mysqli dropdown

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  • 12Strings
    New Member
    • May 2014
    • 12
    #1

    mysqli dropdown

    I'm attempting a mysqli dropdown to select and print one record from table and update one
    field, ("lastused" ($currdate))in that record. I get the following errors: Any help?

    Notice: Undefined index: lastused in C:\xampp\htdocs \home\emaildrop .php on line 109

    Notice: Undefined index: target in C:\xampp\htdocs \home\emaildrop .php on line 110
    target Total results: 0

    Warning: mysql_fetch_arr ay() expects parameter 1 to be resource, object given in C:\xampp\htdocs \home\emaildrop .php on line 141
    target username password emailused lastused purpose saved
    could not retrieve data from database

    Code:
    <?php
 $db = new mysqli('localhost', 'root', 'cookie', 'homedb');
if($db->connect_errno > 0)
    {die('Unable to connect to database [' . $db->connect_error . ']');}

$lastused = $_POST['lastused']; // 108 ****************************
$target = $_POST['target']; // 109 *********************************

$currdate = date('Y-m-d');	
$lastused = $currdate;
if($target=="email"){$target=$username;}
//echo "lastused $lastused";
echo "target $target";
$sql = <<<SQL
SELECT * FROM `oocust` WHERE 'target' = '$target'
SQL;
if(!$result = $db->query($sql))
    {die('There was an error running the query [' . $db->error . ']');}
while($row = $result->fetch_assoc()) 
    {echo $row['username'] . '<br />';}
echo 'Total results: ' . $result->num_rows;

$result->free(); 
$db->escape_string('This is an unescape "string"');
$db->close();
?>
<center>
  <table cellspacing=2 cellpadding=0 border=1>
   <tr> 
    <th bgcolor="#ccffff">target</TH>      
    <th bgcolor="#ccffff">username</TH> 
    <th bgcolor="#ccffff">password</TH>
    <th bgcolor="#ccffff">emailused</TH>
    <th bgcolor="#ccffff">lastused</TH>
    <th bgcolor="#ccffff">purpose</TH>
    <th bgcolor="#ccffff">saved</TH>
    <?php
while($data = mysql_fetch_array($result)) // 139 ******************************* 
    {
echo'<tr>'; 
// printing table row
echo '
<td>'.$data['target'].'</td>
<td>'.$data['username'].'</td>
<td>'.$data['password'].'</td>
<td>'.$data['emailused'].'</td>
<td>'.$data['lastused'].'</td>
<td>'.$data['purpose'].'</td>
<td>'.$data['saved'].'</td>'; 
// looping all data to be printed till last row in the table
echo'</tr>'; 
// closing table row
   }
echo '</table>';  //closing table tag
mysql_query("UPDATE emailtbl SET lastused=$lastused"); 
$result=mysql_query("select lastused from emailtbl") or die ("could not retrieve data from database");
$data=mysql_fetch_assoc($result);
echo 'Total rows updated: ' . $db->affected_rows;
echo "lastused ".$data['lastused'];
mysql_close();
   ?>
    </table></center></body></html>
    Tags: None
    • Dormilich
      Recognized Expert Expert
      • Aug 2008
      • 8694
      #2
      error #1: if you request the script using GET, you don’t have POST data. hence never assume that the data are always there.

      tip: use filter_input()

      error #2: cf. error #1

      error #3: you cannot mix mysql_* functions and MySQLi.

        Comment

        • 12Strings
          New Member
          • May 2014
          • 12
          #3
          this is my current code. clicking "submit" gets no result?

          Code:
          <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" 
    "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> 
    <html xmlns="http://www.w3.org/1999/xhtml"> 
<html><body>
<form name="form" action="" method="post">
Select email to view<SELECT name=target>
<OPTION class=highlight value=email selected>owenspaulette59@yahoo
<OPTION class=highlight value=paypal>paypal 
<OPTION class=highlight value=ebay>ebay
</SELECT><p>
<CENTER><input type="submit" name="submit" value="THEN submit choice"><p>
          Code:
          <?php
     $target = (isset($_POST['target']))?1:0;    // this is my target
     $payrec = (isset($_POST['payrec']))?1:0;
 $lastused = (isset($_POST['lastused']))?1:0;
   

