Mamo Posted April 17, 2013 Share Posted April 17, 2013 if (mysql_connect('localhost', 'root', 'mamo12121') && mysql_select_db("project")){ $select = mysql_query("SELECT title, time, question FROM ask ORDER BY time"); if (mysql_num_rows($select) == 0) { echo "No Entries Has been found"; } else { while ($entries = mysql_fetch_assoc($select)) { $entries_time = date('D:M:Y @h:i:s', $entries['time']); $entries_title = $entries['title']; echo '<br >' . $entries_time . '    ' . ?> <a href="ask_3.php"> <?php $entries_title ?> </a ><?php . '<br >'; } } } else { echo "Something went worng"; } the browser tells me that there a parse error Quote Link to comment https://forums.phpfreaks.com/topic/277030-how-to-make-php-variable-a-html-link/ Share on other sites More sharing options...
jcbones Posted April 17, 2013 Share Posted April 17, 2013 (edited) You can't jump out of php and back into php inside of an echo statement. PHP doesn't know what is going on. Change: echo '<br >' . $entries_time . '    ' . ?> <a href="ask_3.php"> <?php $entries_title ?> </a ><?php . '<br >'; To: echo '<br >' . $entries_time . '    <a href="ask_3.php">' . $entries_title . '</a ><br >'; Edited April 17, 2013 by jcbones Quote Link to comment https://forums.phpfreaks.com/topic/277030-how-to-make-php-variable-a-html-link/#findComment-1425237 Share on other sites More sharing options...
Mamo Posted April 17, 2013 Author Share Posted April 17, 2013 You can't jump out of php and back into php inside of an echo statement. PHP doesn't know what is going on. Change: echo '<br >' . $entries_time . '    ' . ?> <a href="ask_3.php"> <?php $entries_title ?> </a ><?php . '<br >'; To: echo '<br >' . $entries_time . '    <a href="ask_3.php">' . $entries_title . '</a ><br >'; Thank you so much jchones. Quote Link to comment https://forums.phpfreaks.com/topic/277030-how-to-make-php-variable-a-html-link/#findComment-1425241 Share on other sites More sharing options...
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