[SOLVED] mysql_fetch_array(): supplied argument is not a valid MySQL result resource

jnerotrix · December 7, 2008

Error i Get:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home3/sex1800/public_html/demos/search/search.php on line 13

Heres the code

<?php
$hostname_php_result_conn = "localhost";
$database_php_result_conn = "sex1800_phpresult";
$username_php_result_conn = "sex1800_demo";
$password_php_result_conn = "demotable";   
$php_result_conn = mysql_connect($hostname_php_result_conn, $username_php_result_conn, $password_php_result_conn);
mysql_select_db($database_php_result_conn,$php_result_conn);

if(isset($_POST['submit'])) {
   $roll_no = $_POST['roll_no'];
   
   $sql = mysql_query("SELECT * FROM result_table WHERE roll_no = $roll_no");
$row1 = mysql_fetch_array($sql);
$name = $row1['name'];
$marks = $row1['marks'];
   
   if($row1 == 0) {
      echo 'Error, ID does not exist';
   } else {

echo 'Name = '.$name.'<br />
Marks = '.$marks;

   }   
} else {
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html>
<head>

<style type="text/css">
#header { width:100%;
background:lightblue;
margin:0px;
padding:0px;
}

#leftbar { float:left;
width:40%;
height:30px;
background:lightblue;
text-align:right;
}

#rightbar { float:left;
width:60%;
height:30px;
background:lightblue;
text-align:center;
}

#bar { width:100%;
background:lightblue;
}

#center {float:center;
width:60%;
height:10px;
background:lightblue;
align:center;
}

a { font-family:Arial;
font-size:16px;
background-color:aqua;
color:blue;
font-weight:bold;
vertical-align:text-bottom;
}

a:hover { font-family:Lucida Calligraphy;
font-size:ahoversize1;
background-color:yellow;
color:darkblue;
}
</style>

</head>
<center>
<body>
<br><Br><Br>
<h1> Search Roll Number </h1>

<Br><br>
<div id="header">
<div id="center">
<table>
<form action="search.php" method="post">
<tr>
<td>Roll Number:</td><td><input name="roll_no" type="text" id="roll_no" /></td>
</tr>
<tr>
<td align="center"><input type="submit" name="submit" value="search" /></td>
</tr>
</form>
</table>
</div>
<div id="leftbar">
</div>
<div id="rightbar">
</div>
</div>
<div id="bar">
</div>
<div id="header">
<div id="leftbar">
</div>
<div id="rightbar">
</div>
</div>


</body>
</html>
<?php
}
?>

MatthewJ · December 7, 2008

$sql = mysql_query("SELECT * FROM result_table WHERE roll_no = $roll_no") or die (mysql_error());

May shed some light on the issue. You could also do the same for the db selection and connection if the query looks okay.

jnerotrix · December 7, 2008

Now it Just Says "No database selected"

----------

Heres The Code:

Added Your Line and error_reporting(E_ALL);

<?php
error_reporting(E_ALL);

$hostname_php_result_conn = "localhost";
$database_php_result_conn = "sex1800_phpresult";
$username_php_result_conn = "sex1800_demo";
$password_php_result_conn = "demotable";   
$php_result_conn = mysql_connect($hostname_php_result_conn, $username_php_result_conn, $password_php_result_conn);
mysql_select_db($database_php_result_conn,$php_result_conn);

if(isset($_POST['submit'])) {
   $roll_no = $_POST['roll_no'];
   
   $sql = mysql_query("SELECT * FROM result_table WHERE roll_no = $roll_no") or die (mysql_error());
$row1 = mysql_fetch_array($sql);
$name = $row1['name'];
$marks = $row1['marks'];
   
   if($row1 == 0) {
      echo 'Error, ID does not exist';
   } else {

echo 'Name = '.$name.'<br />
Marks = '.$marks;

   }   
} else {
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html>
<head>

<style type="text/css">
#header { width:100%;
background:lightblue;
margin:0px;
padding:0px;
}

#leftbar { float:left;
width:40%;
height:30px;
background:lightblue;
text-align:right;
}

#rightbar { float:left;
width:60%;
height:30px;
background:lightblue;
text-align:center;
}

#bar { width:100%;
background:lightblue;
}

#center {float:center;
width:60%;
height:10px;
background:lightblue;
align:center;
}

a { font-family:Arial;
font-size:16px;
background-color:aqua;
color:blue;
font-weight:bold;
vertical-align:text-bottom;
}

a:hover { font-family:Lucida Calligraphy;
font-size:ahoversize1;
background-color:yellow;
color:darkblue;
}
</style>

