How to make link from database result

[Hikaru12]

Here's my code for returning results from the database. I'd like the URL which is saved as a VARCHAR in the database to be displayed as a link. I can't seem to figure out how to embed the <a href="" element without creating a syntax error. Help is appreciated. Thanks!

 

$result = mysql_query("SELECT * FROM bookmarkstable WHERE (`BookmarkTitle` LIKE '%".$query."%') OR (`Category` LIKE '%".$query."%')ORDER BY BookmarkTitle ASC")or die(mysql_error());         if(mysql_num_rows($result) > 0)  {            while($results = mysql_fetch_array($result))   {                 echo "<tr align='center'><td>".$results['BookmarkTitle']."</td> <td>".$results['URL']."</td><td>".$results['Category']."</td> <td>".$results['Notes']."</td></tr>" ;            }

 

[Ciccioo]

well... knowing that a link is in the form

<a href = "URL"> text </a>

you just have to print that, using your URL instead of the placeholder

[Hikaru12]

well... knowing that a link is in the form

<a href = "URL"> text </a>

you just have to print that, using your URL instead of the placeholder

 

I tried using this in the echo to print it but it doesn't work:

<td>"<a href = ".$results['URL'].">Bookmark Link</a>"""</td>

[SSL]

 

I tried using this in the echo to print it but it doesn't work:

<td>"<a href = ".$results['URL'].">Bookmark Link</a>"""</td>

 

What are all those double quotes doing in there? Hard to tell without seeing your complete code or at least a functional fragment.

[Hikaru12]

What are all those double quotes doing in there? Hard to tell without seeing your complete code or at least a functional fragment.

 

The syntax would break without wrapping it that many quotes. Here's what I have:

    $query = $_GET['query'];     $min_length = 3;  if(strlen($query) >= $min_length)  {         $query = htmlspecialchars($query);         $query = mysql_real_escape_string($query);  echo "<div class='container'>";  echo "<div class='table-responsive'>";  echo "<table class='table'>";  echo "<thead>";  echo "<br />";  echo "<br />";  echo "<br />";  echo "<tr><th>Bookmark Title</th><th>URL</th><th>Category</th><th>Notes</th></tr></thead>";  echo "<tbody>";             $result = mysql_query("SELECT * FROM bookmarkstable WHERE (`BookmarkTitle` LIKE '%".$query."%') OR (`Category` LIKE '%".$query."%')ORDER BY BookmarkTitle ASC")or die(mysql_error());         if(mysql_num_rows($result) > 0)  {            while($results = mysql_fetch_array($result))   {                 echo "<tr align='center'>				<td>".$results['BookmarkTitle']."</td> 				<td>".$results['URL']."</td>				<td>".$results['Category']."</td> 				<td>".$results['Notes']."</td>				</tr>" ;            }    echo "</tbody>"; 	

URL field is a VARCHAR(50) in the database. 

 

The HTML form looks like so, the URL field is entered in as plain text:

<form action="bookmarks.php" method="post">Bookmark Name: <input type="text" name="title"><br /><br/>Bookmark URL: <input type="text" name="url"><br /><br/>Category: <input type="text" name="category"><br /><br />Notes: <textarea name="notes" rows="3" cols="25"></textarea><br /><br/><input type="submit" name="submit" value="Save bookmark"></form>

As you can see, it's a very simple bookmarks script. I want the result to return as a link in that particular field. Makes it easier to click on then copying/pasting the returned field. Thanks for the help.

[SSL]

snip

 

So what isn't working? Obviously I can't run this code without your database schema, so you'll have to work with me here.

[Hikaru12]

So what isn't working? Obviously I can't run this code without your database schema, so you'll have to work with me here.

 

The URL is returned simply as a table data element in plain text. If I try to encapsulate it like I did above in a a href element it either complains about the syntax because there's not enough quotes or when it is quoted correctly, it still continues to display the element as text rather than a link. 

 

I would think it would be simple enough to wrap the result in a <a href element> so that it could be clicked as a link but something with the syntax is iffy which is preventing it from displaying as such.

[SSL]

The URL is returned simply as a table data element in plain text. If I try to encapsulate it like I did above in a a href element it either complains about the syntax because there's not enough quotes or when it is quoted correctly, it still continues to display the element as text rather than a link. 

 

I would think it would be simple enough to wrap the result in a <a href element> so that it could be clicked as a link but something with the syntax is iffy which is preventing it from displaying as such.

 

First thing I would do is stop trying to echo the content of the html page, which is leading to your quoting woes. Instead, include the html as a separate php file and echo the values of each variable where you want it in the layout like

<?php// Database query?><table><?php while ($row = mysql_fetch_assoc($result)): ?><tr>  <td><a href="<?php print $row['URL'] ?>"><?php print $row['BookmarkTitle'] ?></a></td></tr><?php endwhile ?></table>

[Muzzle]

Just escape the double quotes inside the html elements:  

<?php$query = $_GET['query'];$min_length = 3;  if(strlen($query) >= $min_length)  {    $query = htmlspecialchars($query);    $query = mysql_real_escape_string($query);    echo "<div class='container'>";    echo "<div class='table-responsive'>";    echo "<table class='table'>";    echo "<thead>";    echo "<br />";    echo "<br />";    echo "<br />";    echo "<tr><th>Bookmark Title</th><th>URL</th><th>Category</th><th>Notes</th></tr></thead>";    echo "<tbody>";    echo "<tr align='center'>";    $result = mysql_query("SELECT * FROM bookmarkstable WHERE (`BookmarkTitle` LIKE '%".$query."%') OR (`Category` LIKE '%".$query."%')ORDER BY BookmarkTitle ASC")or die(mysql_error());    if(mysql_num_rows($result) > 0) {        while($results = mysql_fetch_array($result)) {                echo "<tr align='center'><td><a href=\"$results[url]\">$results[BookmarkTitle]</a></td>			    <td><a href=\"$results[url]\">$results[url]</a></td>				<td><a href=\"$results[url]\">$results[Category]</a></td>				<td><a href=\"$results[url]\">$results[Notes]</a></td>				</tr>" ;        }    }        echo "</tbody>";  }?>

[SSL]

snip

 

Variable interpolation doesn't work in singly-quoted strings in PHP.

[Muzzle]

Meant to edit another post not start a new one, delete this please kthnx.

[SSL]

Actually, I think that was the original problem the OP was having, trying to concatenate the variable inside a string with single quotes.

 

Do it like I suggested and all will be well.

[Muzzle]

Actually, I think that was the original problem the OP was having, trying to concatenate the variable inside a string with single quotes.

 

Do it like I suggested and all will be well.

 

You are right i did not test my code. I have updated my original post with one that should work by simply ignoring the double quotes in the html elements.