rickxphp
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Posts posted by rickxphp
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Right on Thanks Everybody!!!!!!!!!!!
It is passing the id and not the table values. ex company name for for the next query.
dropdown is populated from mysql
You select a name and click submit
script does a query finds that company name in company_name table and displays it.
The script will grow from here, but I need to start somewhere.
query.php
$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); $dbname = 'webform'; @mysql_select_db($dbname) or die( "Unable to select database"); $query="SELECT id,company_name FROM webprojects"; $result = mysql_query ($query); echo "<form name=form method=post action='data.php'>"; echo "<select name=\"company_name\">"; while($nt=mysql_fetch_assoc($result)){ echo "<option value=$nt[id]>$nt[company_name]</option>"; } echo "</select>";// Closing of list box echo " <input type=submit value=submit name=button> </form>"; ?>
data.php
$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); $dbname = 'webform'; mysql_select_db($dbname) or die( "Unable to select database"); $company_name=$_POST['company_name']; if(isset($_POST['button'])) { echo "<tr> <th>Company Name ".$company_name. "</th></tr>"; $query = "SELECT 'company_name' FROM webprojects WHERE 'company_name'='" . $company_name ."'"; $result = mysql_query($query) or die(mysql_error()); echo "<table border='1'>"; echo "<tr> <th>Company Name</th></tr>"; while($row = mysql_fetch_assoc($result)) { echo "<tr><td>"; echo $row['company_name']; echo "</td></tr>"; } echo "</table>"; } else{ } ?>
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Hi thanks I've done some more work on my scripts.
I'm not even sure if this should be posted here now!
but here it goes.
I have a working dropdown, being populated from mysql.
The issue I have is when I try and query to display my choice.
query.php
$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); $dbname = 'webform'; mysql_select_db($dbname); @mysql_select_db($dbname) or die( "Unable to select database"); $query="SELECT id,company_name FROM webprojects"; $result = mysql_query ($query); echo "<form name=form method=post action='data.php'> <select company_name=company_name>Company Name"; while($nt=mysql_fetch_array($result)){ echo "<option value=$nt[id]>$nt[company_name]</option>"; } echo "</select>";// Closing of list box echo " <input type=submit value=submit name=button> </form>"; ?> [code] data.php [code] mysql_select_db($dbname); @mysql_select_db($dbname) or die( "Unable to select database"); if(isset($_POST['button'])) { $query = "SELECT 'company_name' FROM webprojects WHERE 'company_name'='" . $_POST['company_name'] ."'"; $result = mysql_query($query) or die(mysql_error()); echo "<table border='1'>"; echo "<tr> <th>Company Name</th></tr>"; while($row = mysql_fetch_array($result)) { echo "<tr><td>"; echo $row['company_name']; echo "</td></tr>"; } echo "</table>"; } else{ } ?> [code] I'm not sure where to go from here Thanks.
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Hi all, I'm trying to display a form based on a dropd own selection. The drop down is propagated from a database query. I have the drop down list displaying, but when I click submit, I can't get the result to display.
$sql="SELECT id, company_name FROM webprojects ORDER BY company_name ASC"; $result=mysql_query($sql); $options=""; while ($row=mysql_fetch_array($result)) { $id=$row["id"]; $company_name=$row["company_name"]; $options.="<OPTION VALUE=\"$id\">".$company_name; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>PHP Dynamic Drop Down Menu</title> </head> <body> <form id="form1" name="form1" method="post" action="selected.php"> <SELECT NAME=id> <OPTION VALUE=0>Choose <?=$options.="<OPTION VALUE=\"$id\">".$company_name.'</option>';?> </SELECT> <label> <input type="submit" name="submit" id="submit" value="Submit" /> </label> </form> </body> </html> Any help is greatly appreciated
Displaying data from a mysql dropdown selection
in PHP Coding Help
Posted
ALRIGHTY, Thanks for nothing everybody, I figured it out myself. I have a real sense of accomplishment and an increased feeling that this forum is good for nothing.