MULTI SEARCH PHP

Question

Can anyone help me to figure out the error in my code? i want to display data through two input search..my code goes like these.

<table>
    <tr>
        <td>Studid</td>
        <td>Course</td>
    </tr>

 <?php
    include ("connect.php");


    if(isset($_POST['submit']))

    {
    $studno=$_POST['idsearch'];
    $scourse=$_POST['coursesearch'];
    $sql=mysql_query("SELECT * FROM cfnr WHERE studid= ".$studno." AND course=".$scourse."");

    }

    ?>


    <?php
    while($row=mysql_fetch_array($sql))

    {


 }  

?> 
  <tr>
        <td><?php echo $row['studid'];?></td>
        <td><?php echo $row['course'];?></td>
    </tr>



</table>

i got this error in my screen.

"mysql_fetch_array() expects parameter 1 to be resource, boolean given in "

thanks! :)

ignoring the deprecated library (use `mysqli_*` or PDO) and the SQL Injection vulnerability the error tells you the problem. `$sql` is a boolean. that means there's something wrong with your connection or your query. — Memor-X, Aug 31 '16 at 04:00
Check $result before passing it to mysql_fetch_array. You'll find that it's false because the query failed — Drop Shadow, Aug 31 '16 at 04:00
Possible duplicate of [mysqli\_fetch\_array()/mysqli\_fetch\_assoc()/mysqli\_fetch\_row() expects parameter 1 to be resource or mysqli\_result, boolean given](http://stackoverflow.com/questions/2973202/mysqli-fetch-array-mysqli-fetch-assoc-mysqli-fetch-row-expects-parameter-1) — scrowler, Aug 31 '16 at 04:01

Indrasis Datta · Answer 1 · 2016-08-31T04:09:00.110

-1

You should use a bit of debugging. Try this:

$sql   =  "SELECT * FROM cfnr WHERE studid= '".$studno."' AND course = '".$scourse."'";
$query =  mysql_query($sql) or die(__LINE__."has an error: ".mysql_error());  // Gives an error if there's a syntax error

$row = mysql_fetch_array($query);

echo "<pre>";
print_r($row);
exit;

while ($row = mysql_fetch_array($query)) {
     /* Other code */
}

edited Aug 31 '16 at 04:09

answered Aug 31 '16 at 03:58

Indrasis Datta

8,692
2
14
32

i tried it. but it says that my "query was empty". what will i do with that? do i have error in my select query? – Oman Reyes Dela Cruz Aug 31 '16 at 04:02
Where exactly is it showing that? Okay, let me modify my answer and let me know what you get now. – Indrasis Datta Aug 31 '16 at 04:07
when i run it in my page. ok tnx. – Oman Reyes Dela Cruz Aug 31 '16 at 04:09

score -1 · Answer 2 · answered Aug 31 '16 at 04:14

-1

Replace your sql with below line :

$sql=mysql_query("SELECT * FROM cfnr WHERE studid= ".$studno." AND course=".$scourse."") or die(mysql_error());

answered Aug 31 '16 at 04:14

Ravi Roshan

570
3
14

i did it sir but it displays the same error, undefined variable in while($row=mysql_fetch_array($sql)) and mysql_fetch_array() expects parameter 1 to be resource, null given in. – Oman Reyes Dela Cruz Aug 31 '16 at 04:18

score -1 · Answer 3 · answered Aug 31 '16 at 04:56

-1

The issue is your `$sql variable is within the if condition and if your if condition fails there is no scope of $sql outside of if condition

Instead of

while($row=mysql_fetch_array($sql))

Use

while($row=@mysql_fetch_array($sql))

you will not get this error.

answered Aug 31 '16 at 04:56

PBajpai

1
2

and also put your section within the while loop and your while loop within the if condition – PBajpai Aug 31 '16 at 04:58
i did it nothing displays in my page. – Oman Reyes Dela Cruz Aug 31 '16 at 05:23

score -1 · Answer 4 · answered Aug 31 '16 at 05:58

-1

("SELECT * FROM cfnr WHERE studid= '$studno' AND course ='$scourse.'") or die(mysql_error());

answered Aug 31 '16 at 05:58

castaway13

33
4

FullStack · Answer 5 · 2016-09-01T05:59:16.760

Change the code :

while($row=mysql_fetch_array($sql))

to

while($row=mysql_fetch_assoc($sql))

and also change this code:

 $sql=mysql_query("SELECT * FROM cfnr WHERE studid= ".$studno." AND course=".$scourse."");

to

$sql=mysql_query("SELECT * FROM cfnr WHERE studid= '$studno' AND course='$scourse'");

Not only that, the close brace of the isset() should be on before your (?>) last closing php tag.

Change the code :

include ("connect.php");

to

include 'connect.php';

MULTI SEARCH PHP

5 Answers5