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Member Avatar for Cristianh21

Hi, i'm trying to select two values from my database. $nc ='North Carolina'; $t ='Hello'; $q = 'SELECT COUNT(state, title) AS c FROM all WHERE state = "' . mysql_real_escape_string($nc) . '" AND title = "' . mysql_real_escape_string($t) . '"'; $rq = mysql_query($q); $fetch = mysql_fetch_assoc($rq); $count = $fetch['c']; echo …

Member Avatar for The Diamonds
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Member Avatar for Cristianh21

Hello, I'm having a problem with my code. It seem to only echo out the picture name from my database and not the picture. This is my code search.php $result = mysql_query("SELECT * FROM all WHERE (`title` LIKE '%".$query."%') OR (`add_type` LIKE '%".$query."%')") or die(mysql_error()); if(mysql_num_rows($result) > 0) { while($results …

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Member Avatar for Cristianh21

Im having a problem with this. I want to echo out the number of post from every city in N.C Example: Wilmington (24) <--*24 is the number of post. Charlotte (35) Raleigh (15) include "connect.php"; $wil ='Wilmington'; $cnt = mysql_num_rows(mysql_query("SELECT add_city FROM dbAdd WHERE $add_city='$wil' ")); echo $cnt; This is …

Member Avatar for Cristianh21
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Member Avatar for Cristianh21

THIS CODE IS TO DISPLAY THE DATA FROM THE DATABASE AFTER THE FORM HAS BEEN SUBMITTED. THE PROBLEM I'M HAVING HERE IS THAT WHEN I SUBMIT THE FORM THE LINK IS NOT WORKING. IT GIVES ME THIS : http://yourweb.com/test2.php?id= INSTEAD OF : http://yourweb.com/test2.php?id=23 HOW CAN I FIX THIS? THANK YOU! …

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