alalj23 Posted December 4, 2012 Share Posted December 4, 2012 Trying to pass my results from MYSQL statement as links to card.php. How do I echo out the results as clickable links to card.php passing a get variable so that I can use this variable in card.php. The links should be playerName. <?php include_once 'header.php'; $con = mysql_connect("localhost","*****","******"); mysql_select_db("cards",$con); if (!$con) { die('Could not connect: ' . mysql_error()); } $sql = "SELECT * FROM cards ORDER BY RAND() LIMIT 4"; $results = mysql_query($sql); $array = mysql_fetch_array($results); $num=mysql_num_rows($results); $i=0; while ($i < $num) { $array = mysql_fetch_array($results); echo '<a href="card.php?cardid=' . $array->id . '">' . $array->name . '</a>'; echo "</br>"; $i++; } include_once 'footer.php'; ?> Link to comment https://forums.phpfreaks.com/topic/271571-displaying-mysql-results-as-hyperlinks-get-variable/ Share on other sites More sharing options...
floridaflatlander Posted December 4, 2012 Share Posted December 4, 2012 Well you already have this "$num=mysql_num_rows($results);" with the LIMIT 4 and because the data is returned in rows I use $row = mysql_fetch_array($results); Then $sql = "SELECT * FROM cards ORDER BY RAND() LIMIT 4"; $results = mysql_query($sql); if ($r && (mysqli_num_rows($r) >= 1)) { // run if there is data while ($row = mysqli_fetch_array($results, MYSQLI_ASSOC)){ echo '<a href="card.php?cardid=' . $row['id'] . '">' .$row['name'].'</a>'; echo "</br>"; } } else {echo 'Something or nothing here';} Link to comment https://forums.phpfreaks.com/topic/271571-displaying-mysql-results-as-hyperlinks-get-variable/#findComment-1397368 Share on other sites More sharing options...
MDCode Posted December 4, 2012 Share Posted December 4, 2012 Well you already have this "$num=mysql_num_rows($results);" with the LIMIT 4 and because the data is returned in rows I use $row = mysql_fetch_array($results); Then $sql = "SELECT * FROM cards ORDER BY RAND() LIMIT 4"; $results = mysql_query($sql); if ($r && (mysqli_num_rows($r) >= 1)) { // run if there is data while ($row = mysqli_fetch_array($results, MYSQLI_ASSOC)){ echo '<a href="card.php?cardid=' . $row['id'] . '">' .$row['name'].'</a>'; echo "</br>"; } } else {echo 'Something or nothing here';} You can not use mysqli and mysql together in the same connection. Link to comment https://forums.phpfreaks.com/topic/271571-displaying-mysql-results-as-hyperlinks-get-variable/#findComment-1397376 Share on other sites More sharing options...
floridaflatlander Posted December 4, 2012 Share Posted December 4, 2012 You can not use mysqli and mysql together in the same connection. Yes,I just copied and pasted and should use mysqli if at all posssible. Link to comment https://forums.phpfreaks.com/topic/271571-displaying-mysql-results-as-hyperlinks-get-variable/#findComment-1397447 Share on other sites More sharing options...
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