NERVi
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Posts posted by NERVi
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$query = "SELECT `name`, `link` FROM `categorys`";
btw, parādi visu failu.
<?php require_once 'connect.inc.php'; $query = "SELECT `name`, `link` FROM `categorys`"; $query = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_row($query)){ echo $row['name']." link: <a href='".$row['link']."'>link</a>"; } ?>
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pareizāk būtu..
$query = "SELECT name, link FROM categorys"; $query = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_row($query)){ echo $row['name']." link: <a href='".$row['link']."'>link</a>"; }
Un jā, vēl tad varētu iesaistīt, mysql num rows un tml..
Nospiežot uz linka, netiek redirektots uz citu lapu.Paliek tajā pašā lokalhostā
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Ko es daru nepareizi?
<?php require_once 'connect.inc.php'; $query = "SELECT `id` FROM `categorys`"; $query_run = mysql_query($query); for($field = 1;$field <= mysql_num_rows($query_run);$field++){ $query2 = "SELECT `title` , `link` FROM `categorys` WHERE `id` = '$field'"; if( $query2_run = mysql_query($query2)){ $query_title = mysql_result($query2_run,`title`); $query_link = mysql_result($query2_run,`link`); echo $query_title." Link: ".$query_link."<br>"; } } ?>
Tabula izskatās šādi:
Galarezultātam jābūt šādam:
category1 Link: http://lol.com
category2 Link: http://lol2.com
category3 Link: http://lol2.com
category4 Link: http://lol3.com
Bet ir šāds:
category1 Link: category1
category2 Link: category2
category3 Link: category3
category4 Link: category4
Datu izvadīšana no datubāzes ar kvēriju - nepareizi izvada
in Iesācējiem
Posted · Edited by NERVi
Esmu visu pareizi savadījis
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