andrej13 Posted March 8, 2011 Share Posted March 8, 2011 As you can see, my array contains drinks $dranken = array("cola", "fanta", "bier", "koffie", "thee"); but how can I call the drinks up from an sql table instead of typing them in like I did now? <?php if (!isset($_POST['submit'])) { ?> <html> <form method="post" action="<?php echo $PHP_SELF; ?>"> <?php $dranken = array("cola", "fanta", "bier", "koffie", "thee"); $prijzen = array("2", "2", "1.80", "2.20", "2.20"); $i = 0; echo "<table>"; while ($dranken[$i]) { $listnaam = $dranken[$i] . "_aantal"; $optionlist = "<select name= '$listnaam'><option>0</option><option>1</option><option>2</option><option>3</option></select>"; echo "<tr><td >" . $dranken[$i] . "</td>"; echo "<td>" . $optionlist . "</td>"; echo "<td>" . $prijzen[$i] . "</td></tr>"; $i++; } echo "</table>"; ?> <input type="submit" value="Toon Output" name="submit"/> </form> <?php } $dranken = array("cola", "fanta", "bier", "koffie", "thee"); $prijzen = array("2", "2", "1.80", "2.20", "2.20"); $i = 0; $totaalPrijs = 0; while ($dranken[$i]) { $aantal = $_POST[$dranken[$i] . "_aantal"]; if ($aantal > 0) { $prijsperDrank = $aantal * $prijzen[$i]; echo $dranken[$i] . ":" . $aantal . "Prijs:" . $prijsperDrank . "</br>"; $totaalPrijs += $prijsperDrank; echo "totaal: $totaalPrijs"; } $i++; } ?> </html> Link to comment https://forums.phpfreaks.com/topic/230001-sql-in-php-form/ Share on other sites More sharing options...
Riparian Posted March 8, 2011 Share Posted March 8, 2011 try this <?php // connect to your database $DrinkArray=array() ; $DrinkResult=mysql_query("select flavour from drinks ")or die(mysql_error()); while($DrinkRow=mysql_fetch_assoc($DrinkResult)){ $DrinkArray[]=$DrinkRow[flavour]; } //check your output with something like this foreach($DrinkArray as $value){ echo $value.'<br />'; } ?> Link to comment https://forums.phpfreaks.com/topic/230001-sql-in-php-form/#findComment-1184761 Share on other sites More sharing options...
andrej13 Posted March 9, 2011 Author Share Posted March 9, 2011 without succes Link to comment https://forums.phpfreaks.com/topic/230001-sql-in-php-form/#findComment-1185025 Share on other sites More sharing options...
Riparian Posted March 10, 2011 Share Posted March 10, 2011 What I have put here is correct ... obviously you have to change the mysql_query("select flavour from drinks ") and $DrinkArray[]=$DrinkRow[flavour]; to suit your table contents... Afraid I only speak English Link to comment https://forums.phpfreaks.com/topic/230001-sql-in-php-form/#findComment-1185471 Share on other sites More sharing options...
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