[SOLVED] Warning: Supplied argument is not a valid MySQL result resource in c:\apache\htd

[beboo002]

i con't understand this problem

 

<?php

if($_POST['Submit2']!="") {

$link=mysql_connect("localhost","root","");

if(!$link) die("can not connect to mysql");

mysql_select_db("timesheet",$link);

/////////////////////////////////////////fetch/////////////////////////////

$sql2="select firstName,lastname,address,officeemail,otheremail,phoneno,mobileno,panno,bloodgroup,department,designation,manager,dateofjoining from employeeinfo where name ='".$_POST['text2']."'";

$result2=mysql_query($sql2,$link);

//$query1=mysql_fetch_array($result1);

echo "<table border=1 align = center>

<tr>

<th>firstName</th>

<th>lastname</th><th>address</th>

<th>officeemail</th><th>otheremail</th>

<th>phoneno</th><th>mobileno</th>

<th>panno</th><th>bloodgroup</th><th>deportment</th>

<th>designation</th><th>manager</th><th>dateofjoining</th></tr>";

while($row1 = mysql_fetch_array($result2))

{

  echo "<tr><td>";

  echo $row1['firstName'];

    echo "</td>";

echo "<td>";

  echo $row1['lastname'];

echo "<td>";

  echo $row1['address'];

echo "<td>";

  echo $row1['officeemail'];

echo "<td>";

  echo $row1['otheremail'];

echo "<td>";

  echo $row1['phoneno'];

echo "<td>";

  echo $row1['mobileno']; 

echo "<td>";

  echo $row1['panno'];

echo "<td>";

echo $row1['bloodgroup'];

 

echo "<td>";

  echo $row1['department'];

echo "<td>";

  echo $row1['designation'];

echo "<td>";

echo $row1['manager'];

 

echo "<td>";

  echo $row1['dateofjoining'];

 

    echo "</td></tr>";

  }

echo "</table>";

 

mysql_close($link);

}

?>

[AndyB]

Change:

$result2=mysql_query($sql2,$link);

 

to:

$result2=mysql_query($sql2,$link) or die("Error: ". mysql_error(). " with query ". $sql2);

 

And if it's not obvious what needs to be fixed, post the result of that change here.

[beboo002]

thanks for reply

warning is remove but

there is no data shown in table.

[AndyB]

Do you mean you no longer get the 'not a valid ...' error message?  I assume means that you corrected something in your script. If so, we need to see what you now have assuming you mean the script works but does not display any data even though you have data in the table.

 

If I misunderstood, what's the present problem?

[beboo002]

i hav emplyee info table in mysql i want to fetch the record in table and store something like this

firstName lastname address officeemail otheremail phoneno mobileno panno bloodgroup deportment designation manager dateofjoining

           

table is printing but record are not found

thats my problem where goes my record for above code i hav change in my where cluse officeemail insted of name.

[AndyB]

After you define your query string ($sql2), add a new line of code:

 

echo $sql2;

 

Then you'll know for certain just what query is being passed to the database.  If that doesn't make the solution obvious, post what output it gives.