Dynamic Link with php content from database doesn't appear

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  • NewBieBR
    New Member
    • May 2015
    • 5
    #1

    Dynamic Link with php content from database doesn't appear

    Hi everyone, i have been working on this for hours but i can't solve the problem myself. Hope someone can help me out.

    So i have a main page with has a search function
    Code:
    <?php 
$search = $_GET['search'];
$terms = explode(" ", $search);
$query = "SELECT * FROM search WHERE ";
foreach ($terms as $each){
$i++;
if ($i == 1) 
$query .= "keywords LIKE '%$each%' ";
else
$query .= "OR keywords LIKE '%$each%' ";
}

$db = mysql_connect("a", "a", "a");
mysql_select_db("a");
mysql_query("SET NAMES UTF8", $db);
$query = mysql_query($query);
$numrows = mysql_num_rows($query);
if ($numrows > 0) {
while ($row = mysql_fetch_array($query)) {
$id = $row['id'];
$title = $row['title'];
$description = $row['description'];
$keywords = $row['keywords'];
$link = $row['link'];

echo "<div id='kq' ><a href='http://bytes.com/a/rs.php?id=$id'>$title</a></div></br >";
}
}
else
echo "<div id='kq1'> '$search'</div>";

mysql_close ();
?>
    as you can see, every result has link rs.php?id=$id specific to its id. Until now it's fine.
    Then i create the rs.php

    Code:
    <?php
$newid = (int) $_GET['id'];

$servername = "a";
$username = "a";
$password = "a";

$conn = new mysqli($servername, $username, $password);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

mysql_select_db("a", $db);

$newid = (int) $_GET['id'];
$result = mysql_query("SELECT * FROM 'search' WHERE id = 1", $db) ;
$myrow = mysql_fetch_array($result);


echo $myrow["title"];
mysql_close();
?> <html> <head> <meta charset='utf-8'> <title>dd</title> </head> <body> <?php echo $myrow['title']; ?> </body> </html>
    When I click on the link of a result, no error found but the title doesn't appear. Ive been looking on this for hours, I found a same question on this website which was posted since 2007. Except the main page, our rs.php pages are exactly the same so i don't know mine doesn't work. Hope someone can help me out. Thank you...
    Tags: None
    • Rabbit
      Recognized Expert MVP
      • Jan 2007
      • 12517
      #2
      On line 16, you're only looking for Id = 1

        Comment

        • NewBieBR
          New Member
          • May 2015
          • 5
          #3
          Yeah I forgot to change that back, I do that to test if there is any problem with the $newid=$_GET['id']; The original code:
          Code:
          $newid = (int) $_GET['id'];
$result = mysql_query("SELECT * FROM 'search' WHERE id = $newid", $db) ;
$myrow = mysql_fetch_array($result);

echo $newid;
echo sdqsd;
echo $myrow["title"];
mysql_close();
?>

<html>
    <head>
        <meta charset='utf-8'>
        <title>dd</title>
    </head>
    
    <body>
        <?php echo $myrow['title']; ?>
    </body>
    
</html>

          Still doesn't work though. I try
          Code:
          echo $newid
          this time, apparently it's fine. It got the right id. But the title doesn't appear.
          I also used
          Code:
          if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
echo "Connected successfully";
?>
          The connection to database seems to be good too. hmmm Still working on this.

            Comment

            • RonB
              Recognized Expert Contributor
              • Jun 2009
              • 589
              #4
              Try dumping $myrow to see if it holds what you expect.

              Change:
              Code:
              <?php echo $myrow['title']; ?>
              To:
              Code:
              <?php print_r($myrow); ?>
              or:
              Code:
              <?php var_dump($myrow); ?>

                Comment

                • Rabbit
                  Recognized Expert MVP
                  • Jan 2007
                  • 12517
                  #5
                  You should check that the table has
                  1) the id you're searching for
                  2) a field named title
                  3) the field is populated

                    Comment

                    • NewBieBR
                      New Member
                      • May 2015
                      • 5
                      #6
                      yes It does. I checked that too. The id is int, auto_incr. the title is varchar limit 250 characters.the field is populated. hmmm.

                        Comment

                        • NewBieBR
                          New Member
                          • May 2015
                          • 5
                          #7
                          With
                          Code:
                          <?php print_r($myrow['title']); ?>
                          nothing change . With
                          Code:
                          <?php var_dump($myrow['title']); ?>
                          i got the "NULL". hmm. I check id, title, ... everything was right but it still doesn't work ...

                            Comment

                            • NewBieBR
                              New Member
                              • May 2015
                              • 5
                              #8
                              Thanks god, finally I work it out, thanks so much for your helps.
                              the problem was
                              Code:
                              $servername = "a";
$username = "a";
$password = "a";
 
$conn = new mysqli($servername, $username, $password);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
                              i changed it to
                              Code:
                              $db = mysql_connect("a", "a", "a");
                              and it worked ^_^. Im so happy, finally i can sleep well tonight.
                              Once again, thanks a lot for your helps :D

                                Comment

                                • RonB
                                  Recognized Expert Contributor
                                  • Jun 2009
                                  • 589
                                  #9
                                  You just took a step backwards. The mysql_ functions have been depreciated for some time and will removed from the next release.

                                  You should be using mysqli_ or PDO.

                                    Comment

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