Error in mysql fetch error in php

Question

Hi when i run my code this proplem apper

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\Program Files (x86)\EasyPHP-5.3.5.0\www\List.php on line 28

this is my code

<?php
session_start();
?>

<html>
<link rel='stylesheet' type='text/css' href='Style.css'>
<title>Registration</title>
<body>



<?php
$S=$_POST['Sec'];
echo "<H4>Web Development a</H4>";
echo "<b>CS  <br>Registration Page </b><br><br><br><br>";
echo "<b>This is the students lists who are registered in section $S: <br><br>";
echo "<table border=1 width=50%><tr><td width='6'></td><td bgcolor='#E66C2C' width='150'><b><center><font 
color='#FFFFFF'>Name</center></td><td bgcolor='#E66C2C' width='100'><b><center><font 
color='#FFFFFF'>ID</center></td>". "<td bgcolor='#E66C2C' width='100'><b><center><font color='#FFFFFF'>E-Mail</center></td></tr>";


include('con_db.php');

$sql = "select * from students where Sec=$S ";
$result = mysql_query($sql);
$i=0;

while ($r = mysql_fetch_array($result)) {
$i++;
echo "<tr><td bgcolor='#eae5a7' width='6'>";
            echo "<b><center>".$i."</center><br>";
echo"</td>";
echo "<td  width='150'>";
            echo "<b><center>".$r[Fname]."</center><br>" ;
echo "</td>";
echo "<td width='100'>";
            echo "<b><center>".$r[ID]."</center><br>" ;
echo "</td>";
echo "<td width='100'>";
            echo "<b><center>".$r[mail]."</center><br>" ;
echo "</td></tr>";
echo "</font>";
}

echo"</table>";
echo "<form name='form' method='post' action='Registration_List.php'>";
?>

<br><br><center><input type='submit' id='send' value='Back'  style="color: #FFFFFF; background-color: #E66C2C; border-width: 1; border-style: 1" ></form>

<?php
echo "<form name='form' method='post' action='ass1.php'>";
?>
<input type='submit' id='send' value='Home Page'  style="color: #FFFFFF; background-color: #E66C2C; border-width: 1; border-style: 1"></form></center></td></tr></table>


</body>
</html>

Don't forget that you can mark an answer as accepted. You only need to click on the tick icon. That shows appreciation for the effort. — Álvaro González
– Álvaro González, Commented May 6, 2011 at 11:35
Homework: Try typing <script>top.location="http://www.google.com"</script> in your form and, once you fix the SQL query, ' OR '1'='1. — Álvaro González
– Álvaro González, Commented May 6, 2011 at 11:37
possible duplicate of mysql_fetch_array() expects parameter 1 to be resource, boolean given in select — Jürgen Thelen
– Jürgen Thelen, Commented Aug 3, 2012 at 9:37
mysql_query has been deprecated since version 5.5.0. Better to use mysqli_query that supports prepared statements and therefore is much more secure. — Paul-Beyond
– Paul-Beyond, Commented Jun 23, 2017 at 12:34

Emyr · Accepted Answer · 2011-05-06 11:26:21Z

2

rewrite so this:

$result = mysql_query($sql);
$i=0;

while ($r = mysql_fetch_array($result)) {

becomes this:

if($result = mysql_query($sql)){
    $i=0;

   while ($r = mysql_fetch_array($result)) {
   ...
   }
} else {
   //do something with mysql_error()
}

answered May 6, 2011 at 11:26

Emyr

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1 Comment

MANAL Over a year ago

while ($r = mysql_fetch_array($result)) { if ($r["Sec"]==1 && $s != 1) $s--; if ($r["Sec"]==2 && $s != 1) $u--; if ($r["Sec"]==3 && $s != 1) $m--; if ($r["Sec"]==4 && $s != 1 ) $t--; if ($r["Sec"]==5 && $s != 1) $w--; } ?>

JohnP · Accepted Answer · 2011-05-06 11:26:20Z

1

Your mysql_query($sql) call is probably returning false. You need to see if your script managed to connect to your mysql server

From the manual : http://php.net/manual/en/function.mysql-query.php

if (!$result) {
    die('Invalid query: ' . mysql_error());
}

answered May 6, 2011 at 11:26

JohnP

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3 Comments

Jeff Parker Over a year ago

+1, but a query error would also cause a false return, so check the query as well as the connection. a swift echo type call to mysql_error() after mysql_query() should do the trick.

JohnP Over a year ago

@Jeff Parker I was actually editing my post when you commented :)

Jeff Parker Over a year ago

No problem ... and you've already had your +1. Don't eat it all at once :)

Robik · Accepted Answer · 2011-05-06 11:26:52Z

0

Basically Error like Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given.. means that your query result is wrong. mysql_query instead of returning result returns false, which means your query is wrong.

Simple check what's wrong using:

$sql = "select * from students where Sec=$S ";
#                               V ADDED
$result = mysql_query($sql) or die( mysql_error() );

answered May 6, 2011 at 11:26

Robik

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Comments

Ibu · Accepted Answer · 2011-05-06 11:56:41Z

0

The error means you had an error in your sql, make sure you escape it before using it because you might become victim of sql injection.

$sql = "SELECT * FROM students WHERE seconds=%d";
$sql = sprintf($sql,mysql_real_escape_string($S));

make sure you have the correct data in $S;

$result = mysql_query($sql) or die("Error: ".mysql_error());

and this way you will know what is your exact error.

edited May 6, 2011 at 11:56

answered May 6, 2011 at 11:30

Ibu

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