karol65 Posted May 22, 2008 Share Posted May 22, 2008 Hi, I am completely newbie. I have created dropdown list, that reads some vars form msql Dbase. The thing is I would like to allow, by clicking on selected position form dropdown, to display all rows (form ID) in new or new window. my code: <?php $query = mysql_query("SELECT * FROM table ORDER BY id"); echo '<select name="ids" style="width :500px">'; while($row = mysql_fetch_array($query)) { echo '<option name="'.$row['id'].'">'.$row['name'].', CMD: '.$row['cmd'].', Date: '.$row['StartDate'].'</option>'; } echo '</select>'; ?> I don't know how to pass those vars (or how to declare it). Thanks for any suggestions. Link to comment https://forums.phpfreaks.com/topic/106799-solved-phpmysql-dropdown-select-id-and-display-its-row-in-div-or-new-page/ Share on other sites More sharing options...
MadTechie Posted May 22, 2008 Share Posted May 22, 2008 your list is probably in a form, read here Link to comment https://forums.phpfreaks.com/topic/106799-solved-phpmysql-dropdown-select-id-and-display-its-row-in-div-or-new-page/#findComment-547553 Share on other sites More sharing options...
karol65 Posted May 23, 2008 Author Share Posted May 23, 2008 OK, thanks for reply. I get the point, but I cannot input select list with all records in one place. Instead of this I have (based on script below): |------------||--------------||--------------||------------------| (numbers of those fields = numbers of IDs in table) And how to pass those vars by submit button? <?php include 'config.php'; ?> <?php $sql = "SELECT name,StartDate,id FROM `my_table` ORDER BY `id` ASC"; $result = mysql_query($sql) or die (mysql_error()); echo "<select name=\"\" id=\"\">"; while ($row = mysql_fetch_array($result)){ extract ($row); echo "<form action=\"sum.php\" method=\"post\">"; echo "<select name=\"genfile\" OnChange=\"submit();\">"; echo "<option>-------</option>"; echo "<option value=\"$id\">$name - $cmd - $StartDate</option>"; }//Ends while //echo "</select>"; echo "<input type=\"submit\">"; echo "</form>"; echo "<br />"; echo "<br />"; ?> Link to comment https://forums.phpfreaks.com/topic/106799-solved-phpmysql-dropdown-select-id-and-display-its-row-in-div-or-new-page/#findComment-547947 Share on other sites More sharing options...
karol65 Posted May 23, 2008 Author Share Posted May 23, 2008 Now I have something like this: <?php $sql = "SELECT * FROM `table` ORDER BY `id` ASC"; $result = mysql_query($sql) or die (mysql_error()); //echo "<select name=\"\" id=\"\">"; while ($row = mysql_fetch_array($result)){ extract ($row); echo "<form action=\"sum.php\" method=\"post\">"; // echo "<select name=\"genfile\" value=\"$id\" OnChange=\"submit();\">"; echo "<option value=\"$id\" OnChange=\"submit();\" > $name - $cmd - $StartDate - $t_letter_code_ef </option>"; // echo "<value=\"$row[id]\"> $row[t_letter_code_ef]</option>"; //echo "<option value=\"$id\">$name - $cmd - $StartDate</option>"; }//Ends while //echo "</select>"; echo "<br />"; echo "<br />"; echo "<input type=\"submit\">"; echo "</form>"; echo "<br />"; echo "<br />"; ?> But this is not a select list, it displays all as a static list, I want to make it clickable, like real select list (dropdonw= and send all row to sum.php. Any idea how to do this? Link to comment https://forums.phpfreaks.com/topic/106799-solved-phpmysql-dropdown-select-id-and-display-its-row-in-div-or-new-page/#findComment-547979 Share on other sites More sharing options...
MadTechie Posted May 28, 2008 Share Posted May 28, 2008 untested <?php if(isset($_POST['Test'])) { echo $_POST['Test']."SELETED<br>"; } $sql = "SELECT * FROM `table` ORDER BY `id` ASC"; $result = mysql_query($sql) or die (mysql_error()); //echo "<select name=\"\" id=\"\">"; echo "<form action=\"sum.php\" method=\"post\">"; echo "<select name=\"Test\">"; //ADDED while ($row = mysql_fetch_array($result)){ extract ($row); echo "<option value=\"$id\" OnChange=\"submit();\" > $name - $cmd - $StartDate - $t_letter_code_ef </option>"; }//Ends while echo "</select>"; echo "<br>"; echo "<br>"; echo "<input type=\"submit\">"; echo "</form>"; echo "<br>"; echo "<br>"; Link to comment https://forums.phpfreaks.com/topic/106799-solved-phpmysql-dropdown-select-id-and-display-its-row-in-div-or-new-page/#findComment-551483 Share on other sites More sharing options...
karol65 Posted October 27, 2008 Author Share Posted October 27, 2008 Thanks. Link to comment https://forums.phpfreaks.com/topic/106799-solved-phpmysql-dropdown-select-id-and-display-its-row-in-div-or-new-page/#findComment-675470 Share on other sites More sharing options...
Recommended Posts
Archived
This topic is now archived and is closed to further replies.