</head>
<center>
<body>
<br><Br><Br>
<h1> Search Roll Number </h1>

<Br><br>
<div id="header">
<div id="center">
<table>
<form action="search.php" method="post">
<tr>
<td>Roll Number:</td><td><input name="roll_no" type="text" id="roll_no" /></td>
</tr>
<tr>
<td align="center"><input type="submit" name="submit" value="search" /></td>
</tr>
</form>
</table>
</div>
<div id="leftbar">
</div>
<div id="rightbar">
</div>
</div>
<div id="bar">
</div>
<div id="header">
<div id="leftbar">
</div>
<div id="rightbar">
</div>
</div>


</body>
</html>
<?php
}
?>

jnerotrix · December 7, 2008

Heres the page

http://1800sexnow.com/demos/search/search.php

type in a number and click search

jnerotrix · December 7, 2008

I have changed the code i added some javascript thats all

it disallows users to type text only numbers in form field

heres the new code but i still get the same error

<?php
error_reporting(E_ALL);

$hostname_php_result_conn = "localhost";
$database_php_result_conn = "sex1800_phpresult";
$username_php_result_conn = "sex1800_demo";
$password_php_result_conn = "demotable";   
$php_result_conn = mysql_connect($hostname_php_result_conn, $username_php_result_conn, $password_php_result_conn);
mysql_select_db($database_php_result_conn,$php_result_conn);

if(isset($_POST['submit'])) {
   $roll_no = $_POST['roll_no'];
   
   $sql = mysql_query("SELECT * FROM result_table WHERE roll_no = $roll_no") or die (mysql_error());
$row1 = mysql_fetch_array($sql);
$name = $row1['name'];
$marks = $row1['marks'];
   
   if($row1 == 0) {
      echo 'Error, ID does not exist';
   } else {

echo 'Name = '.$name.'<br />
Marks = '.$marks;

   }   
} else {
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html>
<head>

<style type="text/css">
#header { width:100%;
background:lightblue;
margin:0px;
padding:0px;
}

#leftbar { float:left;
width:40%;
height:30px;
background:lightblue;
text-align:right;
}

#rightbar { float:left;
width:60%;
height:30px;
background:lightblue;
text-align:center;
}

#bar { width:100%;
background:lightblue;
}

#center {float:center;
width:60%;
height:10px;
background:lightblue;
align:center;
}

a { font-family:Arial;
font-size:16px;
background-color:aqua;
color:blue;
font-weight:bold;
vertical-align:text-bottom;
}

a:hover { font-family:Lucida Calligraphy;
font-size:ahoversize1;
background-color:yellow;
color:darkblue;
}
</style>

<script type="text/javascript">

function numbersonly(myfield, e, dec) {
  var key;
  var keychar;

  if (window.event)
    key = window.event.keyCode;
  else if (e)
    key = e.which;
  else
    return true;
  keychar = String.fromCharCode(key);

  // control keys
  if ((key==null) || (key==0) || (key== || (key==9) || (key==13) || (key==27) )
    return true;

  // numbers
  else if ((("0123456789").indexOf(keychar) > -1))
    return true;

  // decimal point jump
  else if (dec && (keychar == ".")) {
    myfield.form.elements[dec].focus();
    return false;
  } else
    return false;
}

</script>


</head>
<center>
<body>
<br><Br><Br>
<h1> Search Roll Number </h1>

<Br><br>
<div id="header">
<div id="center">
<table>
<form action="search.php" method="post">
<tr>
<td>Roll Number:</td><td><input name="roll_no" type="text" id="roll_no" onkeypress="return numbersonly(this, event)" /></td>
</tr>
<tr>
<td align="center"><input type="submit" name="submit" value="search" /></td>
</tr>
</form>
</table>
</div>
<div id="leftbar">
</div>
<div id="rightbar">
</div>
</div>
<div id="bar">
</div>
<div id="header">
<div id="leftbar">
</div>
<div id="rightbar">
</div>
</div>


</body>
</html>
<?php
}
?>

premiso · December 7, 2008

Are you sure you have a database called sex1800_phpresult ?

jnerotrix · December 7, 2008

i feel stupid i mistyped the database name is

sex1800_phpresults

Philip · December 7, 2008

Also, make sure an ID was input, otherwise instead of a nice error message they'll get this mess:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

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[SOLVED] mysql_fetch_array(): supplied argument is not a valid MySQL result resource

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