Keep getting errmsg array

Discussion in 'PHP' started by mark103, Oct 20, 2010.

  1. #1
    Hey guys,

    I need your help. It took me hours to try and fix the problem as I am still trying to get the data from the mysql database to print into the php page, but I keep getting the error message of the array which it said "data is not found".


    
    <?php
    session_start();
        define('DB_HOST', 'localhost');
        define('DB_USER', 'username');
        define('DB_PASSWORD', 'password');
        define('DB_DATABASE', 'databasename');
           
        $errmsg_arr = array();
        $errflag = false;
    
        $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
        if(!$link) {
      die('Failed to connect to server: ' . mysql_error());
        }
    
        $db = mysql_select_db(DB_DATABASE);
        if(!$db) {
      die("Unable to select database");
        }
    
       function clean($var){
    
    return mysql_real_escape_string(strip_tags($var));
        }
      
        $datastrings = clean($_GET['strings1']);
        if($datastrings== '') {
      $errmsg_arr[] = 'data is not found';
      $errflag = true;
        }
        if($errflag) {
      $_SESSION['ERRMSG_ARR'] = $errmsg_arr;
      echo implode('<br />',$errmsg_arr);
       }
       else {
      $query  = "SELECT data_strings FROM table1";
      $result=mysql_query($query) or die('Error:<br />' . $qry . '<br />' . mysql_error());
    
    
    while ($row = mysql_fetch_array($result)) { 
      echo "<p id='data1'>";
      echo $row['data1'] . "</p>";
      }
     }
    ?>
    
    PHP:

    I am sure that the column in the database are valid. The name of the column is data_strings and the table name is called table1.


    Any idea why I keep getting the error message of the array?
     
    mark103, Oct 20, 2010 IP
  2. ThePHPMaster

    ThePHPMaster Well-Known Member

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    #2
    First, are you passing the string1 variable in the URL, as in:

    http://site.com/page.php?strings1=stringyouwant

    Second, your database query is static, never changes. Is that how you want it or do you actually want to use the variable $datastrings in place of the static field data_strings as in:

    
    $query  = "SELECT $datastrings FROM table1";
    
    PHP:
     
    ThePHPMaster, Oct 20, 2010 IP