015-02-04 13:26:14 +00:00
Commented Feb 4, 2015 at 13:26 — talsibony
– talsibony, Commented Feb 4, 2015 at 13:26

talsibony · Accepted Answer · 2015-02-04 13:26:43Z

2

instead of

<?php mysqli_free_result($result3) ; ?>

do this

<?php mysqli_free_result($row2) ; ?>

answered Feb 4, 2015 at 13:26

talsibony

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Joe Swindell · Accepted Answer · 2015-02-04 13:27:04Z

0

Your mysqli_result should be the variable you are using to store the query. IE:

$result=mysqli_query($con,$sql);

You'd want to

mysqli_free_result($result);

By the looks of your code, you should free $row2

answered Feb 4, 2015 at 13:27

Joe Swindell

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Ghanshyam Dekavadiya · Accepted Answer · 2015-02-04 13:35:24Z

0

Remove extra statement from code and add it inside the loop

<li><a href="#" > 
  <?php while($result3 = mysqli_fetch_array($row2)){ 
       echo $result3['category'] ; 
       mysqli_free_result($row2) ;
   } ?> </a></li>

answered Feb 4, 2015 at 13:35

Ghanshyam Dekavadiya

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