$id = (isset($_POST['id'])) ? mysqli_real_escape_string($dbconnect, $_POST['id']) : '';   
$dbconnect = mysqli_connect('localhost','root','cookie')or die(mysqli_error($dbconnect));
$result = mysqli_query($dbconnect, "SELECT * FROM emailtbl where payrec='P'");
    if($result === FALSE) { die(mysql_error()); }
    echo date('m/d/y');
  ?> 
  <table cellspacing=0 cellpadding=0 border=1>
   <tr>  
  <th bgcolor="#7FFF2A">target</th>
  <th bgcolor="#7FFF2A">username</th>
  <th bgcolor="#7FFF2A">password</th>
  <th bgcolor="#7FFF2A">emailused</th>
  <th bgcolor="#7FFF2A">lastused</th>
  <th bgcolor="#7FFF2A">purpose</th>
  <th bgcolor="#7FFF2A">saved</th> 
  </tr>
<?php
// $lastused = $row['CURDATE()']; 
$lastused = date('Y-m-d');  
 while($row = mysql_fetch_array($result))  
   { 
echo "<tr>";
echo "<td>" . $row['target'] . "</td>";
echo "<td>" . $row['username'] . "</td>";
echo "<td>" . $row['password'] . "</td>";
echo "<td>" . $row['emailused'] . "</td>";
echo "<td>" . $row['lastused'] . "</td>";
echo "<td>" . $row['purpose'] . "</td>";
echo "<td>" . $row['saved'] . "</td>";
 echo "</tr>";
 echo "</table>
<input type='submit' name='update' value='submit' />
</form>";
if (!empty($_POST['update_lastused']))
  { $update = mysqli_query($dbconnect, "UPDATE emailtbl SET lastused = '$lastused' WHERE id ='$id");
       if($update == false)
       { die("UPDATE FAILED: ".mysqli_error($dbconnect)); }
       echo "Success!";
  }
    else { echo "oops!"; }
  }
?>
          Code:
          </body></html>

            Comment

            • Dormilich
              Recognized Expert Expert
              • Aug 2008
              • 8694
              #4
              remember what I said about error #3?

                Comment

                • 12Strings
                  New Member
                  • May 2014
                  • 12
                  #5
                  you mean the fetch? thanks

                    Comment

                    • Dormilich
                      Recognized Expert Expert
                      • Aug 2008
                      • 8694
                      #6
                      among others. you‘re wildly mixing two incompatible systems.

                        Comment

                        • 12Strings
                          New Member
                          • May 2014
                          • 12
                          #7
                          Guess what? I got one of those AHA moments. The following code
                          is what I want in that it creates a menu and I can select and
                          display a table row. I still need to use that selection to update
                          the "lastused". I really appreciate your help.



                          Code:
                          <!DOCTYPE><html><head><title>email menu</title></head>     
    <body><center>
    <form name="form" method="post" action="">
    <?php
    $con=mysqli_connect("localhost","root","cookie","homedb");
    //============== check connection
    if(mysqli_errno($con))
    {echo "Can't Connect to mySQL:".mysqli_connect_error();}
    else
    {echo "Connected to mySQL</br>";}
       //This creates the drop down box
    echo "<select name= 'target'>";
    echo '<option value="">'.'--- Select email account ---'.'</option>';
    $query = mysqli_query($con,"SELECT target FROM emailtbl");
    $query_display = mysqli_query($con,"SELECT * FROM emailtbl");
    while($row=mysqli_fetch_array($query))
    {echo "<option value='". $row['target']."'>".$row['target']
    .'</option>';}
    echo '</select>';
    ?>
    <input type="submit" name="submit" value="Submit"/><!-- update "lastused" using selected "target"-->
    </form></body></html>

<!DOCTYPE><html><head><title>email menu</title></head>    
    <body><center>
    <?php
    $con=mysqli_connect("localhost","root","cookie","homedb");
    if(mysqli_errno($con))
    {echo "Can't Connect to mySQL:".mysqli_connect_error();}
        if(isset($_POST['target']))
      {
    $name = $_POST['target'];
    $fetch="SELECT target,username,password,emailused,lastused, purpose, saved FROM emailtbl WHERE target = '".$name."'";
    $result = mysqli_query($con,$fetch);
    if(!$result)
    {echo "Error:".(mysqli_error($con));}
   $lastused = "CURDATE()"; // update "lastused" using selected "target"
    //display the table
    echo '<table border="1">'.'<tr>'.'<td bgcolor="#ccffff align="center">'. 'Email menu'. '</td>'.'</tr>';
    echo '<tr>'.'<td>'.'<table border="1">'.'<tr>'.'<td bgcolor="#ccffff align="center">'.'target'.'</td>'.'<td bgcolor="#ccffff align="center">'.'username'.'</td>'.'<td bgcolor="#ccffff align="center">'.'password'.'</td>'.'<td bgcolor="#ccffff align="center">'.'emailused'.'</td>'.'<td bgcolor="#ccffff align="center">'.'lastused'.'</td>'.'<td bgcolor="#ccffff align="center">'.'purpose'. '</td>'.'<td bgcolor="#ccffff align="center">'. 'saved' .'</td>'.'</tr>';
    while($data=mysqli_fetch_row($result))
    {echo ("<tr><td>$data[0]</td><td>$data[1]</td><td>$data[2]</td><td>$data[3]</td><td>$data[4]</td><td>$data[5]</td><td>$data[6]</td></tr>");}
    echo '</table>'.'</td>'.'</tr>'.'</table>';
      }
    ?>
       </body></html>

                            Comment

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