why it is not working

Discussion in 'PHP' started by anchitjindal07, Sep 26, 2009.

  1. anchitjindal07

    anchitjindal07 New Member

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    $selct="SELECT * FROM stu_info";

    $result=mysql_query($selct,$status);
    while($row = mysql_fetch_array('$result',MYSQL_BOTH))
    {
    echo $row['Name'],$row['Father s Name'],$row['Address'],$row['Date of birth'],$row['Course'],$row['Branch'],$row['Batch'],$row['Roll No.'],$row['Univ. Reg. No.'],$row['Gender'];
    echo "<br />";
    }

    Above code is giving following Warning:

    Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\EasyPHP 3.0\www\process.php on line 44

    Plz help
     
  2. nimesh

    nimesh New Member

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  3. anchitjindal07

    anchitjindal07 New Member

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    But the same warning is still there
     
  4. nimesh

    nimesh New Member

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    Sorry, I don't know php, so can't help much

    Maybe some expert would be able to help you.

    Shabbir: Can you check this one?
     
  5. shabbir

    shabbir Administrator Staff Member

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    $result is a variable which is passed in quotes for the function. Try removing the single quotes and see what happens.
     

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